3.6.32 \(\int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\) [532]

3.6.32.1 Optimal result

Integrand size = 23, antiderivative size = 339 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=-\frac {\left (12 a^2-54 a b+77 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 (a-b)^{9/2} d}+\frac {\left (12 a^2+54 a b+77 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 (a+b)^{9/2} d}-\frac {b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (3 a^4-16 a^2 b^2-127 b^4\right )}{8 \left (a^2-b^2\right )^4 d \sqrt {a+b \sin (c+d x)}}+\frac {\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}} \]

output

-1/32*(12*a^2-54*a*b+77*b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/( 
a-b)^(9/2)/d+1/32*(12*a^2+54*a*b+77*b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a 
+b)^(1/2))/(a+b)^(9/2)/d-1/48*b*(18*a^4-81*a^2*b^2-77*b^4)/(a^2-b^2)^3/d/( 
a+b*sin(d*x+c))^(3/2)-1/4*sec(d*x+c)^4*(b-a*sin(d*x+c))/(a^2-b^2)/d/(a+b*s 
in(d*x+c))^(3/2)+1/16*sec(d*x+c)^2*(b*(3*a^2+11*b^2)+2*a*(3*a^2-10*b^2)*si 
n(d*x+c))/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^(3/2)-1/8*a*b*(3*a^4-16*a^2*b^2-1 
27*b^4)/(a^2-b^2)^4/d/(a+b*sin(d*x+c))^(1/2)
 

3.6.32.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.11 (sec) , antiderivative size = 296, normalized size of antiderivative = 0.87 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {\frac {1}{2} \left (18 a^4-81 a^2 b^2-77 b^4\right ) \left ((a+b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b}\right )+(-a+b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \sin (c+d x)}{a+b}\right )\right )-12 (a-b)^2 (a+b)^2 \sec ^4(c+d x) (-b+a \sin (c+d x))-15 a \left (3 a^2-10 b^2\right ) \left ((a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b}\right )+(-a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin (c+d x)}{a+b}\right )\right ) (a+b \sin (c+d x))-3 (a-b) (a+b) \sec ^2(c+d x) \left (3 a^2 b+11 b^3+\left (6 a^3-20 a b^2\right ) \sin (c+d x)\right )}{48 \left (a^2-b^2\right )^2 \left (-a^2+b^2\right ) d (a+b \sin (c+d x))^{3/2}} \]

input

Integrate[Sec[c + d*x]^5/(a + b*Sin[c + d*x])^(5/2),x]
 

output

(((18*a^4 - 81*a^2*b^2 - 77*b^4)*((a + b)*Hypergeometric2F1[-3/2, 1, -1/2, 
 (a + b*Sin[c + d*x])/(a - b)] + (-a + b)*Hypergeometric2F1[-3/2, 1, -1/2, 
 (a + b*Sin[c + d*x])/(a + b)]))/2 - 12*(a - b)^2*(a + b)^2*Sec[c + d*x]^4 
*(-b + a*Sin[c + d*x]) - 15*a*(3*a^2 - 10*b^2)*((a + b)*Hypergeometric2F1[ 
-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a - b)] + (-a + b)*Hypergeometric2F1[- 
1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a + b)])*(a + b*Sin[c + d*x]) - 3*(a - 
b)*(a + b)*Sec[c + d*x]^2*(3*a^2*b + 11*b^3 + (6*a^3 - 20*a*b^2)*Sin[c + d 
*x]))/(48*(a^2 - b^2)^2*(-a^2 + b^2)*d*(a + b*Sin[c + d*x])^(3/2))
 

3.6.32.3 Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.30, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 3147, 496, 27, 686, 27, 655, 25, 655, 25, 654, 25, 1480, 220}

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.

\(\displaystyle \int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (c+d x)^5 (a+b \sin (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3147

\(\displaystyle \frac {b^5 \int \frac {1}{(a+b \sin (c+d x))^{5/2} \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 496

\(\displaystyle \frac {b^5 \left (\frac {\int \frac {6 a^2+9 b \sin (c+d x) a-11 b^2}{2 (a+b \sin (c+d x))^{5/2} \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^5 \left (\frac {\int \frac {6 a^2+9 b \sin (c+d x) a-11 b^2}{(a+b \sin (c+d x))^{5/2} \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{8 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 686

