\(\int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\) [359]

Optimal result

Integrand size = 33, antiderivative size = 198 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}-\frac {(i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}-\frac {2 a^2 (A b-a B)}{3 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.73 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.31 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {2 (a-i b) (a+i b) (A b+2 a B)+b (A b-a B) \left (i (a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )-(i a+b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )\right )+6 (a-i b) (a+i b) b B \tan (c+d x)+3 b B \left (i (a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )-(i a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )\right ) (a+b \tan (c+d x))}{3 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}} \] Input:

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

Output:

-1/3*(2*(a - I*b)*(a + I*b)*(A*b + 2*a*B) + b*(A*b - a*B)*(I*(a + I*b)*Hyp 
ergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a - I*b)] - (I*a + b)* 
Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a + I*b)]) + 6*(a - 
 I*b)*(a + I*b)*b*B*Tan[c + d*x] + 3*b*B*(I*(a + I*b)*Hypergeometric2F1[-1 
/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I*b)] - (I*a + b)*Hypergeometric2F1[ 
-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)])*(a + b*Tan[c + d*x]))/(b^2* 
(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 1.10 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4087, 25, 3042, 4111, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4087

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4111

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-2*a^2*(A*b - a*B))/(3*b^2*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) - (( 
((a + I*b)^2*b*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b 
]*d) + ((a - I*b)^2*b*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[ 
a + I*b]*d))/(a^2 + b^2) - (2*a*(2*A*b^3 - a*(a^2 + 3*b^2)*B))/(b*(a^2 + b 
^2)*d*Sqrt[a + b*Tan[c + d*x]]))/(b*(a^2 + b^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 4087

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

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 
]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4486\) vs. \(2(174)=348\).

Time = 0.16 (sec) , antiderivative size = 4487, normalized size of antiderivative = 22.66

method	result	size
parts	\(\text {Expression too large to display}\)	\(4487\)
derivativedivides	\(\text {Expression too large to display}\)	\(12849\)
default	\(\text {Expression too large to display}\)	\(12849\)

Input:

int(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x,method=_RETURNV 
ERBOSE)
 

Output:

A*(-1/d*b^3/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*t 
an(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1 
/2))+2/d*b^5/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b* 
tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^( 
1/2))+1/d*b^3/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^ 
2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^ 
(1/2))-2/d*b^5/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a 
^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a) 
^(1/2))+1/4/d*b^3/(a^2+b^2)^3*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2* 
(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)- 
1/4/d*b^3/(a^2+b^2)^3*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1 
/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+2/d*b/(a 
^2+b^2)^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+( 
2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3-2/d*b/(a^ 
2+b^2)^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/ 
2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3+2/d*b^3/(a 
^2+b^2)^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+( 
2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/4/d/b/(a^ 
2+b^2)^3*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x 
+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^4-2/d*b^3/(a^2+b...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7218 vs. \(2 (168) = 336\).

Time = 2.09 (sec) , antiderivative size = 7218, normalized size of antiderivative = 36.45 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorith 
m="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

integrate(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorith 
m="maxima")
 

Output:

Timed out
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorith 
m="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 = 21.53 (sec) , antiderivative size = 9468, normalized size of antiderivative = 47.82 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

(log((((a + b*tan(c + d*x))^(1/2)*(320*A^2*a^4*b^14*d^3 - 16*A^2*b^18*d^3 
+ 1024*A^2*a^6*b^12*d^3 + 1440*A^2*a^8*b^10*d^3 + 1024*A^2*a^10*b^8*d^3 + 
320*A^2*a^12*b^6*d^3 - 16*A^2*a^16*b^2*d^3) + ((((320*A^4*a^2*b^8*d^4 - 16 
*A^4*b^10*d^4 - 1760*A^4*a^4*b^6*d^4 + 1600*A^4*a^6*b^4*d^4 - 400*A^4*a^8* 
b^2*d^4)^(1/2) - 4*A^2*a^5*d^2 + 40*A^2*a^3*b^2*d^2 - 20*A^2*a*b^4*d^2)/(a 
^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a 
^8*b^2*d^4))^(1/2)*(896*A*a^6*b^15*d^4 - 32*A*b^21*d^4 - 160*A*a^2*b^19*d^ 
4 - 128*A*a^4*b^17*d^4 - ((((320*A^4*a^2*b^8*d^4 - 16*A^4*b^10*d^4 - 1760* 
A^4*a^4*b^6*d^4 + 1600*A^4*a^6*b^4*d^4 - 400*A^4*a^8*b^2*d^4)^(1/2) - 4*A^ 
2*a^5*d^2 + 40*A^2*a^3*b^2*d^2 - 20*A^2*a*b^4*d^2)/(a^10*d^4 + b^10*d^4 + 
5*a^2*b^8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2)*(a 
 + b*tan(c + d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 + 2880*a^5*b^18 
*d^5 + 7680*a^7*b^16*d^5 + 13440*a^9*b^14*d^5 + 16128*a^11*b^12*d^5 + 1344 
0*a^13*b^10*d^5 + 7680*a^15*b^8*d^5 + 2880*a^17*b^6*d^5 + 640*a^19*b^4*d^5 
 + 64*a^21*b^2*d^5))/4 + 3136*A*a^8*b^13*d^4 + 4928*A*a^10*b^11*d^4 + 4480 
*A*a^12*b^9*d^4 + 2432*A*a^14*b^7*d^4 + 736*A*a^16*b^5*d^4 + 96*A*a^18*b^3 
*d^4))/4)*(((320*A^4*a^2*b^8*d^4 - 16*A^4*b^10*d^4 - 1760*A^4*a^4*b^6*d^4 
+ 1600*A^4*a^6*b^4*d^4 - 400*A^4*a^8*b^2*d^4)^(1/2) - 4*A^2*a^5*d^2 + 40*A 
^2*a^3*b^2*d^2 - 20*A^2*a*b^4*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 
10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2))/4 - 96*A^3*a^3...
 

Reduce [F]

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

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

Output:

int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2)/(tan(c + d*x)**2*b**2 + 2*t 
an(c + d*x)*a*b + a**2),x)