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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3394, 12, 3387, 3386, 3432, 3385, 3433} \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{3/2}} \, dx=\frac {2 \sqrt {\pi } \sqrt {b} \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{d^{3/2}}+\frac {2 \sqrt {\pi } \sqrt {b} \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{d^{3/2}}-\frac {2 \sin ^2(a+b x)}{d \sqrt {c+d x}} \]

[In]

Int[Sin[a + b*x]^2/(c + d*x)^(3/2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3385

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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 3386

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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 3387

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) +
f*x]/Sqrt[c + d*x], x], x] + Dist[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 3394

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]^
n/(d*(m + 1))), x] - Dist[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 3432

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 3433

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sin ^2(a+b x)}{d \sqrt {c+d x}}+\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}}+\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}}+\frac {\left (2 b \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}} \, dx}{d}+\frac {\left (2 b \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}} \, dx}{d} \\ & = -\frac {2 \sin ^2(a+b x)}{d \sqrt {c+d x}}+\frac {\left (4 b \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \text {Subst}\left (\int \sin \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{d^2}+\frac {\left (4 b \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \text {Subst}\left (\int \cos \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{d^2} \\ & = \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}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.51 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.30 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{3/2}} \, dx=\frac {e^{-\frac {2 i (a d+b (c+d x))}{d}} \left (-\sqrt {2} e^{2 i (2 a+b x)} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {2 i b (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \left (\left (-1+e^{2 i (a+b x)}\right )^2-\sqrt {2} e^{\frac {2 i b (c+d x)}{d}} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {2 i b (c+d x)}{d}\right )\right )\right )}{2 d \sqrt {c+d x}} \]

[In]

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

[Out]

(-(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])

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.07

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

[In]

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

[Out]

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)

none

Time = 0.30 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.02 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left ({\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + {\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \sqrt {d x + c} {\left (\cos \left (b x + a\right )^{2} - 1\right )}\right )}}{d^{2} x + c d} \]

[In]

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

[Out]

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*sqrt(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]

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

[In]

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

[Out]

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 = 0.38 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{3/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \sqrt {\frac {{\left (d x + c\right )} b}{d}} + 8}{8 \, \sqrt {d x + c} d} \]

[In]

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

[Out]

-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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(sin(a + b*x)^2/(c + d*x)^(3/2),x)

[Out]

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