\(\displaystyle \frac {b^5 \left (\frac {\frac {2 a b \left (3 a^2-10 b^2\right ) \sin (c+d x)+b^2 \left (3 a^2+11 b^2\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}-\frac {\int -\frac {12 a^4-19 b^2 a^2+10 b \left (3 a^2-10 b^2\right ) \sin (c+d x) a+77 b^4}{2 (a+b \sin (c+d x))^{5/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2 \left (a^2-b^2\right )}}{8 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^5 \left (\frac {\frac {\int \frac {12 a^4-19 b^2 a^2+10 b \left (3 a^2-10 b^2\right ) \sin (c+d x) a+77 b^4}{(a+b \sin (c+d x))^{5/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{4 b^2 \left (a^2-b^2\right )}+\frac {2 a b \left (3 a^2-10 b^2\right ) \sin (c+d x)+b^2 \left (3 a^2+11 b^2\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}}{8 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 655

\(\displaystyle \frac {b^5 \left (\frac {\frac {-\frac {\int -\frac {a \left (12 a^4-49 b^2 a^2+177 b^4\right )+b \left (18 a^4-81 b^2 a^2-77 b^4\right ) \sin (c+d x)}{(a+b \sin (c+d x))^{3/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {2 \left (18 a^4-81 a^2 b^2-77 b^4\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}+\frac {2 a b \left (3 a^2-10 b^2\right ) \sin (c+d x)+b^2 \left (3 a^2+11 b^2\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}}{8 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b^5 \left (\frac {\frac {\frac {\int \frac {a \left (12 a^4-49 b^2 a^2+177 b^4\right )+b \left (18 a^4-81 b^2 a^2-77 b^4\right ) \sin (c+d x)}{(a+b \sin (c+d x))^{3/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {2 \left (18 a^4-81 a^2 b^2-77 b^4\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}+\frac {2 a b \left (3 a^2-10 b^2\right ) \sin (c+d x)+b^2 \left (3 a^2+11 b^2\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}}{8 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 655

\(\displaystyle \frac {b^5 \left (\frac {\frac {\frac {-\frac {\int -\frac {12 a^6-67 b^2 a^4+258 b^4 a^2+2 b \left (3 a^4-16 b^2 a^2-127 b^4\right ) \sin (c+d x) a+77 b^6}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {4 a \left (3 a^4-16 a^2 b^2-127 b^4\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (18 a^4-81 a^2 b^2-77 b^4\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}+\frac {2 a b \left (3 a^2-10 b^2\right ) \sin (c+d x)+b^2 \left (3 a^2+11 b^2\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}}{8 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b^5 \left (\frac {\frac {\frac {\frac {\int \frac {12 a^6-67 b^2 a^4+258 b^4 a^2+2 b \left (3 a^4-16 b^2 a^2-127 b^4\right ) \sin (c+d x) a+77 b^6}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {4 a \left (3 a^4-16 a^2 b^2-127 b^4\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (18 a^4-81 a^2 b^2-77 b^4\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}+\frac {2 a b \left (3 a^2-10 b^2\right ) \sin (c+d x)+b^2 \left (3 a^2+11 b^2\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}}{8 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 654

\(\displaystyle \frac {b^5 \left (\frac {\frac {\frac {\frac {2 \int -\frac {6 a^6-35 b^2 a^4+512 b^4 a^2+2 b^2 \left (3 a^4-16 b^2 a^2-127 b^4\right ) \sin ^2(c+d x) a+77 b^6}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{a^2-b^2}-\frac {4 a \left (3 a^4-16 a^2 b^2-127 b^4\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (18 a^4-81 a^2 b^2-77 b^4\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}+\frac {2 a b \left (3 a^2-10 b^2\right ) \sin (c+d x)+b^2 \left (3 a^2+11 b^2\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}}{8 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b^5 \left (\frac {\frac {\frac {-\frac {2 \int \frac {6 a^6-35 b^2 a^4+512 b^4 a^2+2 b^2 \left (3 a^4-16 b^2 a^2-127 b^4\right ) \sin ^2(c+d x) a+77 b^6}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{a^2-b^2}-\frac {4 a \left (3 a^4-16 a^2 b^2-127 b^4\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (18 a^4-81 a^2 b^2-77 b^4\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}+\frac {2 a b \left (3 a^2-10 b^2\right ) \sin (c+d x)+b^2 \left (3 a^2+11 b^2\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}}{8 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 1480

