\(\int \frac {(c+d \tan (e+f x))^{5/2} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(a+b \tan (e+f x))^{5/2}} \, dx\) [144]

Optimal result

Integrand size = 49, antiderivative size = 545 \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx=-\frac {(i A+B-i C) (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{5/2} f}-\frac {(B-i (A-C)) (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2} f}+\frac {d^{3/2} (5 b c C+2 b B d-5 a C d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{7/2} f}-\frac {d \left (2 a^3 b B d-5 a^4 C d-2 a b^3 (2 A c-2 c C-3 B d)+2 a^2 b^2 (B c-5 C d)-b^4 (2 B c+(4 A+C) d)\right ) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right )^2 f}+\frac {2 \left (2 a^3 b B d-5 a^4 C d-b^4 (3 B c+5 A d)-2 a b^3 (3 A c-3 c C-4 B d)+a^2 b^2 (3 B c+(A-11 C) d)\right ) (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right )^2 f \sqrt {a+b \tan (e+f x)}}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}} \] Output:

-(I*A+B-I*C)*(c-I*d)^(5/2)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a 
-I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/(a-I*b)^(5/2)/f-(B-I*(A-C))*(c+I*d)^(5 
/2)*arctanh((c+I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f* 
x+e))^(1/2))/(a+I*b)^(5/2)/f+d^(3/2)*(2*B*b*d-5*C*a*d+5*C*b*c)*arctanh(d^( 
1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan(f*x+e))^(1/2))/b^(7/2)/f-d*(2 
*a^3*b*B*d-5*a^4*C*d-2*a*b^3*(2*A*c-3*B*d-2*C*c)+2*a^2*b^2*(B*c-5*C*d)-b^4 
*(2*B*c+(4*A+C)*d))*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/b^3/(a^2 
+b^2)^2/f+2/3*(2*a^3*b*B*d-5*a^4*C*d-b^4*(5*A*d+3*B*c)-2*a*b^3*(3*A*c-4*B* 
d-3*C*c)+a^2*b^2*(3*B*c+(A-11*C)*d))*(c+d*tan(f*x+e))^(3/2)/b^2/(a^2+b^2)^ 
2/f/(a+b*tan(f*x+e))^(1/2)-2/3*(A*b^2-a*(B*b-C*a))*(c+d*tan(f*x+e))^(5/2)/ 
b/(a^2+b^2)/f/(a+b*tan(f*x+e))^(3/2)
 

Mathematica [C] (verified)

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

Time = 6.99 (sec) , antiderivative size = 802, normalized size of antiderivative = 1.47 \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx=\frac {C (c+d \tan (e+f x))^{5/2}}{b f (a+b \tan (e+f x))^{3/2}}+\frac {-\frac {2 b (A-i B-C) (c-i d) (c+d \tan (e+f x))^{3/2}}{3 (i a+b) f (a+b \tan (e+f x))^{3/2}}+\frac {2 b (A+i B-C) (c+i d) (c+d \tan (e+f x))^{3/2}}{3 (i a-b) f (a+b \tan (e+f x))^{3/2}}-\frac {10 c C (b c-a d) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) \sqrt {c+d \tan (e+f x)}}{3 b f (a+b \tan (e+f x))^{3/2} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}-\frac {4 B d (b c-a d) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) \sqrt {c+d \tan (e+f x)}}{3 b f (a+b \tan (e+f x))^{3/2} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}+\frac {10 a C d (b c-a d) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) \sqrt {c+d \tan (e+f x)}}{3 b^2 f (a+b \tan (e+f x))^{3/2} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}+\frac {2 b (i A+B-i C) (c-i d)^2 \left (\frac {\sqrt {-c+i d} \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-a+i b)^{3/2}}+\frac {\sqrt {c+d \tan (e+f x)}}{(a-i b) \sqrt {a+b \tan (e+f x)}}\right )}{(a-i b) f}-\frac {2 b (A+i B-C) (c+i d)^2 \left (\frac {\sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2}}-\frac {\sqrt {c+d \tan (e+f x)}}{(a+i b) \sqrt {a+b \tan (e+f x)}}\right )}{(i a-b) f}}{2 b} \] Input:

Integrate[((c + d*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x] + C*Tan[e + f*x] 
^2))/(a + b*Tan[e + f*x])^(5/2),x]
 

Output:

