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

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

[Out]

-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)*(sin(a+1/4*
Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(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

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Rubi [A]
time = 0.09, antiderivative size = 112, normalized size of antiderivative = 1.00, 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 = {2676, 2678, 2681, 2653, 2720} \begin {gather*} \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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*Sin[a + b*x]/(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]*Ell
ipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(12*b*d^2)

Rule 2653

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 2676

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 2678

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*
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 2681

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 2720

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]

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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.43, size = 102, normalized size = 0.91 \begin {gather*} -\frac {\csc (a+b x) \left (\sqrt {\sec ^2(a+b x)} \sin (4 (a+b x))+4 \sqrt [4]{-1} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right )\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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-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])

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Maple [A]
time = 0.33, size = 222, normalized size = 1.98

method	result	size
default	\(-\frac {\left (-1+\cos \left (b x +a \right )\right ) \left (\sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sin \left (b x +a \right ) \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+2 \left (\cos ^{4}\left (b x +a \right )\right ) \sqrt {2}-2 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}-\left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}+\cos \left (b x +a \right ) \sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \sqrt {2}}{12 b \sin \left (b x +a \right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \cos \left (b x +a \right )^{2}}\)	\(222\)

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,method=_RETURNVERBOSE)

[Out]

-1/12/b*(-1+cos(b*x+a))*(((-1+cos(b*x+a))/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)*sin(b*x+a)*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*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (a+b\,x\right )}^3}{{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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