Integrand size = 31, antiderivative size = 582 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {\left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) b (a+b)^{3/2} d}-\frac {\left (15 A b^3+a b^2 (5 A-6 B)-3 a^3 (A-4 B)-a^2 b (21 A+2 B)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^3 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {\sqrt {a+b} (5 A b-2 a B) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^4 d}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2 A-5 A b^2+2 a b B\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]
A*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(3/2)+1/3*(3*A*a^4-26*A*a^2*b^2+15*A*b^4 +14*B*a^3*b-6*B*a*b^3)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^( 1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c) )/(a-b))^(1/2)/a^3/(a-b)/b/(a+b)^(3/2)/d-1/3*(15*A*b^3+a*b^2*(5*A-6*B)-3*a ^3*(A-4*B)-a^2*b*(21*A+2*B))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/( a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec( d*x+c))/(a-b))^(1/2)/a^3/(a^2-b^2)/d/(a+b)^(1/2)+(5*A*b-2*B*a)*cot(d*x+c)* EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2)) *(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2 )/a^4/d+1/3*b*(3*A*a^2-5*A*b^2+2*B*a*b)*tan(d*x+c)/a^2/(a^2-b^2)/d/(a+b*se c(d*x+c))^(3/2)+1/3*b*(3*A*a^4-26*A*a^2*b^2+15*A*b^4+14*B*a^3*b-6*B*a*b^3) *tan(d*x+c)/a^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(2366\) vs. \(2(582)=1164\).
Time = 23.84 (sec) , antiderivative size = 2366, normalized size of antiderivative = 4.07 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]
((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((-2*b*(-10*a^2*A*b + 6*A*b^3 + 7*a ^3*B - 3*a*b^2*B)*Sin[c + d*x])/(3*a^3*(-a^2 + b^2)^2) + (2*(A*b^4*Sin[c + d*x] - a*b^3*B*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (2*(-11*a^2*A*b^3*Sin[c + d*x] + 7*A*b^5*Sin[c + d*x] + 8*a^3*b^2*B*Sin[ c + d*x] - 4*a*b^4*B*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)^2*(b + a*Cos[c + d* x]))))/(d*(a + b*Sec[c + d*x])^(5/2)) - ((b + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(5/2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(3*a^5*A*Tan[( c + d*x)/2] + 3*a^4*A*b*Tan[(c + d*x)/2] - 26*a^3*A*b^2*Tan[(c + d*x)/2] - 26*a^2*A*b^3*Tan[(c + d*x)/2] + 15*a*A*b^4*Tan[(c + d*x)/2] + 15*A*b^5*Ta n[(c + d*x)/2] + 14*a^4*b*B*Tan[(c + d*x)/2] + 14*a^3*b^2*B*Tan[(c + d*x)/ 2] - 6*a^2*b^3*B*Tan[(c + d*x)/2] - 6*a*b^4*B*Tan[(c + d*x)/2] - 6*a^5*A*T an[(c + d*x)/2]^3 + 52*a^3*A*b^2*Tan[(c + d*x)/2]^3 - 30*a*A*b^4*Tan[(c + d*x)/2]^3 - 28*a^4*b*B*Tan[(c + d*x)/2]^3 + 12*a^2*b^3*B*Tan[(c + d*x)/2]^ 3 + 3*a^5*A*Tan[(c + d*x)/2]^5 - 3*a^4*A*b*Tan[(c + d*x)/2]^5 - 26*a^3*A*b ^2*Tan[(c + d*x)/2]^5 + 26*a^2*A*b^3*Tan[(c + d*x)/2]^5 + 15*a*A*b^4*Tan[( c + d*x)/2]^5 - 15*A*b^5*Tan[(c + d*x)/2]^5 + 14*a^4*b*B*Tan[(c + d*x)/2]^ 5 - 14*a^3*b^2*B*Tan[(c + d*x)/2]^5 - 6*a^2*b^3*B*Tan[(c + d*x)/2]^5 + 6*a *b^4*B*Tan[(c + d*x)/2]^5 - 30*a^4*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x) /2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan...
Time = 2.60 (sec) , antiderivative size = 617, normalized size of antiderivative = 1.06, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.548, Rules used = {3042, 4522, 27, 3042, 4549, 27, 3042, 4548, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}
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.
