Integrand size = 35, antiderivative size = 856 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a+a \sec (c+d x))^{5/3}}-\frac {3 (2 A-2 B-5 C) \tan (c+d x)}{7 a d (a+a \sec (c+d x))^{2/3}}-\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (-\frac {1}{6},\frac {1}{2},1,\frac {5}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{a d \sqrt {1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}-\frac {3 \left (1+\sqrt {3}\right ) (2 A-2 B-5 C) \sqrt [3]{1+\sec (c+d x)} \tan (c+d x)}{7 a d (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )}+\frac {3 \sqrt [3]{2} \sqrt [4]{3} (2 A-2 B-5 C) E\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{7 a d (1-\sec (c+d x)) (a+a \sec (c+d x))^{2/3} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}}+\frac {3^{3/4} \left (1-\sqrt {3}\right ) (2 A-2 B-5 C) \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{7\ 2^{2/3} a d (1-\sec (c+d x)) (a+a \sec (c+d x))^{2/3} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]
-3/7*(A-B+C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/3)-3/7*(2*A-2*B-5*C)*tan(d*x +c)/a/d/(a+a*sec(d*x+c))^(2/3)-3/7*(2*A-2*B-5*C)*(1+sec(d*x+c))^(1/3)*(1+3 ^(1/2))*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(2/3)/(2^(1/3)-(1+sec(d*x+c))^(1/3 )*(1+3^(1/2)))-3*A*AppellF1(-1/6,1,1/2,5/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c) )*2^(1/2)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(2/3)/(1-sec(d*x+c))^(1/2)+3/7*2 ^(1/3)*3^(1/4)*(2*A-2*B-5*C)*((2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))^2 /(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)/(2^(1/3)-(1+sec(d*x+c ))^(1/3)*(1-3^(1/2)))*(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))*EllipticE ((1-(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c))^( 1/3)*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(1+sec(d*x+c))^(1/3)*( 2^(1/3)-(1+sec(d*x+c))^(1/3))*((2^(2/3)+2^(1/3)*(1+sec(d*x+c))^(1/3)+(1+se c(d*x+c))^(2/3))/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)*tan(d *x+c)/a/d/(1-sec(d*x+c))/(a+a*sec(d*x+c))^(2/3)/(-(1+sec(d*x+c))^(1/3)*(2^ (1/3)-(1+sec(d*x+c))^(1/3))/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^ (1/2)+1/14*3^(3/4)*(2*A-2*B-5*C)*((2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2) ))^2/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)/(2^(1/3)-(1+sec(d *x+c))^(1/3)*(1-3^(1/2)))*(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))*Ellip ticF((1-(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c ))^(1/3)*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(1+sec(d*x+c))^(1/ 3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))*(1-3^(1/2))*((2^(2/3)+2^(1/3)*(1+sec(...
Leaf count is larger than twice the leaf count of optimal. \(4383\) vs. \(2(856)=1712\).
Time = 22.20 (sec) , antiderivative size = 4383, normalized size of antiderivative = 5.12 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\text {Result too large to show} \]
(Cos[c + d*x]^2*((1 + Cos[c + d*x])*Sec[c + d*x])^(1/3)*(1 + Sec[c + d*x]) ^(5/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-6*Sec[(c + d*x)/2]*(10*A *Sin[(c + d*x)/2] - 3*B*Sin[(c + d*x)/2] - 4*C*Sin[(c + d*x)/2]))/7 + (3*S ec[(c + d*x)/2]^3*(A*Sin[(c + d*x)/2] - B*Sin[(c + d*x)/2] + C*Sin[(c + d* x)/2]))/7 + (6*(9*A - 2*B - 5*C)*Sin[c + d*x])/7))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a*(1 + Sec[c + d*x]))^(5/3)) + (2*2^(1/3)*C os[c + d*x]^2*(1 + Sec[c + d*x])^(5/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x ]^2)*((32*A*(1 + Sec[c + d*x])^(1/3))/7 - (4*B*(1 + Sec[c + d*x])^(1/3))/7 - (10*C*(1 + Sec[c + d*x])^(1/3))/7 + Cos[c + d*x]*((-54*A*(1 + Sec[c + d *x])^(1/3))/7 + (12*B*(1 + Sec[c + d*x])^(1/3))/7 + (30*C*(1 + Sec[c + d*x ])^(1/3))/7))*Tan[(c + d*x)/2]*(-(((9*A - 2*B - 5*C)*AppellF1[3/2, 1/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2)/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3)) + (9*(3*AppellF1[1/2, 1/3, 1, 3/2, Tan[( c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(5*A + 2*B + 5*C - 7*(7*A - 2*B - 5*C) *Cos[c + d*x]) + 4*(9*A - 2*B - 5*C)*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d* x)/2]^2, -Tan[(c + d*x)/2]^2])*Cos[c + d*x]*Tan[(c + d*x)/2]^2))/(2*(-1 + Tan[(c + d*x)/2]^2)*(-9*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Ta n[(c + d*x)/2]^2] + 2*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -T an[(c + d*x)/2]^2] - AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Ta...