(C*(c + d*Tan[e + f*x])^(5/2))/(b*f*(a + b*Tan[e + f*x])^(3/2)) + ((-2*b*( 
A - I*B - C)*(c - I*d)*(c + d*Tan[e + f*x])^(3/2))/(3*(I*a + b)*f*(a + b*T 
an[e + f*x])^(3/2)) + (2*b*(A + I*B - C)*(c + I*d)*(c + d*Tan[e + f*x])^(3 
/2))/(3*(I*a - b)*f*(a + b*Tan[e + f*x])^(3/2)) - (10*c*C*(b*c - a*d)*Hype 
rgeometric2F1[-3/2, -3/2, -1/2, -((d*(a + b*Tan[e + f*x]))/(b*c - a*d))]*S 
qrt[c + d*Tan[e + f*x]])/(3*b*f*(a + b*Tan[e + f*x])^(3/2)*Sqrt[(b*(c + d* 
Tan[e + f*x]))/(b*c - a*d)]) - (4*B*d*(b*c - a*d)*Hypergeometric2F1[-3/2, 
-3/2, -1/2, -((d*(a + b*Tan[e + f*x]))/(b*c - a*d))]*Sqrt[c + d*Tan[e + f* 
x]])/(3*b*f*(a + b*Tan[e + f*x])^(3/2)*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c 
- a*d)]) + (10*a*C*d*(b*c - a*d)*Hypergeometric2F1[-3/2, -3/2, -1/2, -((d* 
(a + b*Tan[e + f*x]))/(b*c - a*d))]*Sqrt[c + d*Tan[e + f*x]])/(3*b^2*f*(a 
+ b*Tan[e + f*x])^(3/2)*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)]) + (2*b 
*(I*A + B - I*C)*(c - I*d)^2*((Sqrt[-c + I*d]*ArcTanh[(Sqrt[-c + I*d]*Sqrt 
[a + b*Tan[e + f*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(-a + I* 
b)^(3/2) + Sqrt[c + d*Tan[e + f*x]]/((a - I*b)*Sqrt[a + b*Tan[e + f*x]]))) 
/((a - I*b)*f) - (2*b*(A + I*B - C)*(c + I*d)^2*((Sqrt[c + I*d]*ArcTanh[(S 
qrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f 
*x]])])/(a + I*b)^(3/2) - Sqrt[c + d*Tan[e + f*x]]/((a + I*b)*Sqrt[a + b*T 
an[e + f*x]])))/((I*a - b)*f))/(2*b)
 

Rubi [A] (verified)

Time = 8.33 (sec) , antiderivative size = 603, normalized size of antiderivative = 1.11, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 4128, 27, 3042, 4128, 27, 3042, 4130, 27, 3042, 4138, 2348, 2009}

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.

\(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan (e+f x)^2\right )}{(a+b \tan (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 4128

\(\displaystyle \frac {2 \int \frac {(c+d \tan (e+f x))^{3/2} \left (\left (5 C a^2-2 b B a+2 A b^2+3 b^2 C\right ) d \tan ^2(e+f x)-3 b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)+(b B-a C) (3 b c-5 a d)+A b (3 a c+5 b d)\right )}{2 (a+b \tan (e+f x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(c+d \tan (e+f x))^{3/2} \left (\left (5 C a^2-2 b B a+2 A b^2+3 b^2 C\right ) d \tan ^2(e+f x)-3 b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)+(b B-a C) (3 b c-5 a d)+A b (3 a c+5 b d)\right )}{(a+b \tan (e+f x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(c+d \tan (e+f x))^{3/2} \left (\left (5 C a^2-2 b B a+2 A b^2+3 b^2 C\right ) d \tan (e+f x)^2-3 b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)+(b B-a C) (3 b c-5 a d)+A b (3 a c+5 b d)\right )}{(a+b \tan (e+f x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4128

