\(\int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2} \, dx\) [1236]

Optimal result

Integrand size = 27, antiderivative size = 195 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2} \, dx=-\frac {i (a-i b)^2 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {i (a+i b)^2 (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (2 a b c+a^2 d-b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a b (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b^2 (c+d \tan (e+f x))^{5/2}}{5 d f} \] Output:

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

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.04 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2} \, dx=\frac {\frac {6 b^2 (c+d \tan (e+f x))^{5/2}}{d}+5 i (a-i b)^2 \left (-3 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c-3 i d+d \tan (e+f x))\right )-5 i (a+i b)^2 \left (-3 (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c+3 i d+d \tan (e+f x))\right )}{15 f} \] Input:

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

Output:

((6*b^2*(c + d*Tan[e + f*x])^(5/2))/d + (5*I)*(a - I*b)^2*(-3*(c - I*d)^(3 
/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + Sqrt[c + d*Tan[e + f 
*x]]*(4*c - (3*I)*d + d*Tan[e + f*x])) - (5*I)*(a + I*b)^2*(-3*(c + I*d)^( 
3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + Sqrt[c + d*Tan[e + 
f*x]]*(4*c + (3*I)*d + d*Tan[e + f*x])))/(15*f)
 

Rubi [A] (warning: unable to verify)

Time = 1.07 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.89, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {3042, 4026, 3042, 4011, 3042, 4011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

\(\displaystyle \frac {1}{2} (a+i b)^2 (c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b)^2 (c-i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a b (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b^2 (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} (a+i b)^2 (c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b)^2 (c-i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a b (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b^2 (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {i (a-i b)^2 (c-i d)^2 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a+i b)^2 (c+i d)^2 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a b (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b^2 (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {i (a-i b)^2 (c-i d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {i (a+i b)^2 (c+i d)^2 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a b (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b^2 (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a+i b)^2 (c+i d)^2 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a-i b)^2 (c-i d)^2 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a b (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b^2 (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \left (a^2 d+2 a b c-b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {(a-i b)^2 (c-i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a+i b)^2 (c+i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}+\frac {4 a b (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b^2 (c+d \tan (e+f x))^{5/2}}{5 d f}\)

Input:

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

Output:

((a - I*b)^2*(c - I*d)^(3/2)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f + ((a + 
 I*b)^2*(c + I*d)^(3/2)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f + (2*(2*a*b* 
c + a^2*d - b^2*d)*Sqrt[c + d*Tan[e + f*x]])/f + (4*a*b*(c + d*Tan[e + f*x 
])^(3/2))/(3*f) + (2*b^2*(c + d*Tan[e + f*x])^(5/2))/(5*d*f)
 

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 4011

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

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 4026

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2489\) vs. \(2(165)=330\).

Time = 0.42 (sec) , antiderivative size = 2490, normalized size of antiderivative = 12.77

method	result	size
parts	\(\text {Expression too large to display}\)	\(2490\)
derivativedivides	\(\text {Expression too large to display}\)	\(2559\)
default	\(\text {Expression too large to display}\)	\(2559\)

Input:

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

Output:

a^2*(2/f*d*(c+d*tan(f*x+e))^(1/2)-1/4/f/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e 
))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2) 
+2*c)^(1/2)*(c^2+d^2)^(1/2)*c+1/4/f/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^( 
1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c 
)^(1/2)*c^2-1/4/f*d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^ 
(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+2/f*d/(2*( 
c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1 
/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-1/f*d/(2*(c^2+d^2)^(1/2)- 
2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)) 
/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)+1/4/f/d*ln(d*tan(f*x+e)+c- 
(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(c^2 
+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-1/4/f/d*ln(d*tan(f*x+e)+c-(c+d 
*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+ 
d^2)^(1/2)+2*c)^(1/2)*c^2+1/4/f*d*ln(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2) 
*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1 
/2)+2/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-( 
2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-1/f*d/(2*(c 
^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^(1/ 
2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2))+b^2*(2/5/d/ 
f*(c+d*tan(f*x+e))^(5/2)-2/f*d*(c+d*tan(f*x+e))^(1/2)+1/4/f/d*ln(d*tan(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5471 vs. \(2 (157) = 314\).

