\(\int (b \tan ^4(c+d x))^{5/2} \, dx\) [36]

Optimal result

Integrand size = 14, antiderivative size = 182 \[ \int \left (b \tan ^4(c+d x)\right )^{5/2} \, dx=\frac {b^2 \cot (c+d x) \sqrt {b \tan ^4(c+d x)}}{d}-b^2 x \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)}-\frac {b^2 \tan (c+d x) \sqrt {b \tan ^4(c+d x)}}{3 d}+\frac {b^2 \tan ^3(c+d x) \sqrt {b \tan ^4(c+d x)}}{5 d}-\frac {b^2 \tan ^5(c+d x) \sqrt {b \tan ^4(c+d x)}}{7 d}+\frac {b^2 \tan ^7(c+d x) \sqrt {b \tan ^4(c+d x)}}{9 d} \] Output:

b^2*cot(d*x+c)*(tan(d*x+c)^4*b)^(1/2)/d-b^2*x*cot(d*x+c)^2*(tan(d*x+c)^4*b 
)^(1/2)-1/3*b^2*tan(d*x+c)*(tan(d*x+c)^4*b)^(1/2)/d+1/5*b^2*tan(d*x+c)^3*( 
tan(d*x+c)^4*b)^(1/2)/d-1/7*b^2*tan(d*x+c)^5*(tan(d*x+c)^4*b)^(1/2)/d+1/9* 
b^2*tan(d*x+c)^7*(tan(d*x+c)^4*b)^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.47 \[ \int \left (b \tan ^4(c+d x)\right )^{5/2} \, dx=\frac {\cot (c+d x) \left (35-45 \cot ^2(c+d x)+63 \cot ^4(c+d x)-105 \cot ^6(c+d x)+315 \cot ^8(c+d x)-315 \arctan (\tan (c+d x)) \cot ^9(c+d x)\right ) \left (b \tan ^4(c+d x)\right )^{5/2}}{315 d} \] Input:

Integrate[(b*Tan[c + d*x]^4)^(5/2),x]
 

Output:

(Cot[c + d*x]*(35 - 45*Cot[c + d*x]^2 + 63*Cot[c + d*x]^4 - 105*Cot[c + d* 
x]^6 + 315*Cot[c + d*x]^8 - 315*ArcTan[Tan[c + d*x]]*Cot[c + d*x]^9)*(b*Ta 
n[c + d*x]^4)^(5/2))/(315*d)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.55, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {3042, 4141, 3042, 3954, 3042, 3954, 3042, 3954, 3042, 3954, 3042, 3954, 24}

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.

\(\displaystyle \int \left (b \tan ^4(c+d x)\right )^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (b \tan (c+d x)^4\right )^{5/2}dx\)

\(\Big \downarrow \) 4141

\(\displaystyle b^2 \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \int \tan ^{10}(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \int \tan (c+d x)^{10}dx\)

\(\Big \downarrow \) 3954

\(\displaystyle b^2 \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (\frac {\tan ^9(c+d x)}{9 d}-\int \tan ^8(c+d x)dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (\frac {\tan ^9(c+d x)}{9 d}-\int \tan (c+d x)^8dx\right )\)

\(\Big \downarrow \) 3954

\(\displaystyle b^2 \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (\int \tan ^6(c+d x)dx+\frac {\tan ^9(c+d x)}{9 d}-\frac {\tan ^7(c+d x)}{7 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (\int \tan (c+d x)^6dx+\frac {\tan ^9(c+d x)}{9 d}-\frac {\tan ^7(c+d x)}{7 d}\right )\)

\(\Big \downarrow \) 3954

\(\displaystyle b^2 \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (-\int \tan ^4(c+d x)dx+\frac {\tan ^9(c+d x)}{9 d}-\frac {\tan ^7(c+d x)}{7 d}+\frac {\tan ^5(c+d x)}{5 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (-\int \tan (c+d x)^4dx+\frac {\tan ^9(c+d x)}{9 d}-\frac {\tan ^7(c+d x)}{7 d}+\frac {\tan ^5(c+d x)}{5 d}\right )\)