\(\displaystyle \frac {\frac {2 \int \frac {\sqrt {c+d \tan (e+f x)} \left (-3 ((a c+b d) ((A-C) (b c-a d)-B (a c+b d))+(b c-a d) (b B c+b (A-C) d+a (A c-C c-B d))) \tan (e+f x) b^2+(a c+3 b d) ((b B-a C) (3 b c-5 a d)+A b (3 a c+5 b d)) b-3 d \left (-5 C d a^4+2 b B d a^3+2 b^2 (B c-5 C d) a^2-2 b^3 (2 A c-2 C c-3 B d) a-b^4 (2 B c+(4 A+C) d)\right ) \tan ^2(e+f x)+(b c-3 a d) \left (-5 C d a^3+2 b B d a^2+3 b^2 (B c-2 C d) a-A b^2 (3 b c-a d)+3 b^3 (c C+B d)\right )\right )}{2 \sqrt {a+b \tan (e+f x)}}dx}{b \left (a^2+b^2\right )}+\frac {2 (c+d \tan (e+f x))^{3/2} \left (-5 a^4 C d+2 a^3 b B d+a^2 b^2 (d (A-11 C)+3 B c)-2 a b^3 (3 A c-4 B d-3 c C)-b^4 (5 A d+3 B c)\right )}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sqrt {c+d \tan (e+f x)} \left (-3 ((a c+b d) ((A-C) (b c-a d)-B (a c+b d))+(b c-a d) (b B c+b (A-C) d+a (A c-C c-B d))) \tan (e+f x) b^2+(a c+3 b d) ((b B-a C) (3 b c-5 a d)+A b (3 a c+5 b d)) b-3 d \left (-5 C d a^4+2 b B d a^3+2 b^2 (B c-5 C d) a^2-2 b^3 (2 A c-2 C c-3 B d) a-b^4 (2 B c+(4 A+C) d)\right ) \tan ^2(e+f x)+(b c-3 a d) \left (-5 C d a^3+2 b B d a^2+3 b^2 (B c-2 C d) a-A b^2 (3 b c-a d)+3 b^3 (c C+B d)\right )\right )}{\sqrt {a+b \tan (e+f x)}}dx}{b \left (a^2+b^2\right )}+\frac {2 (c+d \tan (e+f x))^{3/2} \left (-5 a^4 C d+2 a^3 b B d+a^2 b^2 (d (A-11 C)+3 B c)-2 a b^3 (3 A c-4 B d-3 c C)-b^4 (5 A d+3 B c)\right )}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\sqrt {c+d \tan (e+f x)} \left (-3 ((a c+b d) ((A-C) (b c-a d)-B (a c+b d))+(b c-a d) (b B c+b (A-C) d+a (A c-C c-B d))) \tan (e+f x) b^2+(a c+3 b d) ((b B-a C) (3 b c-5 a d)+A b (3 a c+5 b d)) b-3 d \left (-5 C d a^4+2 b B d a^3+2 b^2 (B c-5 C d) a^2-2 b^3 (2 A c-2 C c-3 B d) a-b^4 (2 B c+(4 A+C) d)\right ) \tan (e+f x)^2+(b c-3 a d) \left (-5 C d a^3+2 b B d a^2+3 b^2 (B c-2 C d) a-A b^2 (3 b c-a d)+3 b^3 (c C+B d)\right )\right )}{\sqrt {a+b \tan (e+f x)}}dx}{b \left (a^2+b^2\right )}+\frac {2 (c+d \tan (e+f x))^{3/2} \left (-5 a^4 C d+2 a^3 b B d+a^2 b^2 (d (A-11 C)+3 B c)-2 a b^3 (3 A c-4 B d-3 c C)-b^4 (5 A d+3 B c)\right )}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {\frac {\frac {\int -\frac {3 \left (5 C d^3 a^5-b d^2 (5 c C+2 B d) a^4+10 b^2 C d^3 a^3-2 b^3 \left (A c^3-C c^3-3 B d c^2-3 A d^2 c+8 C d^2 c+3 B d^3\right ) a^2-b^4 \left (4 A d \left (3 c^2-d^2\right )-C d \left (12 c^2+d^2\right )+4 B \left (c^3-3 c d^2\right )\right ) a-\left (a^2+b^2\right )^2 d^2 (5 b c C-5 a d C+2 b B d) \tan ^2(e+f x)-b^5 c \left (2 C c^2+6 B d c-C d^2-2 A \left (c^2-3 d^2\right )\right )+2 b^3 \left (-\left (\left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right ) a^2\right )+2 b \left (A c^3-C c^3-3 B d c^2-3 A d^2 c+3 C d^2 c+B d^3\right ) a+b^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)\right )}{2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{b}-\frac {3 d \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (-5 a^4 C d+2 a^3 b B d+2 a^2 b^2 (B c-5 C d)-2 a b^3 (2 A c-3 B d-2 c C)-b^4 (d (4 A+C)+2 B c)\right )}{b f}}{b \left (a^2+b^2\right )}+\frac {2 (c+d \tan (e+f x))^{3/2} \left (-5 a^4 C d+2 a^3 b B d+a^2 b^2 (d (A-11 C)+3 B c)-2 a b^3 (3 A c-4 B d-3 c C)-b^4 (5 A d+3 B c)\right )}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {3 \int \frac {5 C d^3 a^5-b d^2 (5 c C+2 B d) a^4+10 b^2 C d^3 a^3-2 b^3 \left (A c^3-C c^3-3 B d c^2-3 A d^2 c+8 C d^2 c+3 B d^3\right ) a^2-b^4 \left (4 A d \left (3 c^2-d^2\right )-C d \left (12 c^2+d^2\right )+4 B \left (c^3-3 c d^2\right )\right ) a-\left (a^2+b^2\right )^2 d^2 (5 b c C-5 a d C+2 b B d) \tan ^2(e+f x)-b^5 c \left (2 C c^2+6 B d c-C d^2-2 A \left (c^2-3 d^2\right )\right )-2 b^3 \left (\left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right ) a^2+2 b \left (C c^3+3 B d c^2-3 C d^2 c-B d^3-A \left (c^3-3 c d^2\right )\right ) a-b^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{2 b}-\frac {3 d \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (-5 a^4 C d+2 a^3 b B d+2 a^2 b^2 (B c-5 C d)-2 a b^3 (2 A c-3 B d-2 c C)-b^4 (d (4 A+C)+2 B c)\right )}{b f}}{b \left (a^2+b^2\right )}+\frac {2 (c+d \tan (e+f x))^{3/2} \left (-5 a^4 C d+2 a^3 b B d+a^2 b^2 (d (A-11 C)+3 B c)-2 a b^3 (3 A c-4 B d-3 c C)-b^4 (5 A d+3 B c)\right )}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {3 \int \frac {5 C d^3 a^5-b d^2 (5 c C+2 B d) a^4+10 b^2 C d^3 a^3-2 b^3 \left (A c^3-C c^3-3 B d c^2-3 A d^2 c+8 C d^2 c+3 B d^3\right ) a^2-b^4 \left (4 A d \left (3 c^2-d^2\right )-C d \left (12 c^2+d^2\right )+4 B \left (c^3-3 c d^2\right )\right ) a-\left (a^2+b^2\right )^2 d^2 (5 b c C-5 a d C+2 b B d) \tan (e+f x)^2-b^5 c \left (2 C c^2+6 B d c-C d^2-2 A \left (c^2-3 d^2\right )\right )-2 b^3 \left (\left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right ) a^2+2 b \left (C c^3+3 B d c^2-3 C d^2 c-B d^3-A \left (c^3-3 c d^2\right )\right ) a-b^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{2 b}-\frac {3 d \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (-5 a^4 C d+2 a^3 b B d+2 a^2 b^2 (B c-5 C d)-2 a b^3 (2 A c-3 B d-2 c C)-b^4 (d (4 A+C)+2 B c)\right )}{b f}}{b \left (a^2+b^2\right )}+\frac {2 (c+d \tan (e+f x))^{3/2} \left (-5 a^4 C d+2 a^3 b B d+a^2 b^2 (d (A-11 C)+3 B c)-2 a b^3 (3 A c-4 B d-3 c C)-b^4 (5 A d+3 B c)\right )}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4138

