3.50 \(\int \frac{\sin ^2(a+b x)}{(c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=170 \[ \frac{8 \sqrt{\pi } b^{3/2} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{3 d^{5/2}}-\frac{8 \sqrt{\pi } b^{3/2} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{3 d^{5/2}}-\frac{8 b \sin (a+b x) \cos (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{3 d (c+d x)^{3/2}} \]

[Out]

(8*b^(3/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(3*d^(5/2)) -
(8*b^(3/2)*Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(3*d^(5/2))
- (8*b*Cos[a + b*x]*Sin[a + b*x])/(3*d^2*Sqrt[c + d*x]) - (2*Sin[a + b*x]^2)/(3*d*(c + d*x)^(3/2))

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Rubi [A]  time = 0.328234, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3314, 32, 3312, 3306, 3305, 3351, 3304, 3352} \[ \frac{8 \sqrt{\pi } b^{3/2} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{3 d^{5/2}}-\frac{8 \sqrt{\pi } b^{3/2} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{3 d^{5/2}}-\frac{8 b \sin (a+b x) \cos (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{3 d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2/(c + d*x)^(5/2),x]

[Out]

(8*b^(3/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(3*d^(5/2)) -
(8*b^(3/2)*Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(3*d^(5/2))
- (8*b*Cos[a + b*x]*Sin[a + b*x])/(3*d^2*Sqrt[c + d*x]) - (2*Sin[a + b*x]^2)/(3*d*(c + d*x)^(3/2))

Rule 3314

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(b*Si
n[e + f*x])^n)/(d*(m + 1)), x] + (Dist[(b^2*f^2*n*(n - 1))/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin
[e + f*x])^(n - 2), x], x] - Dist[(f^2*n^2)/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x
], x] - Simp[(b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*(b*Sin[e + f*x])^(n - 1))/(d^2*(m + 1)*(m + 2)), x]) /; Fre
eQ[{b, c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3312

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

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)^{5/2}} \, dx &=-\frac{8 b \cos (a+b x) \sin (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{3 d (c+d x)^{3/2}}+\frac{\left (8 b^2\right ) \int \frac{1}{\sqrt{c+d x}} \, dx}{3 d^2}-\frac{\left (16 b^2\right ) \int \frac{\sin ^2(a+b x)}{\sqrt{c+d x}} \, dx}{3 d^2}\\ &=\frac{16 b^2 \sqrt{c+d x}}{3 d^3}-\frac{8 b \cos (a+b x) \sin (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac{\left (16 b^2\right ) \int \left (\frac{1}{2 \sqrt{c+d x}}-\frac{\cos (2 a+2 b x)}{2 \sqrt{c+d x}}\right ) \, dx}{3 d^2}\\ &=-\frac{8 b \cos (a+b x) \sin (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{3 d (c+d x)^{3/2}}+\frac{\left (8 b^2\right ) \int \frac{\cos (2 a+2 b x)}{\sqrt{c+d x}} \, dx}{3 d^2}\\ &=-\frac{8 b \cos (a+b x) \sin (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{3 d (c+d x)^{3/2}}+\frac{\left (8 b^2 \cos \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}{3 d^2}-\frac{\left (8 b^2 \sin \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}{3 d^2}\\ &=-\frac{8 b \cos (a+b x) \sin (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{3 d (c+d x)^{3/2}}+\frac{\left (16 b^2 \cos \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 )}{3 d^3}-\frac{\left (16 b^2 \sin \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 )}{3 d^3}\\ &=\frac{8 b^{3/2} \sqrt{\pi } \cos \left (2 a-\frac{2 b c}{d}\right ) C\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{3 d^{5/2}}-\frac{8 b^{3/2} \sqrt{\pi } S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right ) \sin \left (2 a-\frac{2 b c}{d}\right )}{3 d^{5/2}}-\frac{8 b \cos (a+b x) \sin (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{3 d (c+d x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.39855, size = 158, normalized size = 0.93 \[ \frac{2 \left (4 \sqrt{\pi } b \sqrt{\frac{b}{d}} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-4 \sqrt{\pi } b \sqrt{\frac{b}{d}} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-\frac{\sin (a+b x) (4 b (c+d x) \cos (a+b x)+d \sin (a+b x))}{(c+d x)^{3/2}}\right )}{3 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2/(c + d*x)^(5/2),x]

[Out]

(2*(4*b*Sqrt[b/d]*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] - 4*b*Sqrt[b/d]
*Sqrt[Pi]*FresnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]]*Sin[2*a - (2*b*c)/d] - (Sin[a + b*x]*(4*b*(c + d*x)*C
os[a + b*x] + d*Sin[a + b*x]))/(c + d*x)^(3/2)))/(3*d^2)

