\(\int \frac {(a+b \tan (e+f x)) (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(c+d \tan (e+f x))^{5/2}} \, dx\) [124]

Optimal result

Integrand size = 45, antiderivative size = 273 \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {(a-i b) (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}+\frac {(i a-b) (A+i B-C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{5/2} f}+\frac {2 (b c-a d) \left (c^2 C-B c d+A d^2\right )}{3 d^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (b \left (c^4 C-c^2 (A-3 C) d^2-2 B c d^3+A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{d^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \] Output:

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

Mathematica [C] (verified)

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

Time = 2.00 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {2 (c-i d) (c+i d) (2 b c C+b B d-2 a C d)+d (A b c+a B c-b c C-a A d+b B d+a C d) \left (i (c+i d) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )-(i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )\right )+6 C (c-i d) (c+i d) d (a+b \tan (e+f x))-3 (A b+a B-b C) d \left (i (c+i d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )-(i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )\right ) (c+d \tan (e+f x))}{3 d^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}} \] Input:

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

Output:

-1/3*(2*(c - I*d)*(c + I*d)*(2*b*c*C + b*B*d - 2*a*C*d) + d*(A*b*c + a*B*c 
 - b*c*C - a*A*d + b*B*d + a*C*d)*(I*(c + I*d)*Hypergeometric2F1[-3/2, 1, 
-1/2, (c + d*Tan[e + f*x])/(c - I*d)] - (I*c + d)*Hypergeometric2F1[-3/2, 
1, -1/2, (c + d*Tan[e + f*x])/(c + I*d)]) + 6*C*(c - I*d)*(c + I*d)*d*(a + 
 b*Tan[e + f*x]) - 3*(A*b + a*B - b*C)*d*(I*(c + I*d)*Hypergeometric2F1[-1 
/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)] - (I*c + d)*Hypergeometric2F1[ 
-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)])*(c + d*Tan[e + f*x]))/(d^2* 
(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 2.51 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.10, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 4118, 3042, 4111, 25, 3042, 4022, 3042, 4020, 25, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4118

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4111

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{3 d^2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (-c^2 d^2 (A-3 C)+A d^4-2 B c d^3+c^4 C\right )\right )}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {1}{2} d (a+i b) (c-i d)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {1}{2} d (a-i b) (c+i d)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{3 d^2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (-c^2 d^2 (A-3 C)+A d^4-2 B c d^3+c^4 C\right )\right )}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {1}{2} d (a+i b) (c-i d)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {1}{2} d (a-i b) (c+i d)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(2*(b*c - a*d)*(c^2*C - B*c*d + A*d^2))/(3*d^2*(c^2 + d^2)*f*(c + d*Tan[e 
+ f*x])^(3/2)) + (-((-(((a - I*b)*(A - I*B - C)*(c + I*d)^2*d*ArcTan[Tan[e 
 + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f)) - ((a + I*b)*(A + I*B - C)*(c - 
 I*d)^2*d*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f))/(c^2 + d^ 
2)) - (2*(b*(c^4*C - c^2*(A - 3*C)*d^2 - 2*B*c*d^3 + A*d^4) + a*d^2*(2*c*( 
A - C)*d - B*(c^2 - d^2))))/(d*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]))/(d 
*(c^2 + d^2))
 

Defintions of rubi rules used

rule 25

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

rule 73

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]
 

rule 221

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

rule 3042

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

rule 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

rule 4111

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

rule 4118

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(8982\) vs. \(2(246)=492\).

Time = 0.25 (sec) , antiderivative size = 8983, normalized size of antiderivative = 32.90

method	result	size
parts	\(\text {Expression too large to display}\)	\(8983\)
derivativedivides	\(\text {Expression too large to display}\)	\(40201\)
default	\(\text {Expression too large to display}\)	\(40201\)

Input:

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

Output:

result too large to display
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[6,14,5]%%%}+%%%{6,[6,12,5]%%%}+%%%{15,[6,10,5]%%%}+ 
%%%{20,[6
 

Mupad [B] (verification not implemented)

