\(\int \frac {\sin (a+b x)}{(c+d x)^{3/2}} \, dx\) [42]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [C] (verification not implemented)
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 16, antiderivative size = 139 \[ \int \frac {\sin (a+b x)}{(c+d x)^{3/2}} \, dx=\frac {2 \sqrt {b} \sqrt {2 \pi } \cos \left (a-\frac {b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {2 \sqrt {b} \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (a-\frac {b c}{d}\right )}{d^{3/2}}-\frac {2 \sin (a+b x)}{d \sqrt {c+d x}} \] Output:

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)-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) 
/d/(d*x+c)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06 \[ \int \frac {\sin (a+b x)}{(c+d x)^{3/2}} \, dx=\frac {i e^{-\frac {i (b c+a d)}{d}} \left (-e^{2 i a} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {i b (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {i b (c+d x)}{d}\right )+2 i e^{\frac {i (b c+a d)}{d}} \sin (a+b x)\right )}{d \sqrt {c+d x}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {3042, 3778, 3042, 3787, 3042, 3785, 3786, 3832, 3833}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

\(\displaystyle \frac {2 b \int \frac {\cos (a+b x)}{\sqrt {c+d x}}dx}{d}-\frac {2 \sin (a+b x)}{d \sqrt {c+d x}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b \int \frac {\sin \left (a+b x+\frac {\pi }{2}\right )}{\sqrt {c+d x}}dx}{d}-\frac {2 \sin (a+b x)}{d \sqrt {c+d x}}\)

\(\Big \downarrow \) 3787

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3785

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

\(\Big \downarrow \) 3786

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

\(\Big \downarrow \) 3832

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

\(\Big \downarrow \) 3833

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

Input:

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

Output:

(2*b*((Sqrt[2*Pi]*Cos[a - (b*c)/d]*FresnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d 
*x])/Sqrt[d]])/(Sqrt[b]*Sqrt[d]) - (Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi 
]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(Sqrt[b]*Sqrt[d])))/d - (2*Sin 
[a + b*x])/(d*Sqrt[c + d*x])
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3778
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c 
 + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m + 1))), x] - Simp[f/(d*(m + 1))   Int[( 
c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[m, - 
1]
 

rule 3785
Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S 
imp[2/d   Subst[Int[Cos[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, 
d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
 

rule 3786
Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[2/d 
   Subst[Int[Sin[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f 
}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
 

rule 3787
Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Cos 
[(d*e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] + Simp[Sin[( 
d*e - c*f)/d]   Int[Cos[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c, d 
, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]
 

rule 3832
Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 3833
Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 
Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.01

method result size
derivativedivides \(\frac {-\frac {2 \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{\sqrt {d x +c}}+\frac {2 b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}}{d}\) \(140\)
default \(\frac {-\frac {2 \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{\sqrt {d x +c}}+\frac {2 b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}}{d}\) \(140\)

Input:

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

Output:

2/d*(-1/(d*x+c)^(1/2)*sin(b*(d*x+c)/d+(a*d-b*c)/d)+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)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.05 \[ \int \frac {\sin (a+b x)}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (\sqrt {2} {\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - \sqrt {2} {\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) - \sqrt {d x + c} \sin \left (b x + a\right )\right )}}{d^{2} x + c d} \] Input:

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

Output:

2*(sqrt(2)*(pi*d*x + pi*c)*sqrt(b/(pi*d))*cos(-(b*c - a*d)/d)*fresnel_cos( 
sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) - sqrt(2)*(pi*d*x + pi*c)*sqrt(b/(pi 
*d))*fresnel_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c - a*d)/d) 
 - sqrt(d*x + c)*sin(b*x + a))/(d^2*x + c*d)
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.93 \[ \int \frac {\sin (a+b x)}{(c+d x)^{3/2}} \, dx=-\frac {{\left ({\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )} \sqrt {\frac {{\left (d x + c\right )} b}{d}}}{4 \, \sqrt {d x + c} d} \] Input:

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

Output:

-1/4*(((I - 1)*sqrt(2)*gamma(-1/2, I*(d*x + c)*b/d) - (I + 1)*sqrt(2)*gamm 
a(-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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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