Time = 1.36 (sec) , antiderivative size = 844, normalized size of antiderivative = 0.99, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.629, Rules used = {3042, 4540, 27, 3042, 4412, 3042, 4266, 3042, 4265, 149, 25, 1012, 4315, 3042, 4314, 61, 73, 837, 25, 27, 766, 2420}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a \sec (c+d x)+a)^{5/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/3}}dx\) |
\(\Big \downarrow \) 4540 |
\(\displaystyle -\frac {3 \int -\frac {7 a A-a (2 A-2 B-5 C) \sec (c+d x)}{3 (\sec (c+d x) a+a)^{2/3}}dx}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {7 a A-a (2 A-2 B-5 C) \sec (c+d x)}{(\sec (c+d x) a+a)^{2/3}}dx}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {7 a A-a (2 A-2 B-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}}dx}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 4412 |
\(\displaystyle \frac {7 a A \int \frac {1}{(\sec (c+d x) a+a)^{2/3}}dx-a (2 A-2 B-5 C) \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^{2/3}}dx}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {7 a A \int \frac {1}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}}dx-a (2 A-2 B-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}}dx}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 4266 |
\(\displaystyle \frac {\frac {7 a A (\sec (c+d x)+1)^{2/3} \int \frac {1}{(\sec (c+d x)+1)^{2/3}}dx}{(a \sec (c+d x)+a)^{2/3}}-a (2 A-2 B-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}}dx}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {7 a A (\sec (c+d x)+1)^{2/3} \int \frac {1}{\left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{2/3}}dx}{(a \sec (c+d x)+a)^{2/3}}-a (2 A-2 B-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}}dx}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 4265 |
\(\displaystyle \frac {-a (2 A-2 B-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}}dx-\frac {7 a A \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \int \frac {\cos (c+d x)}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 149 |
\(\displaystyle \frac {-a (2 A-2 B-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}}dx-\frac {42 a A \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \int \frac {\cos ^3(c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {42 a A \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \int -\frac {\cos ^3(c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}-a (2 A-2 B-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}}dx}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {-a (2 A-2 B-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}}dx-\frac {21 \sqrt {2} a A \sin (c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {1}{6},1,\frac {1}{2},\frac {5}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle \frac {-\frac {a (2 A-2 B-5 C) (\sec (c+d x)+1)^{2/3} \int \frac {\sec (c+d x)}{(\sec (c+d x)+1)^{2/3}}dx}{(a \sec (c+d x)+a)^{2/3}}-\frac {21 \sqrt {2} a A \sin (c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {1}{6},1,\frac {1}{2},\frac {5}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {a (2 A-2 B-5 C) (\sec (c+d x)+1)^{2/3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{2/3}}dx}{(a \sec (c+d x)+a)^{2/3}}-\frac {21 \sqrt {2} a A \sin (c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {1}{6},1,\frac {1}{2},\frac {5}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 4314 |
\(\displaystyle \frac {\frac {a (2 A-2 B-5 C) \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \int \frac {1}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}-\frac {21 \sqrt {2} a A \sin (c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {1}{6},1,\frac {1}{2},\frac {5}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {a (2 A-2 B-5 C) \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \left (-\int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}}d\sec (c+d x)-\frac {3 \sqrt {1-\sec (c+d x)}}{\sqrt [6]{\sec (c+d x)+1}}\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}-\frac {21 \sqrt {2} a A \sin (c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {1}{6},1,\frac {1}{2},\frac {5}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {a (2 A-2 B-5 C) \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \left (-6 \int \frac {(\sec (c+d x)+1)^{2/3}}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}-\frac {3 \sqrt {1-\sec (c+d x)}}{\sqrt [6]{\sec (c+d x)+1}}\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}-\frac {21 \sqrt {2} a A \sin (c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {1}{6},1,\frac {1}{2},\frac {5}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 837 |
