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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.08, 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 = {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 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]

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]

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 = 0.45, size = 166, normalized size = 1.44 \[ \frac {2 (c+d x)^{3/2} \sqrt [3]{\frac {b^2}{(c+d x)^3}} \sin \left (a+\frac {b}{(c+d x)^{3/2}}\right )+b (\cos (a)-i \sin (a)) \sqrt [3]{-\frac {i b}{(c+d x)^{3/2}}} \Gamma \left (\frac {1}{3},\frac {i b}{(c+d x)^{3/2}}\right )+b (\cos (a)+i \sin (a)) \sqrt [3]{\frac {i b}{(c+d x)^{3/2}}} \Gamma \left (\frac {1}{3},-\frac {i b}{(c+d x)^{3/2}}\right )}{2 d \sqrt {c+d x} \sqrt [3]{\frac {b^2}{(c+d x)^3}}} \]

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])

________________________________________________________________________________________

fricas [A]  time = 0.89, size = 144, normalized size = 1.25 \[ \frac {-i \, \left (i \, b\right )^{\frac {2}{3}} e^{\left (-i \, a\right )} \Gamma \left (\frac {1}{3}, \frac {i \, \sqrt {d x + c} b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + i \, \left (-i \, b\right )^{\frac {2}{3}} e^{\left (i \, a\right )} \Gamma \left (\frac {1}{3}, -\frac {i \, \sqrt {d x + c} b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 2 \, {\left (d x + c\right )} \sin \left (\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + \sqrt {d x + c} b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \, d} \]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {3}{2}}}\right )\, dx \]

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)

________________________________________________________________________________________

maxima [A]  time = 0.60, size = 151, normalized size = 1.31 \[ \frac {4 \, {\left (d x + c\right )}^{\frac {3}{2}} \left (\frac {b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right )^{\frac {1}{3}} \sin \left (\frac {{\left (d x + c\right )}^{\frac {3}{2}} a + b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right ) + {\left ({\left ({\left (\sqrt {3} - i\right )} \Gamma \left (\frac {1}{3}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (\frac {1}{3}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right )\right )} \cos \relax (a) + {\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right )\right )} \sin \relax (a)\right )} b}{4 \, \sqrt {d x + c} d \left (\frac {b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right )^{\frac {1}{3}}} \]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________