\(\int \frac {\sin ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) [99]

Optimal result

Integrand size = 21, antiderivative size = 112 \[ \int \frac {\sin ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=-\frac {\sin (a+b x)}{6 b d \sqrt {d \tan (a+b x)}}+\frac {\sin ^3(a+b x)}{3 b d \sqrt {d \tan (a+b x)}}+\frac {\csc (a+b x) \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{12 b d^2} \] Output:

-1/6*sin(b*x+a)/b/d/(d*tan(b*x+a))^(1/2)+1/3*sin(b*x+a)^3/b/d/(d*tan(b*x+a 
))^(1/2)+1/12*csc(b*x+a)*InverseJacobiAM(a-1/4*Pi+b*x,2^(1/2))*sin(2*b*x+2 
*a)^(1/2)*(d*tan(b*x+a))^(1/2)/b/d^2
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.40 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.91 \[ \int \frac {\sin ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=-\frac {\csc (a+b x) \left (\sqrt {\sec ^2(a+b x)} \sin (4 (a+b x))+4 \sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \sqrt {\tan (a+b x)}\right ) \sqrt {d \tan (a+b x)}}{24 b d^2 \sqrt {\sec ^2(a+b x)}} \] Input:

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

Output:

-1/24*(Csc[a + b*x]*(Sqrt[Sec[a + b*x]^2]*Sin[4*(a + b*x)] + 4*(-1)^(1/4)* 
EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Sqrt[Tan[a + b*x]] 
)*Sqrt[d*Tan[a + b*x]])/(b*d^2*Sqrt[Sec[a + b*x]^2])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.01, 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, 3053, 3042, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3076

\(\displaystyle \frac {\int \sin (a+b x) \sqrt {d \tan (a+b x)}dx}{6 d^2}+\frac {\sin ^3(a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sin (a+b x) \sqrt {d \tan (a+b x)}dx}{6 d^2}+\frac {\sin ^3(a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3078

\(\displaystyle \frac {\frac {1}{2} \int \csc (a+b x) \sqrt {d \tan (a+b x)}dx-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}}{6 d^2}+\frac {\sin ^3(a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)}dx-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}}{6 d^2}+\frac {\sin ^3(a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3081

\(\displaystyle \frac {\frac {\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{2 \sqrt {\sin (a+b x)}}-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}}{6 d^2}+\frac {\sin ^3(a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{2 \sqrt {\sin (a+b x)}}-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}}{6 d^2}+\frac {\sin ^3(a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3053

\(\displaystyle \frac {\frac {1}{2} \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}}{6 d^2}+\frac {\sin ^3(a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}}{6 d^2}+\frac {\sin ^3(a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right ) \sqrt {d \tan (a+b x)}}{2 b}-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}}{6 d^2}+\frac {\sin ^3(a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\)

Input:

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

Output:

Sin[a + b*x]^3/(3*b*d*Sqrt[d*Tan[a + b*x]]) + (-((d*Sin[a + b*x])/(b*Sqrt[ 
d*Tan[a + b*x]])) + (Csc[a + b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2* 
a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(2*b))/(6*d^2)
 

Defintions of rubi rules used

rule 3042

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

rule 3053

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ 
)]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b 
*Cos[e + f*x]])   Int[1/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 3120

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

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.21 (sec) , antiderivative size = 1510, normalized size of antiderivative = 13.48

method	result	size
default	\(\text {Expression too large to display}\)	\(1510\)

Input:

int(sin(b*x+a)^3/(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/96/b/(d*tan(b*x+a))^(1/2)/d*(2^(1/2)*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b 
*x+a)+2*sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cos 
(b*x+a)+csc(b*x+a)-sin(b*x+a)+2)/(-1+cos(b*x+a)))*(-2*sin(b*x+a)*cos(b*x+a 
)/(cos(b*x+a)+1)^2)^(1/2)*(sin(b*x+a)*(-6*cos(b*x+a)-6)+3+3*sec(b*x+a))+si 
n(b*x+a)*(12*cos(b*x+a)+12)*(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2 
)*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)+2*sin(b*x+a)*(-2*sin(b*x+a)*cos( 
b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)+csc(b*x+a)-sin(b*x+a)+2)/(-1+c 
os(b*x+a)))+2^(1/2)*ln((2*sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a) 
+1)^2)^(1/2)-cot(b*x+a)*cos(b*x+a)+sin(b*x+a)+2*cos(b*x+a)-csc(b*x+a)+2*co 
t(b*x+a)-2)/(-1+cos(b*x+a)))*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^( 
1/2)*(sin(b*x+a)*(6*cos(b*x+a)+6)-3-3*sec(b*x+a))+sin(b*x+a)*(-12*cos(b*x+ 
a)-12)*(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln((2*sin(b*x+a)*(- 
2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-cot(b*x+a)*cos(b*x+a)+sin( 
b*x+a)+2*cos(b*x+a)-csc(b*x+a)+2*cot(b*x+a)-2)/(-1+cos(b*x+a)))+2^(1/2)*ar 
ctan((sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-cos(b*x 
+a)+1)/(-1+cos(b*x+a)))*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)* 
(sin(b*x+a)*(-12*cos(b*x+a)-12)+6+6*sec(b*x+a))+sin(b*x+a)*(24*cos(b*x+a)+ 
24)*(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((sin(b*x+a)*(-2 
*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-cos(b*x+a)+1)/(-1+cos(b*x+a 
)))+2^(1/2)*arctan((sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(sin(b*x+a)**3/(d*tan(b*x+a))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate(sin(b*x + a)^3/(d*tan(b*x + a))^(3/2), x)
 

Giac [F]

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

integrate(sin(b*x+a)^3/(d*tan(b*x+a))^(3/2),x, algorithm="giac")
 

Output:

integrate(sin(b*x + a)^3/(d*tan(b*x + a))^(3/2), x)
 

Mupad [F(-1)]

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

int(sin(a + b*x)^3/(d*tan(a + b*x))^(3/2),x)
 

Output:

int(sin(a + b*x)^3/(d*tan(a + b*x))^(3/2), x)
 

Reduce [F]

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

int(sin(b*x+a)^3/(d*tan(b*x+a))^(3/2),x)
 

Output:

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