\(\int \sin (a+\frac {b}{(c+d x)^{3/2}}) \, dx\) [204]

Optimal result

Integrand size = 14, antiderivative size = 115 \[ \int \sin \left (a+\frac {b}{(c+d x)^{3/2}}\right ) \, dx=-\frac {i e^{i a} \left (-\frac {i b}{(c+d x)^{3/2}}\right )^{2/3} (c+d x) \Gamma \left (-\frac {2}{3},-\frac {i b}{(c+d x)^{3/2}}\right )}{3 d}+\frac {i e^{-i a} \left (\frac {i b}{(c+d x)^{3/2}}\right )^{2/3} (c+d x) \Gamma \left (-\frac {2}{3},\frac {i b}{(c+d x)^{3/2}}\right )}{3 d} \] Output:

-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)
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.44 \[ \int \sin \left (a+\frac {b}{(c+d x)^{3/2}}\right ) \, dx=\frac {b \sqrt [3]{-\frac {i b}{(c+d x)^{3/2}}} \Gamma \left (\frac {1}{3},\frac {i b}{(c+d x)^{3/2}}\right ) (\cos (a)-i \sin (a))+b \sqrt [3]{\frac {i b}{(c+d x)^{3/2}}} \Gamma \left (\frac {1}{3},-\frac {i b}{(c+d x)^{3/2}}\right ) (\cos (a)+i \sin (a))+2 \sqrt [3]{\frac {b^2}{(c+d x)^3}} (c+d x)^{3/2} \sin \left (a+\frac {b}{(c+d x)^{3/2}}\right )}{2 d \sqrt [3]{\frac {b^2}{(c+d x)^3}} \sqrt {c+d x}} \] Input:

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

Output:

(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])
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3844, 3904, 2648}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3844

\(\displaystyle \frac {2 \int \sqrt {c+d x} \sin \left (a+\frac {b}{(c+d x)^{3/2}}\right )d\sqrt {c+d x}}{d}\)

\(\Big \downarrow \) 3904

\(\displaystyle \frac {2 \left (\frac {1}{2} i \int e^{-i a-\frac {i b}{(c+d x)^{3/2}}} \sqrt {c+d x}d\sqrt {c+d x}-\frac {1}{2} i \int e^{i a+\frac {i b}{(c+d x)^{3/2}}} \sqrt {c+d x}d\sqrt {c+d x}\right )}{d}\)

\(\Big \downarrow \) 2648

\(\displaystyle \frac {2 \left (\frac {1}{6} i e^{-i a} (c+d x) \left (\frac {i b}{(c+d x)^{3/2}}\right )^{2/3} \Gamma \left (-\frac {2}{3},\frac {i b}{(c+d x)^{3/2}}\right )-\frac {1}{6} i e^{i a} (c+d x) \left (-\frac {i b}{(c+d x)^{3/2}}\right )^{2/3} \Gamma \left (-\frac {2}{3},-\frac {i b}{(c+d x)^{3/2}}\right )\right )}{d}\)

Input:

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

Output:

(2*((-1/6*I)*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)] + ((I/6)*((I*b)/(c + d*x)^(3/2))^(2/3)*(c + d* 
x)*Gamma[-2/3, (I*b)/(c + d*x)^(3/2)])/E^(I*a)))/d
 

Defintions of rubi rules used

rule 2648

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] /; FreeQ[{F 
, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]
 

rule 3844

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_S 
ymbol] :> Module[{k = Denominator[n]}, Simp[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 3904

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

Maple [F]

\[\int \sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {3}{2}}}\right )d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.32 \[ \int \sin \left (a+\frac {b}{(c+d x)^{3/2}}\right ) \, dx=\frac {\left (i \, b\right )^{\frac {2}{3}} {\left (-i \, \cos \left (a\right ) - \sin \left (a\right )\right )} \Gamma \left (\frac {1}{3}, \frac {i \, \sqrt {d x + c} b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \left (-i \, b\right )^{\frac {2}{3}} {\left (i \, \cos \left (a\right ) - \sin \left (a\right )\right )} \Gamma \left (\frac {1}{3}, -\frac {i \, \sqrt {d x + c} b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 2 \, {\left (d x + c\right )} \sin \left (\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + \sqrt {d x + c} b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \, d} \] Input:

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

Output:

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

Sympy [F]

\[ \int \sin \left (a+\frac {b}{(c+d x)^{3/2}}\right ) \, dx=\int \sin {\left (a + \frac {b}{\left (c + d x\right )^{\frac {3}{2}}} \right )}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.31 \[ \int \sin \left (a+\frac {b}{(c+d x)^{3/2}}\right ) \, dx=\frac {4 \, {\left (d x + c\right )}^{\frac {3}{2}} \left (\frac {b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right )^{\frac {1}{3}} \sin \left (\frac {{\left (d x + c\right )}^{\frac {3}{2}} a + b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right ) + {\left ({\left ({\left (\sqrt {3} - i\right )} \Gamma \left (\frac {1}{3}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (\frac {1}{3}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right )\right )} \cos \left (a\right ) + {\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right )\right )} \sin \left (a\right )\right )} b}{4 \, \sqrt {d x + c} d \left (\frac {b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right )^{\frac {1}{3}}} \] Input:

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

Output:

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

Giac [F]

\[ \int \sin \left (a+\frac {b}{(c+d x)^{3/2}}\right ) \, dx=\int { \sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {3}{2}}}\right ) \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \sin \left (a+\frac {b}{(c+d x)^{3/2}}\right ) \, dx=\int \sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{3/2}}\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \sin \left (a+\frac {b}{(c+d x)^{3/2}}\right ) \, dx=\int \sin \left (\frac {\sqrt {d x +c}\, a c +\sqrt {d x +c}\, a d x +b}{\sqrt {d x +c}\, c +\sqrt {d x +c}\, d x}\right )d x \] Input:

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

Output:

int(sin((sqrt(c + d*x)*a*c + sqrt(c + d*x)*a*d*x + b)/(sqrt(c + d*x)*c + s 
qrt(c + d*x)*d*x)),x)