\(\int \frac {\sin ^5(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\) [89]

Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int \frac {\sin ^5(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=-\frac {7 d \sin ^3(a+b x)}{30 b (d \tan (a+b x))^{3/2}}-\frac {d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}+\frac {7 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{20 b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}} \] Output:

-7/30*d*sin(b*x+a)^3/b/(d*tan(b*x+a))^(3/2)-1/5*d*sin(b*x+a)^5/b/(d*tan(b* 
x+a))^(3/2)-7/20*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))*sin(b*x+a)/b/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 = 0.77 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \frac {\sin ^5(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\frac {\sin (a+b x) \left (-20 \sin (2 (a+b x))+3 \sin (4 (a+b x))+28 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)} \tan (a+b x)\right )}{120 b \sqrt {d \tan (a+b x)}} \] Input:

Integrate[Sin[a + b*x]^5/Sqrt[d*Tan[a + b*x]],x]
 

Output:

(Sin[a + b*x]*(-20*Sin[2*(a + b*x)] + 3*Sin[4*(a + b*x)] + 28*Hypergeometr 
ic2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2]*Tan[a + b*x]))/ 
(120*b*Sqrt[d*Tan[a + b*x]])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3078, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3078

\(\displaystyle \frac {7}{10} \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}}dx-\frac {d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} \int \frac {\sin (a+b x)^3}{\sqrt {d \tan (a+b x)}}dx-\frac {d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3078

\(\displaystyle \frac {7}{10} \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )-\frac {d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )-\frac {d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3081

\(\displaystyle \frac {7}{10} \left (\frac {\sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )-\frac {d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} \left (\frac {\sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )-\frac {d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3052

\(\displaystyle \frac {7}{10} \left (\frac {\sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )-\frac {d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} \left (\frac {\sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )-\frac {d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {7}{10} \left (\frac {\sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )-\frac {d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}\)

Input:

Int[Sin[a + b*x]^5/Sqrt[d*Tan[a + b*x]],x]
 

Output:

-1/5*(d*Sin[a + b*x]^5)/(b*(d*Tan[a + b*x])^(3/2)) + (7*(-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
 

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 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. \(226\) vs. \(2(94)=188\).

Time = 1.26 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.12

method	result	size
default	\(\frac {-\frac {\sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \left (-21-21 \sec \left (b x +a \right )\right )}{120}-\frac {\sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (42+42 \sec \left (b x +a \right )\right )}{120}-\frac {\cos \left (b x +a \right )^{5}}{5}+\frac {19 \cos \left (b x +a \right )^{3}}{30}-\frac {47 \cos \left (b x +a \right )}{60}+\frac {7}{20}}{b \sqrt {d \tan \left (b x +a \right )}}\)	\(227\)

Input:

int(sin(b*x+a)^5/(d*tan(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/b*(-1/120*(-2*csc(b*x+a)+2*cot(b*x+a)+2)^(1/2)*(-csc(b*x+a)+cot(b*x+a))^ 
(1/2)*EllipticF((csc(b*x+a)-cot(b*x+a)+1)^(1/2),1/2*2^(1/2))*(csc(b*x+a)-c 
ot(b*x+a)+1)^(1/2)*(-21-21*sec(b*x+a))-1/120*(-2*csc(b*x+a)+2*cot(b*x+a)+2 
)^(1/2)*(-csc(b*x+a)+cot(b*x+a))^(1/2)*(csc(b*x+a)-cot(b*x+a)+1)^(1/2)*Ell 
ipticE((csc(b*x+a)-cot(b*x+a)+1)^(1/2),1/2*2^(1/2))*(42+42*sec(b*x+a))-1/5 
*cos(b*x+a)^5+19/30*cos(b*x+a)^3-47/60*cos(b*x+a)+7/20)/(d*tan(b*x+a))^(1/ 
2)
 

Fricas [F]

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

integrate(sin(b*x+a)^5/(d*tan(b*x+a))^(1/2),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*tan(b*x + a)), x)
 

Sympy [F(-1)]

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

integrate(sin(b*x+a)**5/(d*tan(b*x+a))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(sin(b*x+a)^5/(d*tan(b*x+a))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(sin(b*x+a)^5/(d*tan(b*x+a))^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(sin(a + b*x)^5/(d*tan(a + b*x))^(1/2),x)
 

Output:

int(sin(a + b*x)^5/(d*tan(a + b*x))^(1/2), x)
 

Reduce [F]

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

int(sin(b*x+a)^5/(d*tan(b*x+a))^(1/2),x)
 

Output:

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