\(\displaystyle \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4522 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {-3 A b \sec ^2(c+d x)+5 A b-2 a B}{2 (a+b \sec (c+d x))^{5/2}}dx}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {-3 A b \sec ^2(c+d x)+5 A b-2 a B}{(a+b \sec (c+d x))^{5/2}}dx}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {-3 A b \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 A b-2 a B}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{2 a}\) |
\(\Big \downarrow \) 4549 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {-\frac {2 \int -\frac {-b \left (3 A a^2+2 b B a-5 A b^2\right ) \sec ^2(c+d x)-6 a b (A b-a B) \sec (c+d x)+3 \left (a^2-b^2\right ) (5 A b-2 a B)}{2 (a+b \sec (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\int \frac {-b \left (3 A a^2+2 b B a-5 A b^2\right ) \sec ^2(c+d x)-6 a b (A b-a B) \sec (c+d x)+3 \left (a^2-b^2\right ) (5 A b-2 a B)}{(a+b \sec (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\int \frac {-b \left (3 A a^2+2 b B a-5 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-6 a b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )+3 \left (a^2-b^2\right ) (5 A b-2 a B)}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 4548 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {-\frac {2 \int -\frac {3 (5 A b-2 a B) \left (a^2-b^2\right )^2+b \left (3 A a^4+14 b B a^3-26 A b^2 a^2-6 b^3 B a+15 A b^4\right ) \sec ^2(c+d x)-2 a b \left (-6 B a^3+9 A b a^2+2 b^2 B a-5 A b^3\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\frac {\int \frac {3 (5 A b-2 a B) \left (a^2-b^2\right )^2+b \left (3 A a^4+14 b B a^3-26 A b^2 a^2-6 b^3 B a+15 A b^4\right ) \sec ^2(c+d x)-2 a b \left (-6 B a^3+9 A b a^2+2 b^2 B a-5 A b^3\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\frac {\int \frac {3 (5 A b-2 a B) \left (a^2-b^2\right )^2+b \left (3 A a^4+14 b B a^3-26 A b^2 a^2-6 b^3 B a+15 A b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 a b \left (-6 B a^3+9 A b a^2+2 b^2 B a-5 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 4546 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\frac {b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {3 (5 A b-2 a B) \left (a^2-b^2\right )^2+\left (-2 a b \left (-6 B a^3+9 A b a^2+2 b^2 B a-5 A b^3\right )-b \left (3 A a^4+14 b B a^3-26 A b^2 a^2-6 b^3 B a+15 A b^4\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\frac {b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {3 (5 A b-2 a B) \left (a^2-b^2\right )^2+\left (-2 a b \left (-6 B a^3+9 A b a^2+2 b^2 B a-5 A b^3\right )-b \left (3 A a^4+14 b B a^3-26 A b^2 a^2-6 b^3 B a+15 A b^4\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 4409 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\frac {3 \left (a^2-b^2\right )^2 (5 A b-2 a B) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+b (a-b) \left (-3 a^3 (A-4 B)-a^2 b (21 A+2 B)+a b^2 (5 A-6 B)+15 A b^3\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\frac {3 \left (a^2-b^2\right )^2 (5 A b-2 a B) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (a-b) \left (-3 a^3 (A-4 B)-a^2 b (21 A+2 B)+a b^2 (5 A-6 B)+15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 4271 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\frac {b (a-b) \left (-3 a^3 (A-4 B)-a^2 b (21 A+2 B)+a b^2 (5 A-6 B)+15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 \sqrt {a+b} \left (a^2-b^2\right )^2 (5 A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\frac {b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 \sqrt {a+b} \left (a^2-b^2\right )^2 (5 A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}+\frac {2 (a-b) \sqrt {a+b} \left (-3 a^3 (A-4 B)-a^2 b (21 A+2 B)+a b^2 (5 A-6 B)+15 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\frac {-\frac {6 \sqrt {a+b} \left (a^2-b^2\right )^2 (5 A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}+\frac {2 (a-b) \sqrt {a+b} \left (-3 a^3 (A-4 B)-a^2 b (21 A+2 B)+a b^2 (5 A-6 B)+15 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{2 a}\) |
(A*Sin[c + d*x])/(a*d*(a + b*Sec[c + d*x])^(3/2)) - ((-2*b*(3*a^2*A - 5*A* b^2 + 2*a*b*B)*Tan[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2) ) + (((-2*(a - b)*Sqrt[a + b]*(3*a^4*A - 26*a^2*A*b^2 + 15*A*b^4 + 14*a^3* b*B - 6*a*b^3*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sq rt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-(( b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) + (2*(a - b)*Sqrt[a + b]*(15*A*b^3 + a*b^2*(5*A - 6*B) - 3*a^3*(A - 4*B) - a^2*b*(21*A + 2*B))*Cot[c + d*x]*E llipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sq rt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))] )/d - (6*Sqrt[a + b]*(a^2 - b^2)^2*(5*A*b - 2*a*B)*Cot[c + d*x]*EllipticPi [(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)] *Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b ))])/(a*d))/(a*(a^2 - b^2)) - (2*b*(3*a^4*A - 26*a^2*A*b^2 + 15*A*b^4 + 14 *a^3*b*B - 6*a*b^3*B)*Tan[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d* x]]))/(3*a*(a^2 - b^2)))/(2*a)
3.4.90.3.1 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[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) *((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[d Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Sim p[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*B* n - A*b*(m + n + 1) + A*a*(n + 1)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f* x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C )*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C Int[Csc[e + f*x]*(( 1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A , B, C}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^( m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x ] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. ) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 + a^2*C)*Cot[e + f*x]*((a + b*Csc[ e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*b* (A + C)*(m + 1)*Csc[e + f*x] + (A*b^2 + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m ] && LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(10860\) vs. \(2(541)=1082\).
Time = 11.79 (sec) , antiderivative size = 10861, normalized size of antiderivative = 18.66
\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral((B*cos(d*x + c)*sec(d*x + c) + A*cos(d*x + c))*sqrt(b*sec(d*x + c ) + a)/(b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)
\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]