\(\int \frac {\sin ^3(a+b x)}{(c+d x)^{3/2}} \, dx\) [57]

Optimal result

Integrand size = 18, antiderivative size = 270 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^{3/2}} \, dx=\frac {3 \sqrt {b} \sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {\sqrt {b} \sqrt {\frac {3 \pi }{2}} \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {6}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {\sqrt {b} \sqrt {\frac {3 \pi }{2}} \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {6}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (3 a-\frac {3 b c}{d}\right )}{d^{3/2}}-\frac {3 \sqrt {b} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (a-\frac {b c}{d}\right )}{d^{3/2}}-\frac {2 \sin ^3(a+b x)}{d \sqrt {c+d x}} \] Output:

3/2*b^(1/2)*2^(1/2)*Pi^(1/2)*cos(a-b*c/d)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2 
)*(d*x+c)^(1/2)/d^(1/2))/d^(3/2)-1/2*b^(1/2)*6^(1/2)*Pi^(1/2)*cos(3*a-3*b* 
c/d)*FresnelC(b^(1/2)*6^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^(3/2)+1/2* 
b^(1/2)*6^(1/2)*Pi^(1/2)*FresnelS(b^(1/2)*6^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d 
^(1/2))*sin(3*a-3*b*c/d)/d^(3/2)-3/2*b^(1/2)*2^(1/2)*Pi^(1/2)*FresnelS(b^( 
1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))*sin(a-b*c/d)/d^(3/2)-2*sin(b* 
x+a)^3/d/(d*x+c)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.83 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.09 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^{3/2}} \, dx=-\frac {i e^{-3 i a} \left (-e^{-3 i b x}+3 e^{2 i a-i b x}+e^{3 i (2 a+b x)}-3 e^{i (4 a+b x)}+3 e^{4 i a-\frac {i b c}{d}} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {i b (c+d x)}{d}\right )-3 e^{i \left (2 a+\frac {b c}{d}\right )} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {i b (c+d x)}{d}\right )-\sqrt {3} e^{6 i a-\frac {3 i b c}{d}} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {3 i b (c+d x)}{d}\right )+\sqrt {3} e^{\frac {3 i b c}{d}} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {3 i b (c+d x)}{d}\right )\right )}{4 d \sqrt {c+d x}} \] Input:

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

Output:

((-1/4*I)*(-E^((-3*I)*b*x) + 3*E^((2*I)*a - I*b*x) + E^((3*I)*(2*a + b*x)) 
 - 3*E^(I*(4*a + b*x)) + 3*E^((4*I)*a - (I*b*c)/d)*Sqrt[((-I)*b*(c + d*x)) 
/d]*Gamma[1/2, ((-I)*b*(c + d*x))/d] - 3*E^(I*(2*a + (b*c)/d))*Sqrt[(I*b*( 
c + d*x))/d]*Gamma[1/2, (I*b*(c + d*x))/d] - Sqrt[3]*E^((6*I)*a - ((3*I)*b 
*c)/d)*Sqrt[((-I)*b*(c + d*x))/d]*Gamma[1/2, ((-3*I)*b*(c + d*x))/d] + Sqr 
t[3]*E^(((3*I)*b*c)/d)*Sqrt[(I*b*(c + d*x))/d]*Gamma[1/2, ((3*I)*b*(c + d* 
x))/d]))/(d*E^((3*I)*a)*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3794, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3794

\(\displaystyle \frac {6 b \int \left (\frac {\cos (a+b x)}{4 \sqrt {c+d x}}-\frac {\cos (3 a+3 b x)}{4 \sqrt {c+d x}}\right )dx}{d}-\frac {2 \sin ^3(a+b x)}{d \sqrt {c+d x}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {6 b \left (\frac {\sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {\sqrt {\frac {\pi }{6}} \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {6}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}+\frac {\sqrt {\frac {\pi }{6}} \sin \left (3 a-\frac {3 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {6}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {\sqrt {\frac {\pi }{2}} \sin \left (a-\frac {b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}\right )}{d}-\frac {2 \sin ^3(a+b x)}{d \sqrt {c+d x}}\)

Input:

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

Output:

(6*b*((Sqrt[Pi/2]*Cos[a - (b*c)/d]*FresnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d 
*x])/Sqrt[d]])/(2*Sqrt[b]*Sqrt[d]) - (Sqrt[Pi/6]*Cos[3*a - (3*b*c)/d]*Fres 
nelC[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(2*Sqrt[b]*Sqrt[d]) + (S 
qrt[Pi/6]*FresnelS[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[3*a - ( 
3*b*c)/d])/(2*Sqrt[b]*Sqrt[d]) - (Sqrt[Pi/2]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]* 
Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(2*Sqrt[b]*Sqrt[d])))/d - (2*Sin 
[a + b*x]^3)/(d*Sqrt[c + d*x])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3794

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Si 
mp[(c + d*x)^(m + 1)*(Sin[e + f*x]^n/(d*(m + 1))), x] - Simp[f*(n/(d*(m + 1 
)))   Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]^(n 
- 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] & 
& LtQ[m, -1]
 