\(\Big \downarrow \) 3954

\(\displaystyle b^2 \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (\int \tan ^2(c+d x)dx+\frac {\tan ^9(c+d x)}{9 d}-\frac {\tan ^7(c+d x)}{7 d}+\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (\int \tan (c+d x)^2dx+\frac {\tan ^9(c+d x)}{9 d}-\frac {\tan ^7(c+d x)}{7 d}+\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}\right )\)

\(\Big \downarrow \) 3954

\(\displaystyle b^2 \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (-\int 1dx+\frac {\tan ^9(c+d x)}{9 d}-\frac {\tan ^7(c+d x)}{7 d}+\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}+\frac {\tan (c+d x)}{d}\right )\)

\(\Big \downarrow \) 24

\(\displaystyle b^2 \left (\frac {\tan ^9(c+d x)}{9 d}-\frac {\tan ^7(c+d x)}{7 d}+\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}+\frac {\tan (c+d x)}{d}-x\right ) \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)}\)

Input:

Int[(b*Tan[c + d*x]^4)^(5/2),x]
 

Output:

b^2*Cot[c + d*x]^2*Sqrt[b*Tan[c + d*x]^4]*(-x + Tan[c + d*x]/d - Tan[c + d 
*x]^3/(3*d) + Tan[c + d*x]^5/(5*d) - Tan[c + d*x]^7/(7*d) + Tan[c + d*x]^9 
/(9*d))
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3954

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d 
*x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2   Int[(b*Tan[c + d*x])^(n - 2), x] 
, x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
 

rule 4141

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff 
= FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ 
n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p]))   Int[ActivateTrig[u]*(Ta 
n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] 
 && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / 
; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.46

method	result	size
derivativedivides	\(-\frac {\left (b \tan \left (d x +c \right )^{4}\right )^{\frac {5}{2}} \left (-35 \tan \left (d x +c \right )^{9}+45 \tan \left (d x +c \right )^{7}-63 \tan \left (d x +c \right )^{5}+105 \tan \left (d x +c \right )^{3}+315 \arctan \left (\tan \left (d x +c \right )\right )-315 \tan \left (d x +c \right )\right )}{315 d \tan \left (d x +c \right )^{10}}\)	\(84\)
default	\(-\frac {\left (b \tan \left (d x +c \right )^{4}\right )^{\frac {5}{2}} \left (-35 \tan \left (d x +c \right )^{9}+45 \tan \left (d x +c \right )^{7}-63 \tan \left (d x +c \right )^{5}+105 \tan \left (d x +c \right )^{3}+315 \arctan \left (\tan \left (d x +c \right )\right )-315 \tan \left (d x +c \right )\right )}{315 d \tan \left (d x +c \right )^{10}}\)	\(84\)
risch	\(\frac {b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}\, x}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 i b^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}\, \left (1575 \,{\mathrm e}^{16 i \left (d x +c \right )}+6300 \,{\mathrm e}^{14 i \left (d x +c \right )}+21000 \,{\mathrm e}^{12 i \left (d x +c \right )}+31500 \,{\mathrm e}^{10 i \left (d x +c \right )}+39438 \,{\mathrm e}^{8 i \left (d x +c \right )}+26292 \,{\mathrm e}^{6 i \left (d x +c \right )}+13968 \,{\mathrm e}^{4 i \left (d x +c \right )}+3492 \,{\mathrm e}^{2 i \left (d x +c \right )}+563\right )}{315 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7} d}\)	\(218\)

Input:

int((b*tan(d*x+c)^4)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/315/d*(b*tan(d*x+c)^4)^(5/2)*(-35*tan(d*x+c)^9+45*tan(d*x+c)^7-63*tan(d 
*x+c)^5+105*tan(d*x+c)^3+315*arctan(tan(d*x+c))-315*tan(d*x+c))/tan(d*x+c) 
^10
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.53 \[ \int \left (b \tan ^4(c+d x)\right )^{5/2} \, dx=\frac {{\left (35 \, b^{2} \tan \left (d x + c\right )^{9} - 45 \, b^{2} \tan \left (d x + c\right )^{7} + 63 \, b^{2} \tan \left (d x + c\right )^{5} - 105 \, b^{2} \tan \left (d x + c\right )^{3} - 315 \, b^{2} d x + 315 \, b^{2} \tan \left (d x + c\right )\right )} \sqrt {b \tan \left (d x + c\right )^{4}}}{315 \, d \tan \left (d x + c\right )^{2}} \] Input:

integrate((b*tan(d*x+c)^4)^(5/2),x, algorithm="fricas")
 