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Maple [A]  time = 0.014, size = 189, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( -1/6\, \left ( dx+c \right ) ^{-3/2}+1/6\,{\frac{1}{ \left ( dx+c \right ) ^{3/2}}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }+2/3\,{\frac{b}{d} \left ( -{\frac{1}{\sqrt{dx+c}}\sin \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }+2\,{\frac{b\sqrt{\pi }}{d} \left ( \cos \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 ) -\sin \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 ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^2/(d*x+c)^(5/2),x)

[Out]

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

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Maxima [C]  time = 1.29074, size = 644, normalized size = 3.79 \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)^(5/2),x, algorithm="maxima")

[Out]

1/12*(sqrt(2)*((3*(gamma(-3/2, 2*I*(d*x + c)*b/d) + gamma(-3/2, -2*I*(d*x + c)*b/d))*cos(3/4*pi + 3/2*arctan2(
0, b) + 3/2*arctan2(0, d/sqrt(d^2))) + 3*(gamma(-3/2, 2*I*(d*x + c)*b/d) + gamma(-3/2, -2*I*(d*x + c)*b/d))*co
s(-3/4*pi + 3/2*arctan2(0, b) + 3/2*arctan2(0, d/sqrt(d^2))) + (3*I*gamma(-3/2, 2*I*(d*x + c)*b/d) - 3*I*gamma
(-3/2, -2*I*(d*x + c)*b/d))*sin(3/4*pi + 3/2*arctan2(0, b) + 3/2*arctan2(0, d/sqrt(d^2))) + (-3*I*gamma(-3/2,
2*I*(d*x + c)*b/d) + 3*I*gamma(-3/2, -2*I*(d*x + c)*b/d))*sin(-3/4*pi + 3/2*arctan2(0, b) + 3/2*arctan2(0, d/s
qrt(d^2))))*cos(-2*(b*c - a*d)/d) + ((-3*I*gamma(-3/2, 2*I*(d*x + c)*b/d) + 3*I*gamma(-3/2, -2*I*(d*x + c)*b/d
))*cos(3/4*pi + 3/2*arctan2(0, b) + 3/2*arctan2(0, d/sqrt(d^2))) + (-3*I*gamma(-3/2, 2*I*(d*x + c)*b/d) + 3*I*
gamma(-3/2, -2*I*(d*x + c)*b/d))*cos(-3/4*pi + 3/2*arctan2(0, b) + 3/2*arctan2(0, d/sqrt(d^2))) + 3*(gamma(-3/
2, 2*I*(d*x + c)*b/d) + gamma(-3/2, -2*I*(d*x + c)*b/d))*sin(3/4*pi + 3/2*arctan2(0, b) + 3/2*arctan2(0, d/sqr
t(d^2))) - 3*(gamma(-3/2, 2*I*(d*x + c)*b/d) + gamma(-3/2, -2*I*(d*x + c)*b/d))*sin(-3/4*pi + 3/2*arctan2(0, b
) + 3/2*arctan2(0, d/sqrt(d^2))))*sin(-2*(b*c - a*d)/d))*((d*x + c)*abs(b)/abs(d))^(3/2) - 4)/((d*x + c)^(3/2)
*d)

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Fricas [A]  time = 2.33142, size = 502, normalized size = 2.95 \begin{align*} \frac{2 \,{\left (4 \,{\left (\pi b d^{2} x^{2} + 2 \, \pi b c d x + \pi b c^{2}\right )} \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{C}\left (2 \, \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - 4 \,{\left (\pi b d^{2} x^{2} + 2 \, \pi b c d x + \pi b c^{2}\right )} \sqrt{\frac{b}{\pi d}} \operatorname{S}\left (2 \, \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) +{\left (d \cos \left (b x + a\right )^{2} - 4 \,{\left (b d x + b c\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - d\right )} \sqrt{d x + c}\right )}}{3 \,{\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^2/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

2/3*(4*(pi*b*d^2*x^2 + 2*pi*b*c*d*x + pi*b*c^2)*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x +
c)*sqrt(b/(pi*d))) - 4*(pi*b*d^2*x^2 + 2*pi*b*c*d*x + pi*b*c^2)*sqrt(b/(pi*d))*fresnel_sin(2*sqrt(d*x + c)*sqr
t(b/(pi*d)))*sin(-2*(b*c - a*d)/d) + (d*cos(b*x + a)^2 - 4*(b*d*x + b*c)*cos(b*x + a)*sin(b*x + a) - d)*sqrt(d
*x + c))/(d^4*x^2 + 2*c*d^3*x + c^2*d^2)

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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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**2/(d*x+c)**(5/2),x)

[Out]

Integral(sin(a + b*x)**2/(c + d*x)**(5/2), 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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^2/(d*x + c)^(5/2), x)