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

Optimal. Leaf size=112 \[ \frac{\sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{12 b d^2}+\frac{\sin ^3(a+b x)}{3 b d \sqrt{d \tan (a+b x)}}-\frac{\sin (a+b x)}{6 b d \sqrt{d \tan (a+b x)}} \]

[Out]

-Sin[a + b*x]/(6*b*d*Sqrt[d*Tan[a + b*x]]) + Sin[a + b*x]^3/(3*b*d*Sqrt[d*Tan[a + b*x]]) + (Csc[a + b*x]*Ellip
ticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(12*b*d^2)

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Rubi [A]  time = 0.132623, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2596, 2598, 2601, 2573, 2641} \[ \frac{\sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{12 b d^2}+\frac{\sin ^3(a+b x)}{3 b d \sqrt{d \tan (a+b x)}}-\frac{\sin (a+b x)}{6 b d \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sin[a + b*x]/(6*b*d*Sqrt[d*Tan[a + b*x]]) + Sin[a + b*x]^3/(3*b*d*Sqrt[d*Tan[a + b*x]]) + (Csc[a + b*x]*Ellip
ticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(12*b*d^2)

Rule 2596

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] - Dist[(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 2598

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] + Dist[(a^2*(m + n - 1))/m, Int[(a*Sin[e + f*x])^(m - 2)*(b*Ta
n[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 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(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 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[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 2641

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

Rubi steps

\begin{align*} \int \frac{\sin ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=\frac{\sin ^3(a+b x)}{3 b d \sqrt{d \tan (a+b x)}}+\frac{\int \sin (a+b x) \sqrt{d \tan (a+b x)} \, dx}{6 d^2}\\ &=-\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{\int \csc (a+b x) \sqrt{d \tan (a+b x)} \, dx}{12 d^2}\\ &=-\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{\left (\sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)}} \, dx}{12 d^2 \sqrt{\sin (a+b x)}}\\ &=-\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{\left (\csc (a+b x) \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx}{12 d^2}\\ &=-\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) F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}{12 b d^2}\\ \end{align*}

Mathematica [C]  time = 0.359041, size = 102, normalized size = 0.91 \[ -\frac{\csc (a+b x) \sqrt{d \tan (a+b x)} \left (\sin (4 (a+b x)) \sqrt{\sec ^2(a+b x)}+4 \sqrt [4]{-1} \sqrt{\tan (a+b x)} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (a+b x)}\right )\right |-1\right )\right )}{24 b d^2 \sqrt{\sec ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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]])/(24*b*d^2*Sqrt[Sec[a + b*x]^2])

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Maple [A]  time = 0.15, size = 222, normalized size = 2. \begin{align*} -{\frac{\sqrt{2} \left ( \cos \left ( bx+a \right ) -1 \right ) \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{12\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2} \left ( \cos \left ( bx+a \right ) \right ) ^{2}} \left ( \sin \left ( bx+a \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{\sqrt{2}}{2}} \right ) +2\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}\sqrt{2}-2\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sqrt{2}- \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}+\cos \left ( bx+a \right ) \sqrt{2} \right ) \left ({\frac{\sin \left ( bx+a \right ) d}{\cos \left ( bx+a \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/b*2^(1/2)*(cos(b*x+a)-1)*(sin(b*x+a)*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*
x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2
),1/2*2^(1/2))+2*cos(b*x+a)^4*2^(1/2)-2*cos(b*x+a)^3*2^(1/2)-cos(b*x+a)^2*2^(1/2)+cos(b*x+a)*2^(1/2))*(cos(b*x
+a)+1)^2/sin(b*x+a)^2/(d*sin(b*x+a)/cos(b*x+a))^(3/2)/cos(b*x+a)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )^{3}}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt{d \tan \left (b x + a\right )} \sin \left (b x + a\right )}{d^{2} \tan \left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )^{3}}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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