\(\displaystyle \frac {\frac {a (2 A-2 B-5 C) \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \left (-6 \left (-\frac {\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{\sqrt [3]{2}}-\frac {1}{2} \int -\frac {2 (\sec (c+d x)+1)^{2/3}+2^{2/3} \left (1-\sqrt {3}\right )}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}\right )-\frac {3 \sqrt {1-\sec (c+d x)}}{\sqrt [6]{\sec (c+d x)+1}}\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}-\frac {21 \sqrt {2} a A \sin (c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {1}{6},1,\frac {1}{2},\frac {5}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {a (2 A-2 B-5 C) \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \left (-6 \left (\frac {1}{2} \int \frac {2^{2/3} \left (\sqrt [3]{2} (\sec (c+d x)+1)^{2/3}-\sqrt {3}+1\right )}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}-\frac {\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{\sqrt [3]{2}}\right )-\frac {3 \sqrt {1-\sec (c+d x)}}{\sqrt [6]{\sec (c+d x)+1}}\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}-\frac {21 \sqrt {2} a A \sin (c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {1}{6},1,\frac {1}{2},\frac {5}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {a (2 A-2 B-5 C) \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \left (-6 \left (\frac {\int \frac {\sqrt [3]{2} (\sec (c+d x)+1)^{2/3}-\sqrt {3}+1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{\sqrt [3]{2}}-\frac {\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{\sqrt [3]{2}}\right )-\frac {3 \sqrt {1-\sec (c+d x)}}{\sqrt [6]{\sec (c+d x)+1}}\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}-\frac {21 \sqrt {2} a A \sin (c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {1}{6},1,\frac {1}{2},\frac {5}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 766 |
\(\displaystyle \frac {\frac {a (2 A-2 B-5 C) \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \left (-6 \left (\frac {\int \frac {\sqrt [3]{2} (\sec (c+d x)+1)^{2/3}-\sqrt {3}+1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{\sqrt [3]{2}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [6]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2\ 2^{2/3} \sqrt [4]{3} \sqrt {1-\sec (c+d x)} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}\right )-\frac {3 \sqrt {1-\sec (c+d x)}}{\sqrt [6]{\sec (c+d x)+1}}\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}-\frac {21 \sqrt {2} a A \sin (c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {1}{6},1,\frac {1}{2},\frac {5}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}\) |
\(\Big \downarrow \) 2420 |
\(\displaystyle \frac {\frac {a (2 A-2 B-5 C) \sqrt [6]{\sec (c+d x)+1} \left (-6 \left (\frac {\frac {\left (1+\sqrt {3}\right ) \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}}{2^{2/3} \left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )}-\frac {\sqrt [4]{3} E\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \sqrt [6]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}{\sqrt [3]{2} \sqrt {1-\sec (c+d x)} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}}{\sqrt [3]{2}}-\frac {\left (1-\sqrt {3}\right ) \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \sqrt [6]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}{2\ 2^{2/3} \sqrt [4]{3} \sqrt {1-\sec (c+d x)} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}\right )-\frac {3 \sqrt {1-\sec (c+d x)}}{\sqrt [6]{\sec (c+d x)+1}}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x) a+a)^{2/3}}-\frac {21 \sqrt {2} a A \operatorname {AppellF1}\left (-\frac {1}{6},1,\frac {1}{2},\frac {5}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right ) \sqrt [6]{\sec (c+d x)+1} \sin (c+d x)}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x) a+a)^{2/3}}}{7 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{7 d (\sec (c+d x) a+a)^{5/3}}\) |