\(\displaystyle \frac {b^5 \left (\frac {\frac {\frac {\frac {2 \left (\frac {(a+b)^4 \left (12 a^2-54 a b+77 b^2\right ) \int \frac {1}{b^2 \sin ^2(c+d x)-a+b}d\sqrt {a+b \sin (c+d x)}}{2 b}-\frac {(a-b)^4 \left (12 a^2+54 a b+77 b^2\right ) \int \frac {1}{b^2 \sin ^2(c+d x)-a-b}d\sqrt {a+b \sin (c+d x)}}{2 b}\right )}{a^2-b^2}-\frac {4 a \left (3 a^4-16 a^2 b^2-127 b^4\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (18 a^4-81 a^2 b^2-77 b^4\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}+\frac {2 a b \left (3 a^2-10 b^2\right ) \sin (c+d x)+b^2 \left (3 a^2+11 b^2\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}}{8 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {b^5 \left (\frac {\frac {2 a b \left (3 a^2-10 b^2\right ) \sin (c+d x)+b^2 \left (3 a^2+11 b^2\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}+\frac {\frac {\frac {2 \left (\frac {(a-b)^4 \left (12 a^2+54 a b+77 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{2 b \sqrt {a+b}}-\frac {(a+b)^4 \left (12 a^2-54 a b+77 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{2 b \sqrt {a-b}}\right )}{a^2-b^2}-\frac {4 a \left (3 a^4-16 a^2 b^2-127 b^4\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (18 a^4-81 a^2 b^2-77 b^4\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}}{8 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2 (a+b \sin (c+d x))^{3/2}}\right )}{d}\)

input

Int[Sec[c + d*x]^5/(a + b*Sin[c + d*x])^(5/2),x]
 

output

(b^5*(-1/4*(b^2 - a*b*Sin[c + d*x])/(b^2*(a^2 - b^2)*(a + b*Sin[c + d*x])^ 
(3/2)*(b^2 - b^2*Sin[c + d*x]^2)^2) + ((b^2*(3*a^2 + 11*b^2) + 2*a*b*(3*a^ 
2 - 10*b^2)*Sin[c + d*x])/(2*b^2*(a^2 - b^2)*(a + b*Sin[c + d*x])^(3/2)*(b 
^2 - b^2*Sin[c + d*x]^2)) + ((-2*(18*a^4 - 81*a^2*b^2 - 77*b^4))/(3*(a^2 - 
 b^2)*(a + b*Sin[c + d*x])^(3/2)) + ((2*(-1/2*((a + b)^4*(12*a^2 - 54*a*b 
+ 77*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(Sqrt[a - b]*b) + 
 ((a - b)^4*(12*a^2 + 54*a*b + 77*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sq 
rt[a + b]])/(2*b*Sqrt[a + b])))/(a^2 - b^2) - (4*a*(3*a^4 - 16*a^2*b^2 - 1 
27*b^4))/((a^2 - b^2)*Sqrt[a + b*Sin[c + d*x]]))/(a^2 - b^2))/(4*b^2*(a^2 
- b^2)))/(8*b^2*(a^2 - b^2))))/d
 

3.6.32.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 220

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 
1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && 
 (LtQ[a, 0] || GtQ[b, 0])
 

rule 496

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 
 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a 
 + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 
*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad 
raticQ[a, 0, b, c, d, n, p, x]
 

rule 654

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), 
x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d* 
x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
 

rule 655

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), 
 x_Symbol] :> Simp[(e*f - d*g)*((d + e*x)^(m + 1)/((m + 1)*(c*d^2 + a*e^2)) 
), x] + Simp[1/(c*d^2 + a*e^2)   Int[(d + e*x)^(m + 1)*(Simp[c*d*f + a*e*g 
- c*(e*f - d*g)*x, x]/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] 
&& FractionQ[m] && LtQ[m, -1]
 

rule 686

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + 
a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 
1/(2*a*c*(p + 1)*(c*d^2 + a*e^2))   Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim 
p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f 
+ a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ 
[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3147

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p 
 - 1)/2] && NeQ[a^2 - b^2, 0]
 

3.6.32.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(631\) vs. \(2(307)=614\).

Time = 1.02 (sec) , antiderivative size = 632, normalized size of antiderivative = 1.86

