3.194 \(\int \sin (a+b (c+d x)^{3/2}) \, dx\)

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

[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))

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Rubi [A]  time = 0.0812329, antiderivative size = 115, normalized size of antiderivative = 1., 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 = {3363, 3389, 2218} \[ \frac{i e^{i a} (c+d x) \text{Gamma}\left (\frac{2}{3},-i b (c+d x)^{3/2}\right )}{3 d \left (-i b (c+d x)^{3/2}\right )^{2/3}}-\frac{i e^{-i a} (c+d x) \text{Gamma}\left (\frac{2}{3},i b (c+d x)^{3/2}\right )}{3 d \left (i b (c+d x)^{3/2}\right )^{2/3}} \]

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 3363

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 3389

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]

Rule 2218

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

Rubi steps

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

Mathematica [A]  time = 0.149353, size = 123, normalized size = 1.07 \[ \frac{i (c+d x) \left ((\cos (a)+i \sin (a)) \left (i b (c+d x)^{3/2}\right )^{2/3} \text{Gamma}\left (\frac{2}{3},-i b (c+d x)^{3/2}\right )-(\cos (a)-i \sin (a)) \left (-i b (c+d x)^{3/2}\right )^{2/3} \text{Gamma}\left (\frac{2}{3},i b (c+d x)^{3/2}\right )\right )}{3 d \left (b^2 (c+d x)^3\right )^{2/3}} \]

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 = 0.004, size = 0, normalized size = 0. \begin{align*} \int \sin \left ( a+b \left ( dx+c \right ) ^{{\frac{3}{2}}} \right ) \, dx \end{align*}

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 [B]  time = 1.3706, size = 466, normalized size = 4.05 \begin{align*} \text{result too large to display} \end{align*}

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)*abs(b))^(1/3)*(((-I*gamma(2/3, I*(d*x + c)^(3/2)*b) + I*gamma(2/3, -I*(d*x + c)^(3/2)*b))
*cos(1/3*pi + 2/3*arctan2(0, b)) + (-I*gamma(2/3, I*(d*x + c)^(3/2)*b) + I*gamma(2/3, -I*(d*x + c)^(3/2)*b))*c
os(-1/3*pi + 2/3*arctan2(0, b)) - (gamma(2/3, I*(d*x + c)^(3/2)*b) + gamma(2/3, -I*(d*x + c)^(3/2)*b))*sin(1/3
*pi + 2/3*arctan2(0, b)) + (gamma(2/3, I*(d*x + c)^(3/2)*b) + gamma(2/3, -I*(d*x + c)^(3/2)*b))*sin(-1/3*pi +
2/3*arctan2(0, b)))*cos(a) - ((gamma(2/3, I*(d*x + c)^(3/2)*b) + gamma(2/3, -I*(d*x + c)^(3/2)*b))*cos(1/3*pi
+ 2/3*arctan2(0, b)) + (gamma(2/3, I*(d*x + c)^(3/2)*b) + gamma(2/3, -I*(d*x + c)^(3/2)*b))*cos(-1/3*pi + 2/3*
arctan2(0, b)) - (I*gamma(2/3, I*(d*x + c)^(3/2)*b) - I*gamma(2/3, -I*(d*x + c)^(3/2)*b))*sin(1/3*pi + 2/3*arc
tan2(0, b)) - (-I*gamma(2/3, I*(d*x + c)^(3/2)*b) + I*gamma(2/3, -I*(d*x + c)^(3/2)*b))*sin(-1/3*pi + 2/3*arct
an2(0, b)))*sin(a))/(sqrt(d*x + c)*d*abs(b))

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Fricas [A]  time = 2.14939, size = 198, normalized size = 1.72 \begin{align*} -\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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (a + b \left (c + d x\right )^{\frac{3}{2}} \right )}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \sin \left ({\left (d x + c\right )}^{\frac{3}{2}} b + a\right )\,{d x} \end{align*}

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)