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.07

method	result	size
derivativedivides	\(\frac {-\frac {3 \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{2 \sqrt {d x +c}}+\frac {3 b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{2 d \sqrt {\frac {b}{d}}}+\frac {\sin \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 b c}{d}\right )}{2 \sqrt {d x +c}}-\frac {b \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \left (\cos \left (\frac {3 a d -3 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {3 a d -3 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{2 d \sqrt {\frac {b}{d}}}}{d}\)	\(288\)
default	\(\frac {-\frac {3 \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{2 \sqrt {d x +c}}+\frac {3 b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{2 d \sqrt {\frac {b}{d}}}+\frac {\sin \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 b c}{d}\right )}{2 \sqrt {d x +c}}-\frac {b \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \left (\cos \left (\frac {3 a d -3 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {3 a d -3 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{2 d \sqrt {\frac {b}{d}}}}{d}\)	\(288\)

Input:

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

Output:

2/d*(-3/4/(d*x+c)^(1/2)*sin(b*(d*x+c)/d+(a*d-b*c)/d)+3/4*b/d*2^(1/2)*Pi^(1 
/2)/(b/d)^(1/2)*(cos((a*d-b*c)/d)*FresnelC(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*b* 
(d*x+c)^(1/2)/d)-sin((a*d-b*c)/d)*FresnelS(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*b* 
(d*x+c)^(1/2)/d))+1/4/(d*x+c)^(1/2)*sin(3*b*(d*x+c)/d+3*(a*d-b*c)/d)-1/4*b 
/d*2^(1/2)*Pi^(1/2)*3^(1/2)/(b/d)^(1/2)*(cos(3*(a*d-b*c)/d)*FresnelC(2^(1/ 
2)/Pi^(1/2)*3^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)-sin(3*(a*d-b*c)/d)*Fres 
nelS(2^(1/2)/Pi^(1/2)*3^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.01 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^{3/2}} \, dx=-\frac {\sqrt {6} {\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (\sqrt {6} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 3 \, \sqrt {2} {\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 3 \, \sqrt {2} {\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) - \sqrt {6} {\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (\sqrt {6} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) - 4 \, \sqrt {d x + c} {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )}{2 \, {\left (d^{2} x + c d\right )}} \] Input:

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

Output:

-1/2*(sqrt(6)*(pi*d*x + pi*c)*sqrt(b/(pi*d))*cos(-3*(b*c - a*d)/d)*fresnel 
_cos(sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d))) - 3*sqrt(2)*(pi*d*x + pi*c)*sqr 
t(b/(pi*d))*cos(-(b*c - a*d)/d)*fresnel_cos(sqrt(2)*sqrt(d*x + c)*sqrt(b/( 
pi*d))) + 3*sqrt(2)*(pi*d*x + pi*c)*sqrt(b/(pi*d))*fresnel_sin(sqrt(2)*sqr 
t(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c - a*d)/d) - sqrt(6)*(pi*d*x + pi*c)*s 
qrt(b/(pi*d))*fresnel_sin(sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-3*(b* 
c - a*d)/d) - 4*sqrt(d*x + c)*(cos(b*x + a)^2 - 1)*sin(b*x + a))/(d^2*x + 
c*d)
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.94 \[ \int \frac {\sin ^3(a+b x)}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {3} {\left ({\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {3 i \, {\left (d x + c\right )} b}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {3 i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {3 i \, {\left (d x + c\right )} b}{d}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {3 i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )\right )} \sqrt {\frac {{\left (d x + c\right )} b}{d}} - 3 \, {\left ({\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )} \sqrt {\frac {{\left (d x + c\right )} b}{d}}}{16 \, \sqrt {d x + c} d} \] Input:

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

Output:

1/16*(sqrt(3)*(((I - 1)*sqrt(2)*gamma(-1/2, 3*I*(d*x + c)*b/d) - (I + 1)*s 
qrt(2)*gamma(-1/2, -3*I*(d*x + c)*b/d))*cos(-3*(b*c - a*d)/d) + ((I + 1)*s 
qrt(2)*gamma(-1/2, 3*I*(d*x + c)*b/d) - (I - 1)*sqrt(2)*gamma(-1/2, -3*I*( 
d*x + c)*b/d))*sin(-3*(b*c - a*d)/d))*sqrt((d*x + c)*b/d) - 3*(((I - 1)*sq 
rt(2)*gamma(-1/2, I*(d*x + c)*b/d) - (I + 1)*sqrt(2)*gamma(-1/2, -I*(d*x + 
 c)*b/d))*cos(-(b*c - a*d)/d) + ((I + 1)*sqrt(2)*gamma(-1/2, I*(d*x + c)*b 
/d) - (I - 1)*sqrt(2)*gamma(-1/2, -I*(d*x + c)*b/d))*sin(-(b*c - a*d)/d))* 
sqrt((d*x + c)*b/d))/(sqrt(d*x + c)*d)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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