3.2.94 \(\int \sin (a+b (c+d x)^{3/2}) \, dx\) [194]

Optimal. Leaf size=115 \[ \frac {i e^{i a} (c+d x) \Gamma \left (\frac {2}{3},-i b (c+d x)^{3/2}\right )}{3 d \left (-i b (c+d x)^{3/2}\right )^{2/3}}-\frac {i e^{-i a} (c+d x) \Gamma \left (\frac {2}{3},i b (c+d x)^{3/2}\right )}{3 d \left (i b (c+d x)^{3/2}\right )^{2/3}} \]

[Out]

1/3*I*exp(I*a)*(d*x+c)*GAMMA(2/3,-I*b*(d*x+c)^(3/2))/d/(-I*b*(d*x+c)^(3/2))^(2/3)-1/3*I*(d*x+c)*GAMMA(2/3,I*b*
(d*x+c)^(3/2))/d/exp(I*a)/(I*b*(d*x+c)^(3/2))^(2/3)

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Rubi [A]
time = 0.05, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3444, 3470, 2250} \begin {gather*} \frac {i e^{i a} (c+d x) \text {Gamma}\left (\frac {2}{3},-i b (c+d x)^{3/2}\right )}{3 d \left (-i b (c+d x)^{3/2}\right )^{2/3}}-\frac {i e^{-i a} (c+d x) \text {Gamma}\left (\frac {2}{3},i b (c+d x)^{3/2}\right )}{3 d \left (i b (c+d x)^{3/2}\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*(c + d*x)^(3/2)],x]

[Out]

((I/3)*E^(I*a)*(c + d*x)*Gamma[2/3, (-I)*b*(c + d*x)^(3/2)])/(d*((-I)*b*(c + d*x)^(3/2))^(2/3)) - ((I/3)*(c +
d*x)*Gamma[2/3, I*b*(c + d*x)^(3/2)])/(d*E^(I*a)*(I*b*(c + d*x)^(3/2))^(2/3))

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 3444

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Module[{k = Denominator[n
]}, Dist[k/f, Subst[Int[x^(k - 1)*(a + b*Sin[c + d*x^(k*n)])^p, x], x, (e + f*x)^(1/k)], x]] /; FreeQ[{a, b, c
, d, e, f}, x] && IGtQ[p, 0] && FractionQ[n]

Rule 3470

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[I/2, Int[(e*x)^m*E^((-c)*I - d*I*x^n),
x], x] - Dist[I/2, Int[(e*x)^m*E^(c*I + d*I*x^n), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \sin \left (a+b (c+d x)^{3/2}\right ) \, dx &=\frac {2 \text {Subst}\left (\int x \sin \left (a+b x^3\right ) \, dx,x,\sqrt {c+d x}\right )}{d}\\ &=\frac {i \text {Subst}\left (\int e^{-i a-i b x^3} x \, dx,x,\sqrt {c+d x}\right )}{d}-\frac {i \text {Subst}\left (\int e^{i a+i b x^3} x \, dx,x,\sqrt {c+d x}\right )}{d}\\ &=\frac {i e^{i a} (c+d x) \Gamma \left (\frac {2}{3},-i b (c+d x)^{3/2}\right )}{3 d \left (-i b (c+d x)^{3/2}\right )^{2/3}}-\frac {i e^{-i a} (c+d x) \Gamma \left (\frac {2}{3},i b (c+d x)^{3/2}\right )}{3 d \left (i b (c+d x)^{3/2}\right )^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 123, normalized size = 1.07 \begin {gather*} \frac {i (c+d x) \left (-\left (-i b (c+d x)^{3/2}\right )^{2/3} \Gamma \left (\frac {2}{3},i b (c+d x)^{3/2}\right ) (\cos (a)-i \sin (a))+\left (i b (c+d x)^{3/2}\right )^{2/3} \Gamma \left (\frac {2}{3},-i b (c+d x)^{3/2}\right ) (\cos (a)+i \sin (a))\right )}{3 d \left (b^2 (c+d x)^3\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*(c + d*x)*(-(((-I)*b*(c + d*x)^(3/2))^(2/3)*Gamma[2/3, I*b*(c + d*x)^(3/2)]*(Cos[a] - I*Sin[a])) + (I*b
*(c + d*x)^(3/2))^(2/3)*Gamma[2/3, (-I)*b*(c + d*x)^(3/2)]*(Cos[a] + I*Sin[a])))/(d*(b^2*(c + d*x)^3)^(2/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.35, size = 112, normalized size = 0.97 \begin {gather*} -\frac {\left ({\left (d x + c\right )}^{\frac {3}{2}} b\right )^{\frac {1}{3}} {\left ({\left ({\left (\sqrt {3} + i\right )} \Gamma \left (\frac {2}{3}, i \, {\left (d x + c\right )}^{\frac {3}{2}} b\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (\frac {2}{3}, -i \, {\left (d x + c\right )}^{\frac {3}{2}} b\right )\right )} \cos \left (a\right ) - {\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, i \, {\left (d x + c\right )}^{\frac {3}{2}} b\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, -i \, {\left (d x + c\right )}^{\frac {3}{2}} b\right )\right )} \sin \left (a\right )\right )}}{6 \, \sqrt {d x + c} b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((d*x + c)^(3/2)*b)^(1/3)*(((sqrt(3) + I)*gamma(2/3, I*(d*x + c)^(3/2)*b) + (sqrt(3) - I)*gamma(2/3, -I*(
d*x + c)^(3/2)*b))*cos(a) - ((I*sqrt(3) - 1)*gamma(2/3, I*(d*x + c)^(3/2)*b) + (-I*sqrt(3) - 1)*gamma(2/3, -I*
(d*x + c)^(3/2)*b))*sin(a))/(sqrt(d*x + c)*b*d)

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Fricas [A]
time = 0.10, size = 69, normalized size = 0.60 \begin {gather*} -\frac {\left (i \, b\right )^{\frac {1}{3}} e^{\left (-i \, a\right )} \Gamma \left (\frac {2}{3}, {\left (i \, b d x + i \, b c\right )} \sqrt {d x + c}\right ) + \left (-i \, b\right )^{\frac {1}{3}} e^{\left (i \, a\right )} \Gamma \left (\frac {2}{3}, {\left (-i \, b d x - i \, b c\right )} \sqrt {d x + c}\right )}{3 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((I*b)^(1/3)*e^(-I*a)*gamma(2/3, (I*b*d*x + I*b*c)*sqrt(d*x + c)) + (-I*b)^(1/3)*e^(I*a)*gamma(2/3, (-I*b
*d*x - I*b*c)*sqrt(d*x + c)))/(b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*(c + d*x)^(3/2)),x)

[Out]

int(sin(a + b*(c + d*x)^(3/2)), x)

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