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

Optimal. Leaf size=135 \[ \frac{2 \sqrt{\pi } \sqrt{b} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{d^{3/2}}+\frac{2 \sqrt{\pi } \sqrt{b} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{d^{3/2}}-\frac{2 \sin ^2(a+b x)}{d \sqrt{c+d x}} \]

[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])

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Rubi [A]  time = 0.253951, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {3313, 12, 3306, 3305, 3351, 3304, 3352} \[ \frac{2 \sqrt{\pi } \sqrt{b} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{d^{3/2}}+\frac{2 \sqrt{\pi } \sqrt{b} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{d^{3/2}}-\frac{2 \sin ^2(a+b x)}{d \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[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 3313

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 12

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

Rule 3306

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 3305

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 3351

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

Rule 3304

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 3352

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

Rubi steps

\begin{align*} \int \frac{\sin ^2(a+b x)}{(c+d x)^{3/2}} \, dx &=-\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 ) \operatorname{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 ) \operatorname{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 ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{d^{3/2}}+\frac{2 \sqrt{b} \sqrt{\pi } C\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 [A]  time = 0.383539, size = 149, normalized size = 1.1 \[ \frac{2 \sqrt{\pi } \sqrt{\frac{b}{d}} \sqrt{c+d x} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+2 \sqrt{\pi } \sqrt{\frac{b}{d}} \sqrt{c+d x} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+\cos (2 (a+b x))-1}{d \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1 + Cos[2*(a + b*x)] + 2*Sqrt[b/d]*Sqrt[Pi]*Sqrt[c + d*x]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b/d]*Sqrt[c
+ d*x])/Sqrt[Pi]] + 2*Sqrt[b/d]*Sqrt[Pi]*Sqrt[c + d*x]*FresnelC[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]]*Sin[2*a
- (2*b*c)/d])/(d*Sqrt[c + d*x])

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Maple [A]  time = 0.013, size = 145, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( -1/2\,{\frac{1}{\sqrt{dx+c}}}+1/2\,{\frac{1}{\sqrt{dx+c}}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }+{\frac{b\sqrt{\pi }}{d} \left ( \cos \left ( 2\,{\frac{da-cb}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) +\sin \left ( 2\,{\frac{da-cb}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.3232, size = 640, normalized size = 4.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(2)*(((gamma(-1/2, 2*I*(d*x + c)*b/d) + gamma(-1/2, -2*I*(d*x + c)*b/d))*cos(1/4*pi + 1/2*arctan2(0,
b) + 1/2*arctan2(0, d/sqrt(d^2))) + (gamma(-1/2, 2*I*(d*x + c)*b/d) + gamma(-1/2, -2*I*(d*x + c)*b/d))*cos(-1/
4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + (I*gamma(-1/2, 2*I*(d*x + c)*b/d) - I*gamma(-1/2, -2
*I*(d*x + c)*b/d))*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + (-I*gamma(-1/2, 2*I*(d*x +
c)*b/d) + I*gamma(-1/2, -2*I*(d*x + c)*b/d))*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*c
os(-2*(b*c - a*d)/d) + ((-I*gamma(-1/2, 2*I*(d*x + c)*b/d) + I*gamma(-1/2, -2*I*(d*x + c)*b/d))*cos(1/4*pi + 1
/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + (-I*gamma(-1/2, 2*I*(d*x + c)*b/d) + I*gamma(-1/2, -2*I*(d*x
+ c)*b/d))*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + (gamma(-1/2, 2*I*(d*x + c)*b/d) +
gamma(-1/2, -2*I*(d*x + c)*b/d))*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - (gamma(-1/2,
2*I*(d*x + c)*b/d) + gamma(-1/2, -2*I*(d*x + c)*b/d))*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt
(d^2))))*sin(-2*(b*c - a*d)/d))*sqrt((d*x + c)*abs(b)/abs(d)) - 8)/(sqrt(d*x + c)*d)

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Fricas [A]  time = 2.12082, size = 340, normalized size = 2.52 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)