\(\int \tan (c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [326]

Optimal result

Integrand size = 31, antiderivative size = 175 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {(a-i b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 (a A-b B) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 A (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.10 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {\frac {6 B (a+b \tan (c+d x))^{5/2}}{b}+5 (A-i B) \left (-3 (a-i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+\sqrt {a+b \tan (c+d x)} (4 a-3 i b+b \tan (c+d x))\right )+5 (A+i B) \left (-3 (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+\sqrt {a+b \tan (c+d x)} (4 a+3 i b+b \tan (c+d x))\right )}{15 d} \] Input:

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

Output:

((6*B*(a + b*Tan[c + d*x])^(5/2))/b + 5*(A - I*B)*(-3*(a - I*b)^(3/2)*ArcT 
anh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + Sqrt[a + b*Tan[c + d*x]]*(4* 
a - (3*I)*b + b*Tan[c + d*x])) + 5*(A + I*B)*(-3*(a + I*b)^(3/2)*ArcTanh[S 
qrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + Sqrt[a + b*Tan[c + d*x]]*(4*a + ( 
3*I)*b + b*Tan[c + d*x])))/(15*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.95 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.91, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4075, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4075

\(\displaystyle \int (A \tan (c+d x)-B) (a+b \tan (c+d x))^{3/2}dx+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (A \tan (c+d x)-B) (a+b \tan (c+d x))^{3/2}dx+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \int \sqrt {a+b \tan (c+d x)} (-A b-a B+(a A-b B) \tan (c+d x))dx+\frac {2 A (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {a+b \tan (c+d x)} (-A b-a B+(a A-b B) \tan (c+d x))dx+\frac {2 A (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d}\)

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

\(\displaystyle \frac {1}{2} (a+i b)^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (a-i b)^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 (a A-b B) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 A (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} (a+i b)^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (a-i b)^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 (a A-b B) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 A (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d}\)

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {i (a-i b)^2 (B+i A) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a+i b)^2 (-B+i A) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 (a A-b B) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 A (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {i (a-i b)^2 (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (a+i b)^2 (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 (a A-b B) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 A (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a+i b)^2 (-B+i A) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}-\frac {(a-i b)^2 (B+i A) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 (a A-b B) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 A (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 4075

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1667\) vs. \(2(147)=294\).

Time = 0.15 (sec) , antiderivative size = 1668, normalized size of antiderivative = 9.53

method	result	size
parts	\(\text {Expression too large to display}\)	\(1668\)
derivativedivides	\(\text {Expression too large to display}\)	\(1686\)
default	\(\text {Expression too large to display}\)	\(1686\)

Input:

int(tan(d*x+c)*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x,method=_RETURNVER 
BOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3102 vs. \(2 (141) = 282\).

Time = 0.37 (sec) , antiderivative size = 3102, normalized size of antiderivative = 17.73 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),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 = 28.91 (sec) , antiderivative size = 2868, normalized size of antiderivative = 16.39 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

log((16*B^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((16*b^2*(((-B^4*b^2*d^4*(3*a^2 - 
b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(B*b^3 + B*a^2*b 
 + a*d*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2* 
d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d + (16*B^2*b^2*(a + b*tan(c 
+ d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2)*(((-B^4*b^2*d^4*(3*a^2 - b^2)^ 
2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d^4)^(1/2))/2)*((6*B^4*a^2*b^4*d 
^4 - B^4*b^6*d^4 - 9*B^4*a^4*b^2*d^4)^(1/2)/(4*d^4) - (B^2*a^3)/(4*d^2) + 
(3*B^2*a*b^2)/(4*d^2))^(1/2) - log((16*B^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((1 
6*b^2*(-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2* 
d^2)/d^4)^(1/2)*(B*b^3 + B*a^2*b - a*d*(-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^( 
1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2 
)))/d - (16*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^ 
2)*(-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2 
)/d^4)^(1/2))/2)*(-((6*B^4*a^2*b^4*d^4 - B^4*b^6*d^4 - 9*B^4*a^4*b^2*d^4)^ 
(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/(4*d^4))^(1/2) - log((16*B^3*a*b^3* 
(a^2 + b^2)^2)/d^3 - (((16*b^2*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^ 
2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(B*b^3 + B*a^2*b - a*d*(((-B^4*b^2 
*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(a 
 + b*tan(c + d*x))^(1/2)))/d - (16*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 
 + b^4 - 6*a^2*b^2))/d^2)*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*...
 

Reduce [F]

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

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

Output:

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