method	result	size
default	\(\frac {\frac {2 b^{5}}{3 \left (a +b \right )^{3} \left (a -b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {12 b^{5} a}{\left (a +b \right )^{4} \left (a -b \right )^{4} \sqrt {a +b \sin \left (d x +c \right )}}-\frac {3 b \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a}{16 \left (a -b \right )^{4} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {17 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{32 \left (a -b \right )^{4} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {3 b \sqrt {a +b \sin \left (d x +c \right )}\, a^{2}}{16 \left (a -b \right )^{4} \left (b \sin \left (d x +c \right )+b \right )^{2}}-\frac {25 b^{2} \sqrt {a +b \sin \left (d x +c \right )}\, a}{32 \left (a -b \right )^{4} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {19 b^{3} \sqrt {a +b \sin \left (d x +c \right )}}{32 \left (a -b \right )^{4} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a^{2}}{8 \left (a -b \right )^{4} \sqrt {-a +b}}-\frac {27 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{16 \left (a -b \right )^{4} \sqrt {-a +b}}+\frac {77 b^{2} \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{32 \left (a -b \right )^{4} \sqrt {-a +b}}-\frac {3 b \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a}{16 \left (a +b \right )^{4} \left (b \sin \left (d x +c \right )-b \right )^{2}}-\frac {17 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{32 \left (a +b \right )^{4} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {3 b \sqrt {a +b \sin \left (d x +c \right )}\, a^{2}}{16 \left (a +b \right )^{4} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {25 b^{2} \sqrt {a +b \sin \left (d x +c \right )}\, a}{32 \left (a +b \right )^{4} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {19 b^{3} \sqrt {a +b \sin \left (d x +c \right )}}{32 \left (a +b \right )^{4} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a^{2}}{8 \left (a +b \right )^{\frac {9}{2}}}+\frac {27 b \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{16 \left (a +b \right )^{\frac {9}{2}}}+\frac {77 b^{2} \operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{32 \left (a +b \right )^{\frac {9}{2}}}}{d}\)	\(632\)

input

int(sec(d*x+c)^5/(a+b*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

output

(2/3*b^5/(a+b)^3/(a-b)^3/(a+b*sin(d*x+c))^(3/2)+12*b^5*a/(a+b)^4/(a-b)^4/( 
a+b*sin(d*x+c))^(1/2)-3/16*b/(a-b)^4/(b*sin(d*x+c)+b)^2*(a+b*sin(d*x+c))^( 
3/2)*a+17/32*b^2/(a-b)^4/(b*sin(d*x+c)+b)^2*(a+b*sin(d*x+c))^(3/2)+3/16*b/ 
(a-b)^4/(b*sin(d*x+c)+b)^2*(a+b*sin(d*x+c))^(1/2)*a^2-25/32*b^2/(a-b)^4/(b 
*sin(d*x+c)+b)^2*(a+b*sin(d*x+c))^(1/2)*a+19/32*b^3/(a-b)^4/(b*sin(d*x+c)+ 
b)^2*(a+b*sin(d*x+c))^(1/2)+3/8/(a-b)^4/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c 
))^(1/2)/(-a+b)^(1/2))*a^2-27/16*b/(a-b)^4/(-a+b)^(1/2)*arctan((a+b*sin(d* 
x+c))^(1/2)/(-a+b)^(1/2))*a+77/32*b^2/(a-b)^4/(-a+b)^(1/2)*arctan((a+b*sin 
(d*x+c))^(1/2)/(-a+b)^(1/2))-3/16*b/(a+b)^4/(b*sin(d*x+c)-b)^2*(a+b*sin(d* 
x+c))^(3/2)*a-17/32*b^2/(a+b)^4/(b*sin(d*x+c)-b)^2*(a+b*sin(d*x+c))^(3/2)+ 
3/16*b/(a+b)^4/(b*sin(d*x+c)-b)^2*(a+b*sin(d*x+c))^(1/2)*a^2+25/32*b^2/(a+ 
b)^4/(b*sin(d*x+c)-b)^2*(a+b*sin(d*x+c))^(1/2)*a+19/32*b^3/(a+b)^4/(b*sin( 
d*x+c)-b)^2*(a+b*sin(d*x+c))^(1/2)+3/8/(a+b)^(9/2)*arctanh((a+b*sin(d*x+c) 
)^(1/2)/(a+b)^(1/2))*a^2+27/16*b/(a+b)^(9/2)*arctanh((a+b*sin(d*x+c))^(1/2 
)/(a+b)^(1/2))*a+77/32*b^2/(a+b)^(9/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b 
)^(1/2)))/d
 

3.6.32.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1191 vs. \(2 (308) = 616\).

Time = 1.36 (sec) , antiderivative size = 5313, normalized size of antiderivative = 15.67 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

input

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")
 

output

Too large to include
 

3.6.32.6 Sympy [F]

\[ \int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

input

integrate(sec(d*x+c)**5/(a+b*sin(d*x+c))**(5/2),x)
 

output

Integral(sec(c + d*x)**5/(a + b*sin(c + d*x))**(5/2), x)
 

3.6.32.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Exception raised: ValueError} \]

input

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")
 

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for m 
ore detail
 

3.6.32.8 Giac [F(-1)]

Timed out. \[ \int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]

input

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")
 

output

Timed out
 

3.6.32.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Hanged} \]

input

int(1/(cos(c + d*x)^5*(a + b*sin(c + d*x))^(5/2)),x)
 

output

\text{Hanged}