3.4.45 \(\int x (c \sin ^3(a+b x^2))^{2/3} \, dx\) [345]

Optimal. Leaf size=65 \[ -\frac {\cot \left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 b}+\frac {1}{4} x^2 \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \]

[Out]

-1/4*cot(b*x^2+a)*(c*sin(b*x^2+a)^3)^(2/3)/b+1/4*x^2*csc(b*x^2+a)^2*(c*sin(b*x^2+a)^3)^(2/3)

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Rubi [A]
time = 0.07, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6847, 3286, 2715, 8} \begin {gather*} \frac {1}{4} x^2 \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}-\frac {\cot \left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(c*Sin[a + b*x^2]^3)^(2/3),x]

[Out]

-1/4*(Cot[a + b*x^2]*(c*Sin[a + b*x^2]^3)^(2/3))/b + (x^2*Csc[a + b*x^2]^2*(c*Sin[a + b*x^2]^3)^(2/3))/4

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3286

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

\begin {align*} \int x \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx &=\frac {1}{2} \text {Subst}\left (\int \left (c \sin ^3(a+b x)\right )^{2/3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \left (\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \text {Subst}\left (\int \sin ^2(a+b x) \, dx,x,x^2\right )\\ &=-\frac {\cot \left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 b}+\frac {1}{4} \left (\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \text {Subst}\left (\int 1 \, dx,x,x^2\right )\\ &=-\frac {\cot \left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 b}+\frac {1}{4} x^2 \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 55, normalized size = 0.85 \begin {gather*} \frac {\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (2 \left (a+b x^2\right )-\sin \left (2 \left (a+b x^2\right )\right )\right )}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[a + b*x^2]^2*(c*Sin[a + b*x^2]^3)^(2/3)*(2*(a + b*x^2) - Sin[2*(a + b*x^2)]))/(8*b)

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Maple [C] Result contains complex when optimal does not.
time = 0.20, size = 182, normalized size = 2.80

method	result	size
risch	\(-\frac {x^{2} \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{4 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{4 i \left (b \,x^{2}+a \right )}}{16 b \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}+\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}}}{16 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2} b}\)	\(182\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^2/(exp(2*I*(b*x^2+a))-1)^2*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(2/3)*exp(2*I*(b*x^2+a))-
1/16*I/b/(exp(2*I*(b*x^2+a))-1)^2*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(2/3)*exp(4*I*(b*x^2+a))+
1/16*I*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(2/3)/(exp(2*I*(b*x^2+a))-1)^2/b

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Maxima [A]
time = 0.57, size = 28, normalized size = 0.43 \begin {gather*} -\frac {{\left (2 \, b x^{2} - \sin \left (2 \, b x^{2} + 2 \, a\right )\right )} c^{\frac {2}{3}}}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*b*x^2 - sin(2*b*x^2 + 2*a))*c^(2/3)/b

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Fricas [A]
time = 0.35, size = 72, normalized size = 1.11 \begin {gather*} -\frac {{\left (b x^{2} - \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right )\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac {2}{3}}}{4 \, {\left (b \cos \left (b x^{2} + a\right )^{2} - b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b*x^2 - cos(b*x^2 + a)*sin(b*x^2 + a))*(-(c*cos(b*x^2 + a)^2 - c)*sin(b*x^2 + a))^(2/3)/(b*cos(b*x^2 + a
)^2 - b)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (c \sin ^{3}{\left (a + b x^{2} \right )}\right )^{\frac {2}{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(c*sin(a + b*x**2)**3)**(2/3), x)

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x\,{\left (c\,{\sin \left (b\,x^2+a\right )}^3\right )}^{2/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*sin(a + b*x^2)^3)^(2/3),x)

[Out]

int(x*(c*sin(a + b*x^2)^3)^(2/3), x)

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