(-3*(A - B + C)*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^(5/3)) + ((-21*Sqr t[2]*a*A*AppellF1[-1/6, 1, 1/2, 5/6, 1 + Sec[c + d*x], (1 + Sec[c + d*x])/ 2]*(1 + Sec[c + d*x])^(1/6)*Sin[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(a + a *Sec[c + d*x])^(2/3)) + (a*(2*A - 2*B - 5*C)*(1 + Sec[c + d*x])^(1/6)*((-3 *Sqrt[1 - Sec[c + d*x]])/(1 + Sec[c + d*x])^(1/6) - 6*(-1/2*((1 - Sqrt[3]) *EllipticF[ArcCos[(2^(1/3) - (1 - Sqrt[3])*(1 + Sec[c + d*x])^(1/3))/(2^(1 /3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))], (2 + Sqrt[3])/4]*(1 + Sec[ c + d*x])^(1/6)*(2^(1/3) - (1 + Sec[c + d*x])^(1/3))*Sqrt[(2^(2/3) + 2^(1/ 3)*(1 + Sec[c + d*x])^(1/3) + (1 + Sec[c + d*x])^(2/3))/(2^(1/3) - (1 + Sq rt[3])*(1 + Sec[c + d*x])^(1/3))^2])/(2^(2/3)*3^(1/4)*Sqrt[1 - Sec[c + d*x ]]*Sqrt[-(((1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c + d*x])^(1/3)))/ (2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2)]) + (((1 + Sqrt[3])* Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])^(1/6))/(2^(2/3)*(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))) - (3^(1/4)*EllipticE[ArcCos[(2^(1/3) - (1 - Sqrt[3])*(1 + Sec[c + d*x])^(1/3))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec [c + d*x])^(1/3))], (2 + Sqrt[3])/4]*(1 + Sec[c + d*x])^(1/6)*(2^(1/3) - ( 1 + Sec[c + d*x])^(1/3))*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[c + d*x])^(1/3) + (1 + Sec[c + d*x])^(2/3))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1 /3))^2])/(2^(1/3)*Sqrt[1 - Sec[c + d*x]]*Sqrt[-(((1 + Sec[c + d*x])^(1/3)* (2^(1/3) - (1 + Sec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec...
3.7.32.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1 ) - 1)*(c - a*(d/b) + d*(x^k/b))^n*(e - a*(f/b) + f*(x^k/b))^p, x], x, (a + b*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[2*n] && IntegerQ[p]
Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ (s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x ]
Int[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(Sqrt[3] - 1)*(s^2/(2*r^2)) Int[1/Sqrt[ a + b*x^6], x], x] - Simp[1/(2*r^2) Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4)/S qrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(1 + Sqrt[3])*d*s^3*x*(Sqr t[a + b*x^6]/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2))), x] - Simp[3^(1/4)*d*s*x* (s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2 *r^2*Sqrt[(r*x^2*(s + r*x^2))/(s + (1 + Sqrt[3])*r*x^2)^2]*Sqrt[a + b*x^6]) )*EllipticE[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 - Sqrt[3])*d, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^n*(Cot [c + d*x]/(d*Sqrt[1 + Csc[c + d*x]]*Sqrt[1 - Csc[c + d*x]])) Subst[Int[(1 + b*(x/a))^(n - 1/2)/(x*Sqrt[1 - b*(x/a)]), x], x, Csc[c + d*x]], x] /; Fr eeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && GtQ[a, 0 ]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^IntPar t[n]*((a + b*Csc[c + d*x])^FracPart[n]/(1 + (b/a)*Csc[c + d*x])^FracPart[n] ) Int[(1 + (b/a)*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && !GtQ[a, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a^2*d*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x ]]*Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(d*x)^(n - 1)*((a + b*x)^(m - 1/2 )/Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && GtQ[a, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Csc[e + f*x])^FracPart[m ]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]) Int[(1 + (b/a)*Csc[e + f*x])^m*(d *Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^ 2, 0] && !IntegerQ[m] && !GtQ[a, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[c Int[(a + b*Csc[e + f*x])^m, x], x] + Sim p[d Int[(a + b*Csc[e + f*x])^m*Csc[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[2*m]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Sim p[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*(2*m + 1) + (b*B*(m + 1) - a*(A*(m + 1) - C*m))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
\[\int \frac {A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{3}}}d x\]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{3}}}\, dx \]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/3}} \,d x \]