Time = 1.00 (sec) , antiderivative size = 5471, normalized size of antiderivative = 28.06 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate((a+b*tan(f*x+e))^2*(c+d*tan(f*x+e))^(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,[0,19,7]%%%}+%%%{8,[0,17,7]%%%}+%%%{28,[0,15,7]%%%}+ 
%%%{56,[0
 

Mupad [B] (verification not implemented)

Time = 19.47 (sec) , antiderivative size = 15671, normalized size of antiderivative = 80.36 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*b^2*(c + d*tan(e + f*x))^(5/2))/(5*d*f) - atan(((((16*(2*a^2*d^5*f^2 - 
2*b^2*d^5*f^2 + 2*a^2*c^2*d^3*f^2 - 2*b^2*c^2*d^3*f^2 + 4*a*b*c*d^4*f^2 + 
4*a*b*c^3*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(((8*a^4*c 
^3*f^2 + 8*b^4*c^3*f^2 - 32*a*b^3*d^3*f^2 + 32*a^3*b*d^3*f^2 - 24*a^4*c*d^ 
2*f^2 - 24*b^4*c*d^2*f^2 - 48*a^2*b^2*c^3*f^2 + 96*a*b^3*c^2*d*f^2 - 96*a^ 
3*b*c^2*d*f^2 + 144*a^2*b^2*c*d^2*f^2)^2/64 - f^4*(a^8*c^6 + a^8*d^6 + b^8 
*c^6 + b^8*d^6 + 4*a^2*b^6*c^6 + 6*a^4*b^4*c^6 + 4*a^6*b^2*c^6 + 4*a^2*b^6 
*d^6 + 6*a^4*b^4*d^6 + 4*a^6*b^2*d^6 + 3*a^8*c^2*d^4 + 3*a^8*c^4*d^2 + 3*b 
^8*c^2*d^4 + 3*b^8*c^4*d^2 + 12*a^2*b^6*c^2*d^4 + 12*a^2*b^6*c^4*d^2 + 18* 
a^4*b^4*c^2*d^4 + 18*a^4*b^4*c^4*d^2 + 12*a^6*b^2*c^2*d^4 + 12*a^6*b^2*c^4 
*d^2))^(1/2) + a^4*c^3*f^2 + b^4*c^3*f^2 - 4*a*b^3*d^3*f^2 + 4*a^3*b*d^3*f 
^2 - 3*a^4*c*d^2*f^2 - 3*b^4*c*d^2*f^2 - 6*a^2*b^2*c^3*f^2 + 12*a*b^3*c^2* 
d*f^2 - 12*a^3*b*c^2*d*f^2 + 18*a^2*b^2*c*d^2*f^2)/(4*f^4))^(1/2))*(-(((8* 
a^4*c^3*f^2 + 8*b^4*c^3*f^2 - 32*a*b^3*d^3*f^2 + 32*a^3*b*d^3*f^2 - 24*a^4 
*c*d^2*f^2 - 24*b^4*c*d^2*f^2 - 48*a^2*b^2*c^3*f^2 + 96*a*b^3*c^2*d*f^2 - 
96*a^3*b*c^2*d*f^2 + 144*a^2*b^2*c*d^2*f^2)^2/64 - f^4*(a^8*c^6 + a^8*d^6 
+ b^8*c^6 + b^8*d^6 + 4*a^2*b^6*c^6 + 6*a^4*b^4*c^6 + 4*a^6*b^2*c^6 + 4*a^ 
2*b^6*d^6 + 6*a^4*b^4*d^6 + 4*a^6*b^2*d^6 + 3*a^8*c^2*d^4 + 3*a^8*c^4*d^2 
+ 3*b^8*c^2*d^4 + 3*b^8*c^4*d^2 + 12*a^2*b^6*c^2*d^4 + 12*a^2*b^6*c^4*d^2 
+ 18*a^4*b^4*c^2*d^4 + 18*a^4*b^4*c^4*d^2 + 12*a^6*b^2*c^2*d^4 + 12*a^6...
 

Reduce [F]

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

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

Output:

int(sqrt(tan(e + f*x)*d + c),x)*a**2*c + int(sqrt(tan(e + f*x)*d + c)*tan( 
e + f*x)**3,x)*b**2*d + 2*int(sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2,x)* 
a*b*d + int(sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2,x)*b**2*c + int(sqrt( 
tan(e + f*x)*d + c)*tan(e + f*x),x)*a**2*d + 2*int(sqrt(tan(e + f*x)*d + c 
)*tan(e + f*x),x)*a*b*c