\(\int \frac {\sin ^2(a+b x)}{(c+d x)^{3/2}} \, dx\) [49]

Optimal result

Integrand size = 18, antiderivative size = 135 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{3/2}} \, dx=\frac {2 \sqrt {b} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{d^{3/2}}+\frac {2 \sqrt {b} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{d^{3/2}}-\frac {2 \sin ^2(a+b x)}{d \sqrt {c+d x}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.78 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.30 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{3/2}} \, dx=\frac {e^{-\frac {2 i (a d+b (c+d x))}{d}} \left (-\sqrt {2} e^{2 i (2 a+b x)} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {2 i b (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \left (\left (-1+e^{2 i (a+b x)}\right )^2-\sqrt {2} e^{\frac {2 i b (c+d x)}{d}} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {2 i b (c+d x)}{d}\right )\right )\right )}{2 d \sqrt {c+d x}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 3794, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3794

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3787

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3785

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

\(\Big \downarrow \) 3786

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

\(\Big \downarrow \) 3832

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

\(\Big \downarrow \) 3833

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

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 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]
 

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 = 1.19 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.07

method	result	size
derivativedivides	\(\frac {-\frac {1}{\sqrt {d x +c}}+\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{\sqrt {d x +c}}+\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}}{d}\)	\(145\)
default	\(\frac {-\frac {1}{\sqrt {d x +c}}+\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{\sqrt {d x +c}}+\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}}{d}\)	\(145\)

Input:

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

Output:

2/d*(-1/2/(d*x+c)^(1/2)+1/2/(d*x+c)^(1/2)*cos(2*b*(d*x+c)/d+2*(a*d-b*c)/d) 
+b/d*Pi^(1/2)/(b/d)^(1/2)*(cos(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1 
/2)*b*(d*x+c)^(1/2)/d)+sin(2*(a*d-b*c)/d)*FresnelC(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 = 138, normalized size of antiderivative = 1.02 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left ({\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + {\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \sqrt {d x + c} {\left (\cos \left (b x + a\right )^{2} - 1\right )}\right )}}{d^{2} x + c d} \] Input:

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

Output:

2*((pi*d*x + pi*c)*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_sin(2*sqrt 
(d*x + c)*sqrt(b/(pi*d))) + (pi*d*x + pi*c)*sqrt(b/(pi*d))*fresnel_cos(2*s 
qrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) + sqrt(d*x + c)*(cos(b* 
x + a)^2 - 1))/(d^2*x + c*d)
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{3/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \sqrt {\frac {{\left (d x + c\right )} b}{d}} + 8}{8 \, \sqrt {d x + c} d} \] Input:

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

Output:

-1/8*(sqrt(2)*((-(I + 1)*sqrt(2)*gamma(-1/2, 2*I*(d*x + c)*b/d) + (I - 1)* 
sqrt(2)*gamma(-1/2, -2*I*(d*x + c)*b/d))*cos(-2*(b*c - a*d)/d) + ((I - 1)* 
sqrt(2)*gamma(-1/2, 2*I*(d*x + c)*b/d) - (I + 1)*sqrt(2)*gamma(-1/2, -2*I* 
(d*x + c)*b/d))*sin(-2*(b*c - a*d)/d))*sqrt((d*x + c)*b/d) + 8)/(sqrt(d*x 
+ c)*d)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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