3.4.22 \(\int \sqrt [3]{c \sin ^3(a+b x^2)} \, dx\) [322]

Optimal. Leaf size=117 \[ \frac {\sqrt {\frac {\pi }{2}} \cos (a) \csc \left (a+b x^2\right ) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{\sqrt {b}}+\frac {\sqrt {\frac {\pi }{2}} \csc \left (a+b x^2\right ) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{\sqrt {b}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6852, 3434, 3433, 3432} \begin {gather*} \frac {\sqrt {\frac {\pi }{2}} \sin (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 {b}}+\frac {\sqrt {\frac {\pi }{2}} \cos (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 )}}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x^2]^3)^(1/3),x]

[Out]

(Sqrt[Pi/2]*Cos[a]*Csc[a + b*x^2]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x]*(c*Sin[a + b*x^2]^3)^(1/3))/Sqrt[b] + (Sqrt[P
i/2]*Csc[a + b*x^2]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*x]*Sin[a]*(c*Sin[a + b*x^2]^3)^(1/3))/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 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 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 \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx &=\left (\csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \sin \left (a+b x^2\right ) \, dx\\ &=\left (\cos (a) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \sin \left (b x^2\right ) \, dx+\left (\csc \left (a+b x^2\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \cos \left (b x^2\right ) \, dx\\ &=\frac {\sqrt {\frac {\pi }{2}} \cos (a) \csc \left (a+b x^2\right ) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{\sqrt {b}}+\frac {\sqrt {\frac {\pi }{2}} \csc \left (a+b x^2\right ) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{\sqrt {b}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.08, size = 80, normalized size = 0.68 \begin {gather*} \frac {\sqrt {\frac {\pi }{2}} \csc \left (a+b x^2\right ) \left (\cos (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )+C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a)\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 0.17, size = 157, normalized size = 1.34

method	result	size
risch	\(\frac {\erf \left (\sqrt {-i b}\, x \right ) \sqrt {\pi }\, \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{i \left (b \,x^{2}+2 a \right )}}{4 \sqrt {-i b}\, \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )}-\frac {\left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{i x^{2} b} \sqrt {\pi }\, \erf \left (\sqrt {i b}\, x \right )}{4 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right ) \sqrt {i b}}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*erf((-I*b)^(1/2)*x)/(-I*b)^(1/2)*Pi^(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)*exp(I*(b*x^2+2*a))-1/4*(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)*Pi^(1/2)/(I*b)^(1/2)*erf((I*b)^(1/2)*x)

________________________________________________________________________________________

Maxima [C] Result contains complex when optimal does not.
time = 0.55, size = 51, normalized size = 0.44 \begin {gather*} \frac {\sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i + 1\right ) \, \cos \left (a\right ) + \left (i - 1\right ) \, \sin \left (a\right )\right )} \operatorname {erf}\left (\sqrt {i \, b} x\right ) + {\left (\left (i - 1\right ) \, \cos \left (a\right ) - \left (i + 1\right ) \, \sin \left (a\right )\right )} \operatorname {erf}\left (\sqrt {-i \, b} x\right )\right )} c^{\frac {1}{3}}}{16 \, \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 128, normalized size = 1.09 \begin {gather*} -\frac {4^{\frac {1}{3}} {\left (4^{\frac {2}{3}} \sqrt {2} \pi \sqrt {\frac {b}{\pi }} \cos \left (a\right ) \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) \sin \left (b x^{2} + a\right ) + 4^{\frac {2}{3}} \sqrt {2} \pi \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) \sin \left (b x^{2} + a\right ) \sin \left (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 {1}{3}}}{8 \, {\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)^(1/3),x, algorithm="fricas")

[Out]

-1/8*4^(1/3)*(4^(2/3)*sqrt(2)*pi*sqrt(b/pi)*cos(a)*fresnel_sin(sqrt(2)*x*sqrt(b/pi))*sin(b*x^2 + a) + 4^(2/3)*
sqrt(2)*pi*sqrt(b/pi)*fresnel_cos(sqrt(2)*x*sqrt(b/pi))*sin(b*x^2 + a)*sin(a))*(-(c*cos(b*x^2 + a)^2 - c)*sin(
b*x^2 + a))^(1/3)/(b*cos(b*x^2 + a)^2 - b)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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 )}^{1/3} \,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)

[Out]

int((c*sin(a + b*x^2)^3)^(1/3), x)

________________________________________________________________________________________