\(\int \frac {\tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\) [540]

Optimal result

Integrand size = 23, antiderivative size = 125 \[ \int \frac {\tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}-\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}-\frac {2 a^2}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \] Output:

I*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(3/2)/d-I*arctanh( 
(a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(3/2)/d-2*a^2/b/(a^2+b^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 = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.95 \[ \int \frac {\tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {b (-i a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )-(a-i b) \left (2 a+2 i b-i b \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )\right )}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \] Input:

Integrate[Tan[c + d*x]^2/(a + b*Tan[c + d*x])^(3/2),x]
 

Output:

(b*((-I)*a + b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - 
I*b)] - (a - I*b)*(2*a + (2*I)*b - I*b*Hypergeometric2F1[-1/2, 1, 1/2, (a 
+ b*Tan[c + d*x])/(a + I*b)]))/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.64 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4025, 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.

\(\displaystyle \int \frac {\tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4025

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[Tan[c + d*x]^2/(a + b*Tan[c + d*x])^(3/2),x]
 

Output:

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

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 4025

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*c - a*d)^2*((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[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Ta 
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 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. \(1931\) vs. \(2(105)=210\).

Time = 0.25 (sec) , antiderivative size = 1932, normalized size of antiderivative = 15.46

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1932\)
default	\(\text {Expression too large to display}\)	\(1932\)

Input:

int(tan(d*x+c)^2/(a+b*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2*a^2/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)-1/4/d/b/(a^2+b^2)^2*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)*a^3-1/4/d*b/(a^2+b^2)^2*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)*a+1/4/d/b/(a^2+b^2)^(5/2)*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)*a^4-1/4/d*b^3/(a^2+b^2)^(5/2)*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)-4/d*b/(a^2+b^2)^(5/2)/(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))*a^3-3/d*b^3/(a^2+b^2)^(5/2)/(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))*a+1/d/b/(a^2+b^2)^(3/2)/(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))*a^3+1/d*b/(a^2+b^2)^(3/2)/(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))*a+1/d*b/(a^2+b^2)^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))*a^2-1/d/b/(a^2+b^2)^(5/2)/(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/...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1971 vs. \(2 (99) = 198\).

Time = 0.12 (sec) , antiderivative size = 1971, normalized size of antiderivative = 15.77 \[ \int \frac {\tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^2/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-1/2*(4*sqrt(b*tan(d*x + c) + a)*a^2 - ((a^2*b^2 + b^4)*d*tan(d*x + c) + ( 
a^3*b + a*b^3)*d)*sqrt(-((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9* 
a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 
 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + a^3 - 3*a*b^2)/((a^6 + 3*a^4*b^2 
+ 3*a^2*b^4 + b^6)*d^2))*log(-(3*a^2*b - b^3)*sqrt(b*tan(d*x + c) + a) + ( 
(a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d^3*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6 
)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 
+ b^12)*d^4)) + 2*(3*a^3*b^2 - a*b^4)*d)*sqrt(-((a^6 + 3*a^4*b^2 + 3*a^2*b 
^4 + b^6)*d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15 
*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + a^3 - 3*a* 
b^2)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))) + ((a^2*b^2 + b^4)*d*tan( 
d*x + c) + (a^3*b + a*b^3)*d)*sqrt(-((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d 
^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 
20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + a^3 - 3*a*b^2)/((a^6 
+ 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*log(-(3*a^2*b - b^3)*sqrt(b*tan(d*x + 
 c) + a) - ((a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d^3*sqrt(-(9*a^4*b^2 - 6*a 
^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 
 6*a^2*b^10 + b^12)*d^4)) + 2*(3*a^3*b^2 - a*b^4)*d)*sqrt(-((a^6 + 3*a^4*b 
^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a 
^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4...
 

Sympy [F]

\[ \int \frac {\tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\tan ^{2}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(tan(d*x+c)**2/(a+b*tan(d*x+c))**(3/2),x)
 

Output:

Integral(tan(c + d*x)**2/(a + b*tan(c + d*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {\tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{2}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(tan(d*x+c)^2/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate(tan(d*x + c)^2/(b*tan(d*x + c) + a)^(3/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^2/(a+b*tan(d*x+c))^(3/2),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,[3,9,3]%%%}+%%%{4,[3,7,3]%%%}+%%%{6,[3,5,3]%%%}+%%%{ 
4,[3,3,3]
 

Mupad [B] (verification not implemented)

Time = 3.08 (sec) , antiderivative size = 2239, normalized size of antiderivative = 17.91 \[ \int \frac {\tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^2/(a + b*tan(c + d*x))^(3/2),x)
 

Output:

(log(((((-1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i))^(1/2)*(64 
*a*b^11*d^4 - ((-1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i))^(1 
/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10*d^5 + 640*a^5 
*b^8*d^5 + 640*a^7*b^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2*d^5))/2 + 256*a 
^3*b^9*d^4 + 384*a^5*b^7*d^4 + 256*a^7*b^5*d^4 + 64*a^9*b^3*d^4))/2 + (a + 
 b*tan(c + d*x))^(1/2)*(16*b^10*d^3 + 32*a^2*b^8*d^3 - 32*a^6*b^4*d^3 - 16 
*a^8*b^2*d^3))*(-1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i))^(1 
/2))/2 - 8*b^9*d^2 - 24*a^2*b^7*d^2 - 24*a^4*b^5*d^2 - 8*a^6*b^3*d^2)*(-1/ 
(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i))^(1/2))/2 - log(((-1/( 
4*(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i)))^(1/2)*(64*a*b^11*d 
^4 + (-1/(4*(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i)))^(1/2)*(a 
 + b*tan(c + d*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10*d^5 + 640*a^5*b^8*d 
^5 + 640*a^7*b^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2*d^5) + 256*a^3*b^9*d^ 
4 + 384*a^5*b^7*d^4 + 256*a^7*b^5*d^4 + 64*a^9*b^3*d^4) - (a + b*tan(c + d 
*x))^(1/2)*(16*b^10*d^3 + 32*a^2*b^8*d^3 - 32*a^6*b^4*d^3 - 16*a^8*b^2*d^3 
))*(-1/(4*(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i)))^(1/2) - 8* 
b^9*d^2 - 24*a^2*b^7*d^2 - 24*a^4*b^5*d^2 - 8*a^6*b^3*d^2)*(-1/(4*(a^3*d^2 
 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i)))^(1/2) - atan((((-1i/(4*(a^3* 
d^2*1i + b^3*d^2 - a*b^2*d^2*3i - 3*a^2*b*d^2)))^(1/2)*(64*a*b^11*d^4 + (- 
1i/(4*(a^3*d^2*1i + b^3*d^2 - a*b^2*d^2*3i - 3*a^2*b*d^2)))^(1/2)*(a + ...
 

Reduce [F]

\[ \int \frac {\tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{2} b^{2}+2 \tan \left (d x +c \right ) a b +a^{2}}d x \] Input:

int(tan(d*x+c)^2/(a+b*tan(d*x+c))^(3/2),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)