3.4.48 \(\int \frac {(c \sin ^3(a+b x^2))^{2/3}}{x^2} \, dx\) [348]

Optimal. Leaf size=132 \[ -\frac {\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{x}+\sqrt {b} \sqrt {\pi } \cos (2 a) \csc ^2\left (a+b x^2\right ) S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}+\sqrt {b} \sqrt {\pi } \csc ^2\left (a+b x^2\right ) C\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \sin (2 a) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6852, 3474, 4669, 3454, 3434, 3433, 3432} \begin {gather*} \sqrt {\pi } \sqrt {b} \sin (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}+\sqrt {\pi } \sqrt {b} \cos (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}-\frac {\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x^2]^3)^(2/3)/x^2,x]

[Out]

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

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 3434

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

Rule 3454

Int[((a_.) + (b_.)*Sin[u_])^(p_.), x_Symbol] :> Int[(a + b*Sin[ExpandToSum[u, x]])^p, x] /; FreeQ[{a, b, p}, x
] && BinomialQ[u, x] &&  !BinomialMatchQ[u, x]

Rule 3474

Int[(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_)]^(p_), x_Symbol] :> Simp[x^(m + 1)*(Sin[a + b*x^n]^p/(m + 1)), x] -
Dist[b*n*(p/(m + 1)), Int[Sin[a + b*x^n]^(p - 1)*Cos[a + b*x^n], x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 1] &&
EqQ[m + n, 0] && NeQ[n, 1] && IntegerQ[n]

Rule 4669

Int[Cos[w_]^(p_.)*(u_.)*Sin[v_]^(p_.), x_Symbol] :> Dist[1/2^p, Int[u*Sin[2*v]^p, x], x] /; EqQ[w, v] && Integ
erQ[p]

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

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Mathematica [A]
time = 0.14, size = 107, normalized size = 0.81 \begin {gather*} \frac {\csc ^2\left (a+b x^2\right ) \left (-1+\cos \left (2 \left (a+b x^2\right )\right )+2 \sqrt {b} \sqrt {\pi } x \cos (2 a) S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )+2 \sqrt {b} \sqrt {\pi } x C\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}}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[a + b*x^2]^2*(-1 + Cos[2*(a + b*x^2)] + 2*Sqrt[b]*Sqrt[Pi]*x*Cos[2*a]*FresnelS[(2*Sqrt[b]*x)/Sqrt[Pi]] +
2*Sqrt[b]*Sqrt[Pi]*x*FresnelC[(2*Sqrt[b]*x)/Sqrt[Pi]]*Sin[2*a])*(c*Sin[a + b*x^2]^3)^(2/3))/(2*x)

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Maple [C] Result contains complex when optimal does not.
time = 0.18, size = 301, normalized size = 2.28

method	result	size
risch	\(-\frac {\left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}}}{4 x \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i x^{2} b} b \sqrt {\pi }\, \sqrt {2}\, \erf \left (\sqrt {2}\, \sqrt {i b}\, x \right )}{4 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2} \sqrt {i b}}+\frac {\left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}} \left (-\frac {{\mathrm e}^{4 i \left (b \,x^{2}+a \right )}}{x}+\frac {2 i b \sqrt {\pi }\, \erf \left (\sqrt {-2 i b}\, x \right ) {\mathrm e}^{2 i \left (b \,x^{2}+2 a \right )}}{\sqrt {-2 i b}}\right )}{4 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}+\frac {\left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{2 x \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}\)	\(301\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/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)-1/4*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*exp(2*I*b*x^2)*b*Pi^(1/2)*2^(1/2)/(I*b)^(1/
2)*erf(2^(1/2)*(I*b)^(1/2)*x)+1/4/(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)*(-1/x*exp(4*I*(b*x^2+a))+2*I*b*Pi^(1/2)/(-2*I*b)^(1/2)*erf((-2*I*b)^(1/2)*x)*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 = 90, normalized size = 0.68 \begin {gather*} \frac {\sqrt {2} \sqrt {b x^{2}} {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, 2 i \, b x^{2}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -2 i \, b x^{2}\right )\right )} \cos \left (2 \, a\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, 2 i \, b x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -2 i \, b x^{2}\right )\right )} \sin \left (2 \, a\right )\right )} c^{\frac {2}{3}} + 8 \, c^{\frac {2}{3}}}{32 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(sqrt(2)*sqrt(b*x^2)*((-(I + 1)*sqrt(2)*gamma(-1/2, 2*I*b*x^2) + (I - 1)*sqrt(2)*gamma(-1/2, -2*I*b*x^2))
*cos(2*a) + ((I - 1)*sqrt(2)*gamma(-1/2, 2*I*b*x^2) - (I + 1)*sqrt(2)*gamma(-1/2, -2*I*b*x^2))*sin(2*a))*c^(2/
3) + 8*c^(2/3))/x

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Fricas [A]
time = 0.36, size = 127, normalized size = 0.96 \begin {gather*} -\frac {4^{\frac {2}{3}} {\left (4^{\frac {1}{3}} \pi x \sqrt {\frac {b}{\pi }} \cos \left (2 \, a\right ) \operatorname {S}\left (2 \, x \sqrt {\frac {b}{\pi }}\right ) + 4^{\frac {1}{3}} \pi x \sqrt {\frac {b}{\pi }} \operatorname {C}\left (2 \, x \sqrt {\frac {b}{\pi }}\right ) \sin \left (2 \, a\right ) + 4^{\frac {1}{3}} \cos \left (b x^{2} + a\right )^{2} - 4^{\frac {1}{3}}\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 (x \cos \left (b x^{2} + a\right )^{2} - x\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^2,x, algorithm="fricas")

[Out]

-1/4*4^(2/3)*(4^(1/3)*pi*x*sqrt(b/pi)*cos(2*a)*fresnel_sin(2*x*sqrt(b/pi)) + 4^(1/3)*pi*x*sqrt(b/pi)*fresnel_c
os(2*x*sqrt(b/pi))*sin(2*a) + 4^(1/3)*cos(b*x^2 + a)^2 - 4^(1/3))*(-(c*cos(b*x^2 + a)^2 - c)*sin(b*x^2 + a))^(
2/3)/(x*cos(b*x^2 + a)^2 - x)

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

[Out]

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

[Out]

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

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

[Out]

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

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