\(\int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx\) [1250]

Optimal result

Integrand size = 27, antiderivative size = 178 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {(i a+b)^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}-\frac {(i a-b)^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d} f}-\frac {4 b^2 (b c-4 a d) \sqrt {c+d \tan (e+f x)}}{3 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f} \] Output:

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

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {2 \left (-\frac {3 i (a-i b)^3 d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 \sqrt {c-i d}}+\frac {3 i (a+i b)^3 d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 \sqrt {c+i d}}+\frac {2 b^2 (-b c+4 a d) \sqrt {c+d \tan (e+f x)}}{d}+b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}\right )}{3 d f} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 1.00 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.95, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 4049, 27, 3042, 4113, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

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 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 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((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 - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

rule 4113

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3866\) vs. \(2(152)=304\).

Time = 0.49 (sec) , antiderivative size = 3867, normalized size of antiderivative = 21.72

method	result	size
parts	\(\text {Expression too large to display}\)	\(3867\)
derivativedivides	\(\text {Expression too large to display}\)	\(8262\)
default	\(\text {Expression too large to display}\)	\(8262\)

Input:

int((a+b*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

a^3*(1/4/f/d/(c^2+d^2)*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/(c^2+d^2)*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)-1/4/f/d/(c^2+d^ 
2)^(3/2)*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^3-1/4/f*d/(c^2+d^2) 
^(3/2)*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-1/f/d/(c^2+d^2)^(1/2) 
/(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^2-1/f*d/(c^2+d^2)^(1 
/2)/(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))+1/f/d/(c^2+d^2)^(3/ 
2)/(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^4+3/f*d/(c^2+d^2)^ 
(3/2)/(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+2/f*d^3/(c^2+ 
d^2)^(3/2)/(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))-1/4/f/d/(c^2 
+d^2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e) 
-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-1/4/f*d/(c^2+d^2)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3928 vs. \(2 (146) = 292\).

Time = 0.36 (sec) , antiderivative size = 3928, normalized size of antiderivative = 22.07 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate((a+b*tan(f*x+e))**3/(c+d*tan(f*x+e))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((a+b*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

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

integrate((a+b*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),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,[0,14,5]%%%}+%%%{6,[0,12,5]%%%}+%%%{15,[0,10,5]%%%}+ 
%%%{20,[0
 

Mupad [B] (verification not implemented)

Time = 6.82 (sec) , antiderivative size = 3017, normalized size of antiderivative = 16.95 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

int((a + b*tan(e + f*x))^3/(c + d*tan(e + f*x))^(1/2),x)
 

Output:

atan(((((8*(4*a^3*d^3*f^2 - 12*a*b^2*d^3*f^2 + 4*b^3*c*d^2*f^2 - 12*a^2*b* 
c*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*((6*a*b^5 + 6*a^5*b 
- a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - 
 d*f^2)))^(1/2))*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20* 
a^3*b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f^2)))^(1/2) - (16*(c + d*tan(e + 
f*x))^(1/2)*(a^6*d^2 - b^6*d^2 + 15*a^2*b^4*d^2 - 15*a^4*b^2*d^2))/f^2)*(( 
6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*1 
5i)/(4*(c*f^2*1i - d*f^2)))^(1/2)*1i - (((8*(4*a^3*d^3*f^2 - 12*a*b^2*d^3* 
f^2 + 4*b^3*c*d^2*f^2 - 12*a^2*b*c*d^2*f^2))/f^3 + 64*c*d^2*(c + d*tan(e + 
 f*x))^(1/2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3* 
b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f^2)))^(1/2))*((6*a*b^5 + 6*a^5*b - a^ 
6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f 
^2)))^(1/2) + (16*(c + d*tan(e + f*x))^(1/2)*(a^6*d^2 - b^6*d^2 + 15*a^2*b 
^4*d^2 - 15*a^4*b^2*d^2))/f^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2 
*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f^2)))^(1/2)*1i)/((( 
(8*(4*a^3*d^3*f^2 - 12*a*b^2*d^3*f^2 + 4*b^3*c*d^2*f^2 - 12*a^2*b*c*d^2*f^ 
2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*((6*a*b^5 + 6*a^5*b - a^6*1i 
 + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f^2)) 
)^(1/2))*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 
+ a^4*b^2*15i)/(4*(c*f^2*1i - d*f^2)))^(1/2) - (16*(c + d*tan(e + f*x))...
 

Reduce [F]

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

int((a+b*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),x)
 

Output:

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