Time = 83.07 (sec) , antiderivative size = 64641, normalized size of antiderivative = 236.78 \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*(C*b*c^3 + A*b*c*d^2 - B*b*c^2*d))/(3*(c^2 + d^2)) - (2*(c + d*tan(e + 
 f*x))*(A*b*d^4 + C*b*c^4 - 2*B*b*c*d^3 - A*b*c^2*d^2 + 3*C*b*c^2*d^2))/(c 
^2 + d^2)^2)/(d^2*f*(c + d*tan(e + f*x))^(3/2)) - atan(-(((c + d*tan(e + f 
*x))^(1/2)*(16*A^2*b^2*d^18*f^3 - 16*B^2*b^2*d^18*f^3 + 16*C^2*b^2*d^18*f^ 
3 - 320*A^2*b^2*c^4*d^14*f^3 - 1024*A^2*b^2*c^6*d^12*f^3 - 1440*A^2*b^2*c^ 
8*d^10*f^3 - 1024*A^2*b^2*c^10*d^8*f^3 - 320*A^2*b^2*c^12*d^6*f^3 + 16*A^2 
*b^2*c^16*d^2*f^3 + 320*B^2*b^2*c^4*d^14*f^3 + 1024*B^2*b^2*c^6*d^12*f^3 + 
 1440*B^2*b^2*c^8*d^10*f^3 + 1024*B^2*b^2*c^10*d^8*f^3 + 320*B^2*b^2*c^12* 
d^6*f^3 - 16*B^2*b^2*c^16*d^2*f^3 - 320*C^2*b^2*c^4*d^14*f^3 - 1024*C^2*b^ 
2*c^6*d^12*f^3 - 1440*C^2*b^2*c^8*d^10*f^3 - 1024*C^2*b^2*c^10*d^8*f^3 - 3 
20*C^2*b^2*c^12*d^6*f^3 + 16*C^2*b^2*c^16*d^2*f^3 - 32*A*C*b^2*d^18*f^3 - 
128*A*B*b^2*c*d^17*f^3 + 128*B*C*b^2*c*d^17*f^3 - 640*A*B*b^2*c^3*d^15*f^3 
 - 1152*A*B*b^2*c^5*d^13*f^3 - 640*A*B*b^2*c^7*d^11*f^3 + 640*A*B*b^2*c^9* 
d^9*f^3 + 1152*A*B*b^2*c^11*d^7*f^3 + 640*A*B*b^2*c^13*d^5*f^3 + 128*A*B*b 
^2*c^15*d^3*f^3 + 640*A*C*b^2*c^4*d^14*f^3 + 2048*A*C*b^2*c^6*d^12*f^3 + 2 
880*A*C*b^2*c^8*d^10*f^3 + 2048*A*C*b^2*c^10*d^8*f^3 + 640*A*C*b^2*c^12*d^ 
6*f^3 - 32*A*C*b^2*c^16*d^2*f^3 + 640*B*C*b^2*c^3*d^15*f^3 + 1152*B*C*b^2* 
c^5*d^13*f^3 + 640*B*C*b^2*c^7*d^11*f^3 - 640*B*C*b^2*c^9*d^9*f^3 - 1152*B 
*C*b^2*c^11*d^7*f^3 - 640*B*C*b^2*c^13*d^5*f^3 - 128*B*C*b^2*c^15*d^3*f^3) 
 + ((((8*A^2*b^2*c^5*f^2 - 8*B^2*b^2*c^5*f^2 + 8*C^2*b^2*c^5*f^2 - 80*A...
 

Reduce [F]

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

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

Output:

( - 2*sqrt(tan(e + f*x)*d + c)*a**2 + 3*int((sqrt(tan(e + f*x)*d + c)*tan( 
e + f*x)**3)/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2*c*d**2 + 3*tan(e + 
f*x)*c**2*d + c**3),x)*tan(e + f*x)**2*b*c*d**3*f + 6*int((sqrt(tan(e + f* 
x)*d + c)*tan(e + f*x)**3)/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2*c*d** 
2 + 3*tan(e + f*x)*c**2*d + c**3),x)*tan(e + f*x)*b*c**2*d**2*f + 3*int((s 
qrt(tan(e + f*x)*d + c)*tan(e + f*x)**3)/(tan(e + f*x)**3*d**3 + 3*tan(e + 
 f*x)**2*c*d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)*b*c**3*d*f - 3*int((sqr 
t(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)**3*d**3 + 3*tan(e + f 
*x)**2*c*d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)*tan(e + f*x)**2*a**2*d**3 
*f + 3*int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)**3*d** 
3 + 3*tan(e + f*x)**2*c*d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)*tan(e + f* 
x)**2*a*c*d**3*f + 3*int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e 
 + f*x)**3*d**3 + 3*tan(e + f*x)**2*c*d**2 + 3*tan(e + f*x)*c**2*d + c**3) 
,x)*tan(e + f*x)**2*b**2*d**3*f - 6*int((sqrt(tan(e + f*x)*d + c)*tan(e + 
f*x)**2)/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2*c*d**2 + 3*tan(e + f*x) 
*c**2*d + c**3),x)*tan(e + f*x)*a**2*c*d**2*f + 6*int((sqrt(tan(e + f*x)*d 
 + c)*tan(e + f*x)**2)/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2*c*d**2 + 
3*tan(e + f*x)*c**2*d + c**3),x)*tan(e + f*x)*a*c**2*d**2*f + 6*int((sqrt( 
tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x 
)**2*c*d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)*tan(e + f*x)*b**2*c*d**2...