3.4.24 \(\int \frac {\sqrt [3]{c \sin ^3(a+b x^2)}}{x^2} \, dx\) [324]

Optimal. Leaf size=135 \[ -\frac {\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x}+\sqrt {b} \sqrt {2 \pi } \cos (a) \csc \left (a+b x^2\right ) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\sqrt {b} \sqrt {2 \pi } \csc \left (a+b x^2\right ) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6852, 3468, 3435, 3433, 3432} \begin {gather*} \sqrt {2 \pi } \sqrt {b} \cos (a) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {b} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\sqrt {2 \pi } \sqrt {b} \sin (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\frac {\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x^2]^3)^(1/3)/x^2,x]

[Out]

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

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 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 3468

Int[((e_.)*(x_))^(m_)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(e*x)^(m + 1)*(Sin[c + d*x^n]/(e*(m + 1)
)), x] - Dist[d*(n/(e^n*(m + 1))), Int[(e*x)^(m + n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n,
0] && LtQ[m, -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 \frac {\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x^2} \, dx &=\left (\csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \frac {\sin \left (a+b x^2\right )}{x^2} \, dx\\ &=-\frac {\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x}+\left (2 b \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \cos \left (a+b x^2\right ) \, dx\\ &=-\frac {\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x}+\left (2 b \cos (a) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \cos \left (b x^2\right ) \, dx-\left (2 b \csc \left (a+b x^2\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \sin \left (b x^2\right ) \, dx\\ &=-\frac {\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x}+\sqrt {b} \sqrt {2 \pi } \cos (a) \csc \left (a+b x^2\right ) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\sqrt {b} \sqrt {2 \pi } \csc \left (a+b x^2\right ) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 105, normalized size = 0.78 \begin {gather*} \frac {\left (-1+\sqrt {b} \sqrt {2 \pi } x \cos (a) \csc \left (a+b x^2\right ) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )-\sqrt {b} \sqrt {2 \pi } x \csc \left (a+b x^2\right ) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a)\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x^2]^3)^(1/3)/x^2,x]

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.18, size = 232, normalized size = 1.72

method	result	size
risch	\(\frac {\left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} \left (-\frac {{\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{x}+\frac {i b \sqrt {\pi }\, \erf \left (\sqrt {-i b}\, x \right ) {\mathrm e}^{i \left (b \,x^{2}+2 a \right )}}{\sqrt {-i b}}\right )}{2 \,{\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-2}+\frac {\left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}}}{2 x \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )}+\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{i x^{2} b} b \sqrt {\pi }\, \erf \left (\sqrt {i b}\, x \right )}{2 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right ) \sqrt {i b}}\)	\(232\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x^2+a)^3)^(1/3)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.61, size = 76, normalized size = 0.56 \begin {gather*} \frac {\sqrt {b x^{2}} {\left ({\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, i \, b x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -i \, b x^{2}\right )\right )} \cos \left (a\right ) + {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, i \, b x^{2}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -i \, b x^{2}\right )\right )} \sin \left (a\right )\right )} c^{\frac {1}{3}}}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x^2+a)^3)^(1/3)/x^2,x, algorithm="maxima")

[Out]

1/16*sqrt(b*x^2)*(((I - 1)*sqrt(2)*gamma(-1/2, I*b*x^2) - (I + 1)*sqrt(2)*gamma(-1/2, -I*b*x^2))*cos(a) + ((I
+ 1)*sqrt(2)*gamma(-1/2, I*b*x^2) - (I - 1)*sqrt(2)*gamma(-1/2, -I*b*x^2))*sin(a))*c^(1/3)/x

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Fricas [A]
time = 0.37, size = 150, normalized size = 1.11 \begin {gather*} -\frac {4^{\frac {1}{3}} {\left (4^{\frac {2}{3}} \sqrt {2} \pi x \sqrt {\frac {b}{\pi }} \cos \left (a\right ) \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) \sin \left (b x^{2} + a\right ) - 4^{\frac {2}{3}} \sqrt {2} \pi x \sqrt {\frac {b}{\pi }} \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) \sin \left (b x^{2} + a\right ) \sin \left (a\right ) + 4^{\frac {2}{3}} \cos \left (b x^{2} + a\right )^{2} - 4^{\frac {2}{3}}\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac {1}{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)^(1/3)/x^2,x, algorithm="fricas")

[Out]

-1/4*4^(1/3)*(4^(2/3)*sqrt(2)*pi*x*sqrt(b/pi)*cos(a)*fresnel_cos(sqrt(2)*x*sqrt(b/pi))*sin(b*x^2 + a) - 4^(2/3
)*sqrt(2)*pi*x*sqrt(b/pi)*fresnel_sin(sqrt(2)*x*sqrt(b/pi))*sin(b*x^2 + a)*sin(a) + 4^(2/3)*cos(b*x^2 + a)^2 -
4^(2/3))*(-(c*cos(b*x^2 + a)^2 - c)*sin(b*x^2 + a))^(1/3)/(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 {\sqrt [3]{c \sin ^{3}{\left (a + b x^{2} \right )}}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x**2+a)**3)**(1/3)/x**2,x)

[Out]

Integral((c*sin(a + b*x**2)**3)**(1/3)/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)^(1/3)/x^2,x, algorithm="giac")

[Out]

integrate((c*sin(b*x^2 + a)^3)^(1/3)/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 )}^{1/3}}{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)^(1/3)/x^2,x)

[Out]

int((c*sin(a + b*x^2)^3)^(1/3)/x^2, x)

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