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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0822319, 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, 3423, 2218} \[ \frac{i e^{-i a} (c+d x) \left (\frac{i b}{(c+d x)^{3/2}}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},\frac{i b}{(c+d x)^{3/2}}\right )}{3 d}-\frac{i e^{i a} (c+d x) \left (-\frac{i b}{(c+d x)^{3/2}}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},-\frac{i b}{(c+d x)^{3/2}}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^(3/2)],x]

[Out]

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

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 3423

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, n}, x]

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

Mathematica [A]  time = 0.439614, size = 166, normalized size = 1.44 \[ \frac{b (\cos (a)-i \sin (a)) \sqrt [3]{-\frac{i b}{(c+d x)^{3/2}}} \text{Gamma}\left (\frac{1}{3},\frac{i b}{(c+d x)^{3/2}}\right )+b (\cos (a)+i \sin (a)) \sqrt [3]{\frac{i b}{(c+d x)^{3/2}}} \text{Gamma}\left (\frac{1}{3},-\frac{i b}{(c+d x)^{3/2}}\right )+2 (c+d x)^{3/2} \sqrt [3]{\frac{b^2}{(c+d x)^3}} \sin \left (a+\frac{b}{(c+d x)^{3/2}}\right )}{2 d \sqrt{c+d x} \sqrt [3]{\frac{b^2}{(c+d x)^3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(((-I)*b)/(c + d*x)^(3/2))^(1/3)*Gamma[1/3, (I*b)/(c + d*x)^(3/2)]*(Cos[a] - I*Sin[a]) + b*((I*b)/(c + d*x)
^(3/2))^(1/3)*Gamma[1/3, ((-I)*b)/(c + d*x)^(3/2)]*(Cos[a] + I*Sin[a]) + 2*(b^2/(c + d*x)^3)^(1/3)*(c + d*x)^(
3/2)*Sin[a + b/(c + d*x)^(3/2)])/(2*d*(b^2/(c + d*x)^3)^(1/3)*Sqrt[c + d*x])

________________________________________________________________________________________

Maple [F]  time = 0.012, 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)

________________________________________________________________________________________

Maxima [B]  time = 1.38339, size = 516, normalized size = 4.49 \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]

1/4*(4*(d*x + c)^(3/2)*(abs(b)/(d*x + c)^(3/2))^(1/3)*sin(((d*x + c)^(3/2)*a + b)/(d*x + c)^(3/2)) + (((gamma(
1/3, I*b/(d*x + c)^(3/2)) + gamma(1/3, -I*b/(d*x + c)^(3/2)))*cos(1/6*pi + 1/3*arctan2(0, b)) + (gamma(1/3, I*
b/(d*x + c)^(3/2)) + gamma(1/3, -I*b/(d*x + c)^(3/2)))*cos(-1/6*pi + 1/3*arctan2(0, b)) + (-I*gamma(1/3, I*b/(
d*x + c)^(3/2)) + I*gamma(1/3, -I*b/(d*x + c)^(3/2)))*sin(1/6*pi + 1/3*arctan2(0, b)) + (I*gamma(1/3, I*b/(d*x
+ c)^(3/2)) - I*gamma(1/3, -I*b/(d*x + c)^(3/2)))*sin(-1/6*pi + 1/3*arctan2(0, b)))*cos(a) + ((-I*gamma(1/3,
I*b/(d*x + c)^(3/2)) + I*gamma(1/3, -I*b/(d*x + c)^(3/2)))*cos(1/6*pi + 1/3*arctan2(0, b)) + (-I*gamma(1/3, I*
b/(d*x + c)^(3/2)) + I*gamma(1/3, -I*b/(d*x + c)^(3/2)))*cos(-1/6*pi + 1/3*arctan2(0, b)) - (gamma(1/3, I*b/(d
*x + c)^(3/2)) + gamma(1/3, -I*b/(d*x + c)^(3/2)))*sin(1/6*pi + 1/3*arctan2(0, b)) + (gamma(1/3, I*b/(d*x + c)
^(3/2)) + gamma(1/3, -I*b/(d*x + c)^(3/2)))*sin(-1/6*pi + 1/3*arctan2(0, b)))*sin(a))*b)/(sqrt(d*x + c)*d*(abs
(b)/(d*x + c)^(3/2))^(1/3))

________________________________________________________________________________________

Fricas [A]  time = 1.94411, size = 359, normalized size = 3.12 \begin{align*} \frac{-i \, \left (i \, b\right )^{\frac{2}{3}} e^{\left (-i \, a\right )} \Gamma \left (\frac{1}{3}, \frac{i \, \sqrt{d x + c} b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + i \, \left (-i \, b\right )^{\frac{2}{3}} e^{\left (i \, a\right )} \Gamma \left (\frac{1}{3}, -\frac{i \, \sqrt{d x + c} b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 2 \,{\left (d x + c\right )} \sin \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + \sqrt{d x + c} b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \, 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]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (a + \frac{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)

________________________________________________________________________________________

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