Output:

1/315*(35*b^2*tan(d*x + c)^9 - 45*b^2*tan(d*x + c)^7 + 63*b^2*tan(d*x + c) 
^5 - 105*b^2*tan(d*x + c)^3 - 315*b^2*d*x + 315*b^2*tan(d*x + c))*sqrt(b*t 
an(d*x + c)^4)/(d*tan(d*x + c)^2)
 

Sympy [F]

\[ \int \left (b \tan ^4(c+d x)\right )^{5/2} \, dx=\int \left (b \tan ^{4}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \] Input:

integrate((b*tan(d*x+c)**4)**(5/2),x)
 

Output:

Integral((b*tan(c + d*x)**4)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.43 \[ \int \left (b \tan ^4(c+d x)\right )^{5/2} \, dx=\frac {35 \, b^{\frac {5}{2}} \tan \left (d x + c\right )^{9} - 45 \, b^{\frac {5}{2}} \tan \left (d x + c\right )^{7} + 63 \, b^{\frac {5}{2}} \tan \left (d x + c\right )^{5} - 105 \, b^{\frac {5}{2}} \tan \left (d x + c\right )^{3} - 315 \, {\left (d x + c\right )} b^{\frac {5}{2}} + 315 \, b^{\frac {5}{2}} \tan \left (d x + c\right )}{315 \, d} \] Input:

integrate((b*tan(d*x+c)^4)^(5/2),x, algorithm="maxima")
 

Output:

1/315*(35*b^(5/2)*tan(d*x + c)^9 - 45*b^(5/2)*tan(d*x + c)^7 + 63*b^(5/2)* 
tan(d*x + c)^5 - 105*b^(5/2)*tan(d*x + c)^3 - 315*(d*x + c)*b^(5/2) + 315* 
b^(5/2)*tan(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.47 \[ \int \left (b \tan ^4(c+d x)\right )^{5/2} \, dx=-\frac {1}{315} \, b^{\frac {5}{2}} {\left (\frac {315 \, {\left (d x + c\right )}}{d} - \frac {35 \, d^{8} \tan \left (d x + c\right )^{9} - 45 \, d^{8} \tan \left (d x + c\right )^{7} + 63 \, d^{8} \tan \left (d x + c\right )^{5} - 105 \, d^{8} \tan \left (d x + c\right )^{3} + 315 \, d^{8} \tan \left (d x + c\right )}{d^{9}}\right )} \] Input:

integrate((b*tan(d*x+c)^4)^(5/2),x, algorithm="giac")
 

Output:

-1/315*b^(5/2)*(315*(d*x + c)/d - (35*d^8*tan(d*x + c)^9 - 45*d^8*tan(d*x 
+ c)^7 + 63*d^8*tan(d*x + c)^5 - 105*d^8*tan(d*x + c)^3 + 315*d^8*tan(d*x 
+ c))/d^9)
 

Mupad [F(-1)]

Timed out. \[ \int \left (b \tan ^4(c+d x)\right )^{5/2} \, dx=\int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^4\right )}^{5/2} \,d x \] Input:

int((b*tan(c + d*x)^4)^(5/2),x)
 

Output:

int((b*tan(c + d*x)^4)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.35 \[ \int \left (b \tan ^4(c+d x)\right )^{5/2} \, dx=\frac {\sqrt {b}\, b^{2} \left (35 \tan \left (d x +c \right )^{9}-45 \tan \left (d x +c \right )^{7}+63 \tan \left (d x +c \right )^{5}-105 \tan \left (d x +c \right )^{3}+315 \tan \left (d x +c \right )-315 d x \right )}{315 d} \] Input:

int((b*tan(d*x+c)^4)^(5/2),x)
 

Output:

(sqrt(b)*b**2*(35*tan(c + d*x)**9 - 45*tan(c + d*x)**7 + 63*tan(c + d*x)** 
5 - 105*tan(c + d*x)**3 + 315*tan(c + d*x) - 315*d*x))/(315*d)