Integrand size = 33, antiderivative size = 510 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (8 a^4 A b-15 a^2 A b^3+3 A b^5-16 a^5 B+28 a^3 b^2 B-8 a b^4 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^5 (a+b)^{3/2} d}+\frac {2 \left (9 a b^3 (A-B)+b^4 (3 A-B)+16 a^4 B-2 a^2 b^2 (3 A+8 B)-a^3 (8 A b-12 b B)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {2 a (A b-a B) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 A b-7 A b^3-6 a^3 B+10 a b^2 B\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (a A b-2 a^2 B+b^2 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d} \]
-2/3*(8*A*a^4*b-15*A*a^2*b^3+3*A*b^5-16*B*a^5+28*B*a^3*b^2-8*B*a*b^4)*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-b)/b^5/(a +b)^(3/2)/d+2/3*(9*a*b^3*(A-B)+b^4*(3*A-B)+16*B*a^4-2*a^2*b^2*(3*A+8*B)-a^ 3*(8*A*b-12*B*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)/b^4/(a^2-b^2)/d/(a+b)^(1/2)+2/3*a*(A*b-B*a)*sec(d*x+c)^2*tan(d*x +c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-2/3*a^2*(3*A*a^2*b-7*A*b^3-6*B*a^ 3+10*B*a*b^2)*tan(d*x+c)/b^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)-2/3*(A*a *b-2*B*a^2+B*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d
Leaf count is larger than twice the leaf count of optimal. \(4342\) vs. \(2(510)=1020\).
Time = 32.49 (sec) , antiderivative size = 4342, normalized size of antiderivative = 8.51 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]
((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((2*(8*a^4*A*b - 15*a^2*A*b^3 + 3*A *b^5 - 16*a^5*B + 28*a^3*b^2*B - 8*a*b^4*B)*Sin[c + d*x])/(3*b^4*(-a^2 + b ^2)^2) + (2*(a^2*A*b*Sin[c + d*x] - a^3*B*Sin[c + d*x]))/(3*b^2*(-a^2 + b^ 2)*(b + a*Cos[c + d*x])^2) + (2*(-4*a^4*A*b*Sin[c + d*x] + 8*a^2*A*b^3*Sin [c + d*x] + 7*a^5*B*Sin[c + d*x] - 11*a^3*b^2*B*Sin[c + d*x]))/(3*b^3*(-a^ 2 + b^2)^2*(b + a*Cos[c + d*x])) + (2*B*Tan[c + d*x])/(3*b^3)))/(d*(a + b* Sec[c + d*x])^(5/2)) + (2*(b + a*Cos[c + d*x])^2*((5*a^2*A)/((-a^2 + b^2)^ 2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^4*A)/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (A*b^2)/((-a^2 + b^2 )^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (16*a^5*B)/(3*b^3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (28*a^3*B)/(3*b*( -a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (8*a*b*B)/(3* (-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^5*A*Sqr t[Sec[c + d*x]])/(3*b^3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (17*a^3 *A*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (3* a*A*b*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (5*a ^2*B*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (16*a ^6*B*Sqrt[Sec[c + d*x]])/(3*b^4*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (32*a^4*B*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d* x]]) + (b^2*B*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + ...
Time = 2.37 (sec) , antiderivative size = 526, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4517, 27, 3042, 4578, 27, 3042, 4570, 27, 3042, 4493, 3042, 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.
\(\displaystyle \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4517 |
\(\displaystyle \frac {2 \int \frac {\sec ^2(c+d x) \left (-3 \left (-2 B a^2+A b a+b^2 B\right ) \sec ^2(c+d x)-3 b (A b-a B) \sec (c+d x)+4 a (A b-a B)\right )}{2 (a+b \sec (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sec ^2(c+d x) \left (-3 \left (-2 B a^2+A b a+b^2 B\right ) \sec ^2(c+d x)-3 b (A b-a B) \sec (c+d x)+4 a (A b-a B)\right )}{(a+b \sec (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (-3 \left (-2 B a^2+A b a+b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-3 b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )+4 a (A b-a B)\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4578 |
\(\displaystyle \frac {\frac {2 \int \frac {\sec (c+d x) \left (-3 b \left (a^2-b^2\right ) \left (-2 B a^2+A b a+b^2 B\right ) \sec ^2(c+d x)+\left (-12 B a^5+6 A b a^4+22 b^2 B a^3-13 A b^3 a^2-6 b^4 B a+3 A b^5\right ) \sec (c+d x)+a b \left (-6 B a^3+3 A b a^2+10 b^2 B a-7 A b^3\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 B+3 a^2 A b+10 a b^2 B-7 A b^3\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (-3 b \left (a^2-b^2\right ) \left (-2 B a^2+A b a+b^2 B\right ) \sec ^2(c+d x)+\left (-12 B a^5+6 A b a^4+22 b^2 B a^3-13 A b^3 a^2-6 b^4 B a+3 A b^5\right ) \sec (c+d x)+a b \left (-6 B a^3+3 A b a^2+10 b^2 B a-7 A b^3\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 B+3 a^2 A b+10 a b^2 B-7 A b^3\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-3 b \left (a^2-b^2\right ) \left (-2 B a^2+A b a+b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (-12 B a^5+6 A b a^4+22 b^2 B a^3-13 A b^3 a^2-6 b^4 B a+3 A b^5\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a b \left (-6 B a^3+3 A b a^2+10 b^2 B a-7 A b^3\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 B+3 a^2 A b+10 a b^2 B-7 A b^3\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4570 |
