\(\int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx\) [521]

Optimal result

Integrand size = 21, antiderivative size = 158 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=-\frac {(a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.87 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {-15 (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )-15 (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+2 \sqrt {a+b \tan (c+d x)} \left (23 a^2-15 b^2+11 a b \tan (c+d x)+3 b^2 \tan ^2(c+d x)\right )}{15 d} \] Input:

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

Output:

(-15*(a - I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] - 15* 
(a + I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + 2*Sqrt[a 
 + b*Tan[c + d*x]]*(23*a^2 - 15*b^2 + 11*a*b*Tan[c + d*x] + 3*b^2*Tan[c + 
d*x]^2))/(15*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.90 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1360\) vs. \(2(132)=264\).

Time = 0.21 (sec) , antiderivative size = 1361, normalized size of antiderivative = 8.61

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1361\)
default	\(\text {Expression too large to display}\)	\(1361\)

Input:

int(tan(d*x+c)*(a+b*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1122 vs. \(2 (128) = 256\).

Time = 0.10 (sec) , antiderivative size = 1122, normalized size of antiderivative = 7.10 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/30*(15*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100* 
a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4* 
b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a) + (2*a*d^3*sqrt(-(25*a^8*b 
^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) - (5*a^6 - 15*a^4 
*b^2 + 11*a^2*b^4 - b^6)*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-( 
25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)) - 
15*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^ 
4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 
8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a) - (2*a*d^3*sqrt(-(25*a^8*b^2 - 1 
00*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) - (5*a^6 - 15*a^4*b^2 + 
 11*a^2*b^4 - b^6)*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8 
*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)) - 15*d*s 
qrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 11 
0*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2* 
b^6 + b^8)*sqrt(b*tan(d*x + c) + a) + (2*a*d^3*sqrt(-(25*a^8*b^2 - 100*a^6 
*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) + (5*a^6 - 15*a^4*b^2 + 11*a^ 
2*b^4 - b^6)*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 
 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)) + 15*d*sqrt((a 
^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4* 
b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2*b^6...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more 
details)Is
 

Giac [F(-2)]

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

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^(5/2),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 = 9.57 (sec) , antiderivative size = 2191, normalized size of antiderivative = 13.87 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x) 
)^(1/2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4 
*d^2))^(1/2))*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b 
^2)/(4*d^2))^(1/2) - (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15 
*a^4*b^4 - a^6*b^2))/d^2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10 
i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i - (((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 
+ 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i 
+ a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 - a^4*b*5i + a^5 - b 
^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x) 
)^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*((5*a*b^4 - a^4*b* 
5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i)/((((8*(4* 
b^6*d^2 - 4*a^4*b^2*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a* 
b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))* 
((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^( 
1/2) - (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6 
*b^2))/d^2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2 
)/(4*d^2))^(1/2) + (((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 + 64*a*b^2*(a + b 
*tan(c + d*x))^(1/2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 1 
0*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*1 
0i - 10*a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 -...
 

Reduce [F]

\[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{3}d x \right ) b^{2}+2 \left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}d x \right ) a b +\left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )d x \right ) a^{2} \] Input:

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