\(\int \frac {\sin ^5(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\) [109]

Optimal result

Integrand size = 21, antiderivative size = 114 \[ \int \frac {\sin ^5(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=-\frac {\sin ^3(a+b x)}{10 b d (d \tan (a+b x))^{3/2}}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}+\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{20 b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}} \] Output:

-1/10*sin(b*x+a)^3/b/d/(d*tan(b*x+a))^(3/2)+1/5*sin(b*x+a)^5/b/d/(d*tan(b* 
x+a))^(3/2)-3/20*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))*sin(b*x+a)/b/d^2/sin 
(2*b*x+2*a)^(1/2)/(d*tan(b*x+a))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.67 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {\sin ^5(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\frac {\sqrt {d \tan (a+b x)} \left (-\sqrt {\sec ^2(a+b x)} (\sin (3 (a+b x))+\sin (5 (a+b x)))+8 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(a+b x)\right ) \sec (a+b x) \tan (a+b x)\right )}{80 b d^3 \sqrt {\sec ^2(a+b x)}} \] Input:

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

Output:

(Sqrt[d*Tan[a + b*x]]*(-(Sqrt[Sec[a + b*x]^2]*(Sin[3*(a + b*x)] + Sin[5*(a 
 + b*x)])) + 8*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sec[a + b 
*x]*Tan[a + b*x]))/(80*b*d^3*Sqrt[Sec[a + b*x]^2])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3076, 3042, 3078, 3042, 3081, 3042, 3052, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3076

\(\displaystyle \frac {3 \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}}dx}{10 d^2}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int \frac {\sin (a+b x)^3}{\sqrt {d \tan (a+b x)}}dx}{10 d^2}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3078

\(\displaystyle \frac {3 \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )}{10 d^2}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )}{10 d^2}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3081

\(\displaystyle \frac {3 \left (\frac {\sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )}{10 d^2}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )}{10 d^2}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3052

\(\displaystyle \frac {3 \left (\frac {\sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )}{10 d^2}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )}{10 d^2}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {3 \left (\frac {\sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )}{10 d^2}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}\)

Input:

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

Output:

Sin[a + b*x]^5/(5*b*d*(d*Tan[a + b*x])^(3/2)) + (3*(-1/3*(d*Sin[a + b*x]^3 
)/(b*(d*Tan[a + b*x])^(3/2)) + (EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x]) 
/(2*b*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])))/(10*d^2)
 

Defintions of rubi rules used

rule 3042

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

rule 3052

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] 
, x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e 
 + 2*f*x]])   Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, x]
 

rule 3076

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n + 1)/(b*f*m)) 
, x] - Simp[a^2*((n + 1)/(b^2*m))   Int[(a*Sin[e + f*x])^(m - 2)*(b*Tan[e + 
 f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && GtQ[m, 1] 
 && IntegersQ[2*m, 2*n]
 

rule 3078

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_.), x_Symbol] :> Simp[(-b)*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/( 
f*m)), x] + Simp[a^2*((m + n - 1)/m)   Int[(a*Sin[e + f*x])^(m - 2)*(b*Tan[ 
e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1 
] && EqQ[n, 1/2])) && IntegersQ[2*m, 2*n]
 

rule 3081

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ 
n)   Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, 
 f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 
1)]) || IntegersQ[m - 1/2, n - 1/2])
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(229\) vs. \(2(101)=202\).

Time = 1.41 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.02

method	result	size
default	\(\frac {\frac {\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-6-6 \sec \left (b x +a \right )\right )}{40}+\frac {\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (3+3 \sec \left (b x +a \right )\right )}{40}+\frac {\cos \left (b x +a \right )^{5}}{5}-\frac {3 \cos \left (b x +a \right )^{3}}{10}-\frac {\cos \left (b x +a \right )}{20}+\frac {3}{20}}{b \sqrt {d \tan \left (b x +a \right )}\, d^{2}}\)	\(230\)

Input:

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

Output:

1/b*(1/40*(csc(b*x+a)-cot(b*x+a)+1)^(1/2)*(-2*csc(b*x+a)+2*cot(b*x+a)+2)^( 
1/2)*(-csc(b*x+a)+cot(b*x+a))^(1/2)*EllipticE((csc(b*x+a)-cot(b*x+a)+1)^(1 
/2),1/2*2^(1/2))*(-6-6*sec(b*x+a))+1/40*(csc(b*x+a)-cot(b*x+a)+1)^(1/2)*(- 
2*csc(b*x+a)+2*cot(b*x+a)+2)^(1/2)*(-csc(b*x+a)+cot(b*x+a))^(1/2)*Elliptic 
F((csc(b*x+a)-cot(b*x+a)+1)^(1/2),1/2*2^(1/2))*(3+3*sec(b*x+a))+1/5*cos(b* 
x+a)^5-3/10*cos(b*x+a)^3-1/20*cos(b*x+a)+3/20)/(d*tan(b*x+a))^(1/2)/d^2
 

Fricas [F]

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

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

Output:

integral((cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*sqrt(d*tan(b*x + a))*sin( 
b*x + a)/(d^3*tan(b*x + a)^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^5(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(d)*int((sqrt(tan(a + b*x))*sin(a + b*x)**5)/tan(a + b*x)**3,x))/d**3