\(\displaystyle \frac {\frac {-\frac {3 \int \frac {5 C d^3 a^5-b d^2 (5 c C+2 B d) a^4+10 b^2 C d^3 a^3-2 b^3 \left (A c^3-C c^3-3 B d c^2-3 A d^2 c+8 C d^2 c+3 B d^3\right ) a^2-b^4 \left (4 A d \left (3 c^2-d^2\right )-C d \left (12 c^2+d^2\right )+4 B \left (c^3-3 c d^2\right )\right ) a-\left (a^2+b^2\right )^2 d^2 (5 b c C-5 a d C+2 b B d) \tan ^2(e+f x)-b^5 c \left (2 C c^2+6 B d c-C d^2-2 A \left (c^2-3 d^2\right )\right )+2 b^3 \left (-\left (\left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right ) a^2\right )+2 b \left (A c^3-C c^3-3 B d c^2-3 A d^2 c+3 C d^2 c+B d^3\right ) a+b^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (\tan ^2(e+f x)+1\right )}d\tan (e+f x)}{2 b f}-\frac {3 d \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (-5 a^4 C d+2 a^3 b B d+2 a^2 b^2 (B c-5 C d)-2 a b^3 (2 A c-3 B d-2 c C)-b^4 (d (4 A+C)+2 B c)\right )}{b f}}{b \left (a^2+b^2\right )}+\frac {2 (c+d \tan (e+f x))^{3/2} \left (-5 a^4 C d+2 a^3 b B d+a^2 b^2 (d (A-11 C)+3 B c)-2 a b^3 (3 A c-4 B d-3 c C)-b^4 (5 A d+3 B c)\right )}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 2348

