3.4.46 \(\int (c \sin ^3(a+b x^2))^{2/3} \, dx\) [346]

Optimal. Leaf size=148 \[ \frac {1}{2} x \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}-\frac {\sqrt {\pi } \cos (2 a) \csc ^2\left (a+b x^2\right ) C\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 \sqrt {b}}+\frac {\sqrt {\pi } \csc ^2\left (a+b x^2\right ) S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \sin (2 a) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 \sqrt {b}} \]

[Out]

1/2*x*csc(b*x^2+a)^2*(c*sin(b*x^2+a)^3)^(2/3)-1/4*cos(2*a)*csc(b*x^2+a)^2*FresnelC(2*x*b^(1/2)/Pi^(1/2))*(c*si
n(b*x^2+a)^3)^(2/3)*Pi^(1/2)/b^(1/2)+1/4*csc(b*x^2+a)^2*FresnelS(2*x*b^(1/2)/Pi^(1/2))*sin(2*a)*(c*sin(b*x^2+a
)^3)^(2/3)*Pi^(1/2)/b^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6852, 3438, 3435, 3433, 3432} \begin {gather*} -\frac {\sqrt {\pi } \cos (2 a) \text {FresnelC}\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 \sqrt {b}}+\frac {\sqrt {\pi } \sin (2 a) S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 \sqrt {b}}+\frac {1}{2} x \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x^2]^3)^(2/3),x]

[Out]

(x*Csc[a + b*x^2]^2*(c*Sin[a + b*x^2]^3)^(2/3))/2 - (Sqrt[Pi]*Cos[2*a]*Csc[a + b*x^2]^2*FresnelC[(2*Sqrt[b]*x)
/Sqrt[Pi]]*(c*Sin[a + b*x^2]^3)^(2/3))/(4*Sqrt[b]) + (Sqrt[Pi]*Csc[a + b*x^2]^2*FresnelS[(2*Sqrt[b]*x)/Sqrt[Pi
]]*Sin[2*a]*(c*Sin[a + b*x^2]^3)^(2/3))/(4*Sqrt[b])

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3435

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3438

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Sin[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 1]

Rule 6852

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m)^FracPart[p]/v^(m*FracPart[p])), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx &=\left (\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \int \sin ^2\left (a+b x^2\right ) \, dx\\ &=\left (\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \int \left (\frac {1}{2}-\frac {1}{2} \cos \left (2 a+2 b x^2\right )\right ) \, dx\\ &=\frac {1}{2} x \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}-\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 ) \int \cos \left (2 a+2 b x^2\right ) \, dx\\ &=\frac {1}{2} x \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}-\frac {1}{2} \left (\cos (2 a) \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \int \cos \left (2 b x^2\right ) \, dx+\frac {1}{2} \left (\csc ^2\left (a+b x^2\right ) \sin (2 a) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \int \sin \left (2 b x^2\right ) \, dx\\ &=\frac {1}{2} x \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}-\frac {\sqrt {\pi } \cos (2 a) \csc ^2\left (a+b x^2\right ) C\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 \sqrt {b}}+\frac {\sqrt {\pi } \csc ^2\left (a+b x^2\right ) S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \sin (2 a) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 \sqrt {b}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 93, normalized size = 0.63 \begin {gather*} \frac {\csc ^2\left (a+b x^2\right ) \left (2 \sqrt {b} x-\sqrt {\pi } \cos (2 a) C\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )+\sqrt {\pi } S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \sin (2 a)\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[a + b*x^2]^2*(2*Sqrt[b]*x - Sqrt[Pi]*Cos[2*a]*FresnelC[(2*Sqrt[b]*x)/Sqrt[Pi]] + Sqrt[Pi]*FresnelS[(2*Sqr
t[b]*x)/Sqrt[Pi]]*Sin[2*a])*(c*Sin[a + b*x^2]^3)^(2/3))/(4*Sqrt[b])

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Maple [C] Result contains complex when optimal does not.
time = 0.17, size = 224, normalized size = 1.51

method	result	size
risch	\(\frac {\left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i x^{2} b} \sqrt {\pi }\, \sqrt {2}\, \erf \left (\sqrt {2}\, \sqrt {i b}\, x \right )}{16 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2} \sqrt {i b}}+\frac {\erf \left (\sqrt {-2 i b}\, x \right ) \sqrt {\pi }\, \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b \,x^{2}+2 a \right )}}{8 \sqrt {-2 i b}\, \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}-\frac {x \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{2 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}\)	\(224\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(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*exp(2*I*b*x^2)*Pi^(1/2)
*2^(1/2)/(I*b)^(1/2)*erf(2^(1/2)*(I*b)^(1/2)*x)+1/8*erf((-2*I*b)^(1/2)*x)/(-2*I*b)^(1/2)*Pi^(1/2)/(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+2*a))-1/2*x/(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))

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Maxima [C] Result contains complex when optimal does not.
time = 0.62, size = 76, normalized size = 0.51 \begin {gather*} -\frac {4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i - 1\right ) \, \cos \left (2 \, a\right ) + \left (i + 1\right ) \, \sin \left (2 \, a\right )\right )} \operatorname {erf}\left (\sqrt {2 i \, b} x\right ) + {\left (-\left (i + 1\right ) \, \cos \left (2 \, a\right ) - \left (i - 1\right ) \, \sin \left (2 \, a\right )\right )} \operatorname {erf}\left (\sqrt {-2 i \, b} x\right )\right )} b^{\frac {3}{2}} c^{\frac {2}{3}} + 16 \, b^{2} c^{\frac {2}{3}} x}{64 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(4^(1/4)*sqrt(2)*sqrt(pi)*(((I - 1)*cos(2*a) + (I + 1)*sin(2*a))*erf(sqrt(2*I*b)*x) + (-(I + 1)*cos(2*a)
- (I - 1)*sin(2*a))*erf(sqrt(-2*I*b)*x))*b^(3/2)*c^(2/3) + 16*b^2*c^(2/3)*x)/b^2

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Fricas [A]
time = 0.38, size = 114, normalized size = 0.77 \begin {gather*} \frac {4^{\frac {2}{3}} {\left (4^{\frac {1}{3}} \pi \sqrt {\frac {b}{\pi }} \cos \left (2 \, a\right ) \operatorname {C}\left (2 \, x \sqrt {\frac {b}{\pi }}\right ) - 4^{\frac {1}{3}} \pi \sqrt {\frac {b}{\pi }} \operatorname {S}\left (2 \, x \sqrt {\frac {b}{\pi }}\right ) \sin \left (2 \, a\right ) - 2 \cdot 4^{\frac {1}{3}} b x\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}}}{16 \, {\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((c*sin(b*x^2+a)^3)^(2/3),x, algorithm="fricas")

[Out]

1/16*4^(2/3)*(4^(1/3)*pi*sqrt(b/pi)*cos(2*a)*fresnel_cos(2*x*sqrt(b/pi)) - 4^(1/3)*pi*sqrt(b/pi)*fresnel_sin(2
*x*sqrt(b/pi))*sin(2*a) - 2*4^(1/3)*b*x)*(-(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 \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((c*sin(b*x**2+a)**3)**(2/3),x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\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((c*sin(a + b*x^2)^3)^(2/3),x)

[Out]

int((c*sin(a + b*x^2)^3)^(2/3), x)

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