\(\displaystyle \frac {\frac {\frac {2 \int \frac {3 \sec (c+d x) \left (\left (-4 B a^4+2 A b a^3+7 b^2 B a^2-6 A b^3 a+b^4 B\right ) b^2+\left (-16 B a^5+8 A b a^4+28 b^2 B a^3-15 A b^3 a^2-8 b^4 B a+3 A b^5\right ) \sec (c+d x) b\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 B+a A b+b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 B+3 a^2 A b+10 a b^2 B-7 A b^3\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {\int \frac {\sec (c+d x) \left (\left (-4 B a^4+2 A b a^3+7 b^2 B a^2-6 A b^3 a+b^4 B\right ) b^2+\left (-16 B a^5+8 A b a^4+28 b^2 B a^3-15 A b^3 a^2-8 b^4 B a+3 A b^5\right ) \sec (c+d x) b\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 B+a A b+b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 B+3 a^2 A b+10 a b^2 B-7 A b^3\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (-4 B a^4+2 A b a^3+7 b^2 B a^2-6 A b^3 a+b^4 B\right ) b^2+\left (-16 B a^5+8 A b a^4+28 b^2 B a^3-15 A b^3 a^2-8 b^4 B a+3 A b^5\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 B+a A b+b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 B+3 a^2 A b+10 a b^2 B-7 A b^3\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4493 |
\(\displaystyle \frac {\frac {\frac {b (a-b) \left (16 a^4 B-a^3 (8 A b-12 b B)-2 a^2 b^2 (3 A+8 B)+9 a b^3 (A-B)+b^4 (3 A-B)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+b \left (-16 a^5 B+8 a^4 A b+28 a^3 b^2 B-15 a^2 A b^3-8 a b^4 B+3 A b^5\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 B+a A b+b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 B+3 a^2 A b+10 a b^2 B-7 A b^3\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {b (a-b) \left (16 a^4 B-a^3 (8 A b-12 b B)-2 a^2 b^2 (3 A+8 B)+9 a b^3 (A-B)+b^4 (3 A-B)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (-16 a^5 B+8 a^4 A b+28 a^3 b^2 B-15 a^2 A b^3-8 a b^4 B+3 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 B+a A b+b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 B+3 a^2 A b+10 a b^2 B-7 A b^3\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {\frac {\frac {b \left (-16 a^5 B+8 a^4 A b+28 a^3 b^2 B-15 a^2 A b^3-8 a b^4 B+3 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (16 a^4 B-a^3 (8 A b-12 b B)-2 a^2 b^2 (3 A+8 B)+9 a b^3 (A-B)+b^4 (3 A-B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 B+a A b+b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 B+3 a^2 A b+10 a b^2 B-7 A b^3\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {2 a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {\frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (16 a^4 B-a^3 (8 A b-12 b B)-2 a^2 b^2 (3 A+8 B)+9 a b^3 (A-B)+b^4 (3 A-B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} \left (-16 a^5 B+8 a^4 A b+28 a^3 b^2 B-15 a^2 A b^3-8 a b^4 B+3 A b^5\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 B+a A b+b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 B+3 a^2 A b+10 a b^2 B-7 A b^3\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}\) |
(2*a*(A*b - a*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Se c[c + d*x])^(3/2)) + ((-2*a^2*(3*a^2*A*b - 7*A*b^3 - 6*a^3*B + 10*a*b^2*B) *Tan[c + d*x])/(b^2*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) + (((-2*(a - b )*Sqrt[a + b]*(8*a^4*A*b - 15*a^2*A*b^3 + 3*A*b^5 - 16*a^5*B + 28*a^3*b^2* B - 8*a*b^4*B)*Cot[c + d*x]*EllipticE[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))])/(b*d) + (2*(a - b)*Sqrt[a + b]*(9*a*b^3*(A - B) + b^4*(3*A - B) + 16*a^4*B - 2*a^2*b^2*(3*A + 8*B) - a^3*(8*A*b - 12* b*B))*Cot[c + d*x]*EllipticF[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))])/d)/b - (2*(a^2 - b^2)*(a*A*b - 2*a^2*B + b^2*B)*Sqrt [a + b*Sec[c + d*x]]*Tan[c + d*x])/d)/(b^2*(a^2 - b^2)))/(3*b*(a^2 - b^2))
3.4.85.3.1 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[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_)]*(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_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B) Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B 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} , x] && NeQ[a^2 - b^2, 0] && NeQ[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*d^2*( A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f*(m + 1)*(a^2 - b^2))), x] - Simp[d/(b*(m + 1)*(a^2 - b^2)) Int[( a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*Simp[a*d*(A*b - a*B)*( n - 2) + b*d*(A*b - a*B)*(m + 1)*Csc[e + f*x] - (a*A*b*d*(m + n) - d*B*(a^2 *(n - 1) + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f , A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[ n, 1]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]^2*((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^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x ])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Simp[1/(b^2*(m + 1)*(a^2 - b^ 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(m + 1)*((-a)*(b *B - a*C) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^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 ]
Leaf count of result is larger than twice the leaf count of optimal. \(9479\) vs. \(2(476)=952\).
Time = 32.12 (sec) , antiderivative size = 9480, normalized size of antiderivative = 18.59
method | result | size |
parts | \(\text {Expression too large to display}\) | \(9480\) |
default | \(\text {Expression too large to display}\) | \(9575\) |
\[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral((B*sec(d*x + c)^5 + A*sec(d*x + c)^4)*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)
\[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]