\(\displaystyle \frac {\frac {2 \left (-5 C d a^4+2 b B d a^3+b^2 (3 B c+(A-11 C) d) a^2-2 b^3 (3 A c-3 C c-4 B d) a-b^4 (3 B c+5 A d)\right ) (c+d \tan (e+f x))^{3/2}}{b \left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}}+\frac {-\frac {3 d \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (-5 C d a^4+2 b B d a^3+2 b^2 (B c-5 C d) a^2-2 b^3 (2 A c-2 C c-3 B d) a-b^4 (2 B c+(4 A+C) d)\right )}{b f}-\frac {3 \int \left (\frac {\left (a^2+b^2\right )^2 (-5 b c C+5 a d C-2 b B d) d^2}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-2 B c^3 b^5+2 A d^3 b^5-2 C d^3 b^5+6 B c d^2 b^5-6 A c^2 d b^5+6 c^2 C d b^5-4 a A c^3 b^4-4 a B d^3 b^4+12 a A c d^2 b^4-12 a c C d^2 b^4+4 a c^3 C b^4+12 a B c^2 d b^4+2 a^2 B c^3 b^3-2 a^2 A d^3 b^3+2 a^2 C d^3 b^3-6 a^2 B c d^2 b^3+6 a^2 A c^2 d b^3-6 a^2 c^2 C d b^3+i \left (2 A c^3 b^5+2 B d^3 b^5-6 A c d^2 b^5+6 c C d^2 b^5-2 c^3 C b^5-6 B c^2 d b^5-4 a B c^3 b^4+4 a A d^3 b^4-4 a C d^3 b^4+12 a B c d^2 b^4-12 a A c^2 d b^4+12 a c^2 C d b^4-2 a^2 A c^3 b^3-2 a^2 B d^3 b^3+6 a^2 A c d^2 b^3-6 a^2 c C d^2 b^3+2 a^2 c^3 C b^3+6 a^2 B c^2 d b^3\right )}{2 (i-\tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {2 B c^3 b^5-2 A d^3 b^5+2 C d^3 b^5-6 B c d^2 b^5+6 A c^2 d b^5-6 c^2 C d b^5+4 a A c^3 b^4+4 a B d^3 b^4-12 a A c d^2 b^4+12 a c C d^2 b^4-4 a c^3 C b^4-12 a B c^2 d b^4-2 a^2 B c^3 b^3+2 a^2 A d^3 b^3-2 a^2 C d^3 b^3+6 a^2 B c d^2 b^3-6 a^2 A c^2 d b^3+6 a^2 c^2 C d b^3+i \left (2 A c^3 b^5+2 B d^3 b^5-6 A c d^2 b^5+6 c C d^2 b^5-2 c^3 C b^5-6 B c^2 d b^5-4 a B c^3 b^4+4 a A d^3 b^4-4 a C d^3 b^4+12 a B c d^2 b^4-12 a A c^2 d b^4+12 a c^2 C d b^4-2 a^2 A c^3 b^3-2 a^2 B d^3 b^3+6 a^2 A c d^2 b^3-6 a^2 c C d^2 b^3+2 a^2 c^3 C b^3+6 a^2 B c^2 d b^3\right )}{2 (\tan (e+f x)+i) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\right )d\tan (e+f x)}{2 b f}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}-\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}+\frac {\frac {2 (c+d \tan (e+f x))^{3/2} \left (-5 a^4 C d+2 a^3 b B d+a^2 b^2 (d (A-11 C)+3 B c)-2 a b^3 (3 A c-4 B d-3 c C)-b^4 (5 A d+3 B c)\right )}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}+\frac {-\frac {3 d \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (-5 a^4 C d+2 a^3 b B d+2 a^2 b^2 (B c-5 C d)-2 a b^3 (2 A c-3 B d-2 c C)-b^4 (d (4 A+C)+2 B c)\right )}{b f}-\frac {3 \left (-\frac {2 d^{3/2} \left (a^2+b^2\right )^2 (-5 a C d+2 b B d+5 b c C) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {b}}+\frac {2 b^3 (a+i b)^2 (c-i d)^{5/2} (B+i (A-C)) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a-i b}}-\frac {2 b^3 (a-i b)^2 (c+i d)^{5/2} (i A-B-i C) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b}}\right )}{2 b f}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\)

Input:

Int[((c + d*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/( 
a + b*Tan[e + f*x])^(5/2),x]
 

Output:

(-2*(A*b^2 - a*(b*B - a*C))*(c + d*Tan[e + f*x])^(5/2))/(3*b*(a^2 + b^2)*f 
*(a + b*Tan[e + f*x])^(3/2)) + ((2*(2*a^3*b*B*d - 5*a^4*C*d - b^4*(3*B*c + 
 5*A*d) - 2*a*b^3*(3*A*c - 3*c*C - 4*B*d) + a^2*b^2*(3*B*c + (A - 11*C)*d) 
)*(c + d*Tan[e + f*x])^(3/2))/(b*(a^2 + b^2)*f*Sqrt[a + b*Tan[e + f*x]]) + 
 ((-3*((2*(a + I*b)^2*b^3*(B + I*(A - C))*(c - I*d)^(5/2)*ArcTanh[(Sqrt[c 
- I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])] 
)/Sqrt[a - I*b] - (2*(a - I*b)^2*b^3*(I*A - B - I*C)*(c + I*d)^(5/2)*ArcTa 
nh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[ 
e + f*x]])])/Sqrt[a + I*b] - (2*(a^2 + b^2)^2*d^(3/2)*(5*b*c*C + 2*b*B*d - 
 5*a*C*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*T 
an[e + f*x]])])/Sqrt[b]))/(2*b*f) - (3*d*(2*a^3*b*B*d - 5*a^4*C*d - 2*a*b^ 
3*(2*A*c - 2*c*C - 3*B*d) + 2*a^2*b^2*(B*c - 5*C*d) - b^4*(2*B*c + (4*A + 
C)*d))*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(b*f))/(b*(a^2 + 
 b^2)))/(3*b*(a^2 + b^2))
 

Defintions of rubi rules used

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2348

Int[(Px_)*((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_. 
)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(c + d*x)^m*(e + f*x)^ 
n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[P 
x, x] && (IntegerQ[p] || (IntegerQ[2*p] && IntegerQ[m] && ILtQ[n, 0])) && 
!(IGtQ[m, 0] && IGtQ[n, 0])
 

rule 3042

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

rule 4128

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Tan[e + 
 f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim 
p[1/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e 
 + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* 
(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b 
*(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], 
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ 
[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 4130

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. 
) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, 
b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && 
 NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

rule 4138

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + 
 d^2, 0]
 

Maple [F(-1)]

Timed out.

\[\int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}} \left (A +B \tan \left (f x +e \right )+C \tan \left (f x +e \right )^{2}\right )}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]

Input:

int((c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e) 
)^(5/2),x)
 

Output:

int((c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e) 
)^(5/2),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan( 
f*x+e))^(5/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((c+d*tan(f*x+e))**(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(a+b*ta 
n(f*x+e))**(5/2),x)
 

Output:

Integral((c + d*tan(e + f*x))**(5/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)* 
*2)/(a + b*tan(e + f*x))**(5/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan( 
f*x+e))^(5/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan( 
f*x+e))^(5/2),x, algorithm="giac")
 

Output:

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(d*tan(f*x + e) + c)^(5/ 
2)/(b*tan(f*x + e) + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx=\text {Hanged} \] Input:

int(((c + d*tan(e + f*x))^(5/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/( 
a + b*tan(e + f*x))^(5/2),x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (d \tan \left (f x +e \right )+c \right )^{\frac {5}{2}} \left (A +B \tan \left (f x +e \right )+C \tan \left (f x +e \right )^{2}\right )}{\left (\tan \left (f x +e \right ) b +a \right )^{\frac {5}{2}}}d x \] Input:

int((c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e) 
)^(5/2),x)
 

Output:

int((c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e) 
)^(5/2),x)