3.52 \(\int \frac{\sin ^2(a+b x)}{(c+d x)^{9/2}} \, dx\)

Optimal. Leaf size=247 \[ -\frac{128 \sqrt{\pi } b^{7/2} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{105 d^{9/2}}+\frac{128 \sqrt{\pi } b^{7/2} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{105 d^{9/2}}+\frac{32 b^2 \sin ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}+\frac{128 b^3 \sin (a+b x) \cos (a+b x)}{105 d^4 \sqrt{c+d x}}-\frac{8 b \sin (a+b x) \cos (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac{2 \sin ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac{16 b^2}{105 d^3 (c+d x)^{3/2}} \]

[Out]

(-16*b^2)/(105*d^3*(c + d*x)^(3/2)) - (128*b^(7/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c +
d*x])/(Sqrt[d]*Sqrt[Pi])])/(105*d^(9/2)) + (128*b^(7/2)*Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*S
qrt[Pi])]*Sin[2*a - (2*b*c)/d])/(105*d^(9/2)) - (8*b*Cos[a + b*x]*Sin[a + b*x])/(35*d^2*(c + d*x)^(5/2)) + (12
8*b^3*Cos[a + b*x]*Sin[a + b*x])/(105*d^4*Sqrt[c + d*x]) - (2*Sin[a + b*x]^2)/(7*d*(c + d*x)^(7/2)) + (32*b^2*
Sin[a + b*x]^2)/(105*d^3*(c + d*x)^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.419004, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 11, 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} \[ -\frac{128 \sqrt{\pi } b^{7/2} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{105 d^{9/2}}+\frac{128 \sqrt{\pi } b^{7/2} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{105 d^{9/2}}+\frac{32 b^2 \sin ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}+\frac{128 b^3 \sin (a+b x) \cos (a+b x)}{105 d^4 \sqrt{c+d x}}-\frac{8 b \sin (a+b x) \cos (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac{2 \sin ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac{16 b^2}{105 d^3 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2/(c + d*x)^(9/2),x]

[Out]

(-16*b^2)/(105*d^3*(c + d*x)^(3/2)) - (128*b^(7/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c +
d*x])/(Sqrt[d]*Sqrt[Pi])])/(105*d^(9/2)) + (128*b^(7/2)*Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*S
qrt[Pi])]*Sin[2*a - (2*b*c)/d])/(105*d^(9/2)) - (8*b*Cos[a + b*x]*Sin[a + b*x])/(35*d^2*(c + d*x)^(5/2)) + (12
8*b^3*Cos[a + b*x]*Sin[a + b*x])/(105*d^4*Sqrt[c + d*x]) - (2*Sin[a + b*x]^2)/(7*d*(c + d*x)^(7/2)) + (32*b^2*
Sin[a + b*x]^2)/(105*d^3*(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)^{9/2}} \, dx &=-\frac{8 b \cos (a+b x) \sin (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac{2 \sin ^2(a+b x)}{7 d (c+d x)^{7/2}}+\frac{\left (8 b^2\right ) \int \frac{1}{(c+d x)^{5/2}} \, dx}{35 d^2}-\frac{\left (16 b^2\right ) \int \frac{\sin ^2(a+b x)}{(c+d x)^{5/2}} \, dx}{35 d^2}\\ &=-\frac{16 b^2}{105 d^3 (c+d x)^{3/2}}-\frac{8 b \cos (a+b x) \sin (a+b x)}{35 d^2 (c+d x)^{5/2}}+\frac{128 b^3 \cos (a+b x) \sin (a+b x)}{105 d^4 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{7 d (c+d x)^{7/2}}+\frac{32 b^2 \sin ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}-\frac{\left (128 b^4\right ) \int \frac{1}{\sqrt{c+d x}} \, dx}{105 d^4}+\frac{\left (256 b^4\right ) \int \frac{\sin ^2(a+b x)}{\sqrt{c+d x}} \, dx}{105 d^4}\\ &=-\frac{16 b^2}{105 d^3 (c+d x)^{3/2}}-\frac{256 b^4 \sqrt{c+d x}}{105 d^5}-\frac{8 b \cos (a+b x) \sin (a+b x)}{35 d^2 (c+d x)^{5/2}}+\frac{128 b^3 \cos (a+b x) \sin (a+b x)}{105 d^4 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{7 d (c+d x)^{7/2}}+\frac{32 b^2 \sin ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}+\frac{\left (256 b^4\right ) \int \left (\frac{1}{2 \sqrt{c+d x}}-\frac{\cos (2 a+2 b x)}{2 \sqrt{c+d x}}\right ) \, dx}{105 d^4}\\ &=-\frac{16 b^2}{105 d^3 (c+d x)^{3/2}}-\frac{8 b \cos (a+b x) \sin (a+b x)}{35 d^2 (c+d x)^{5/2}}+\frac{128 b^3 \cos (a+b x) \sin (a+b x)}{105 d^4 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{7 d (c+d x)^{7/2}}+\frac{32 b^2 \sin ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}-\frac{\left (128 b^4\right ) \int \frac{\cos (2 a+2 b x)}{\sqrt{c+d x}} \, dx}{105 d^4}\\ &=-\frac{16 b^2}{105 d^3 (c+d x)^{3/2}}-\frac{8 b \cos (a+b x) \sin (a+b x)}{35 d^2 (c+d x)^{5/2}}+\frac{128 b^3 \cos (a+b x) \sin (a+b x)}{105 d^4 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{7 d (c+d x)^{7/2}}+\frac{32 b^2 \sin ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}-\frac{\left (128 b^4 \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}{105 d^4}+\frac{\left (128 b^4 \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}{105 d^4}\\ &=-\frac{16 b^2}{105 d^3 (c+d x)^{3/2}}-\frac{8 b \cos (a+b x) \sin (a+b x)}{35 d^2 (c+d x)^{5/2}}+\frac{128 b^3 \cos (a+b x) \sin (a+b x)}{105 d^4 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{7 d (c+d x)^{7/2}}+\frac{32 b^2 \sin ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}-\frac{\left (256 b^4 \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 )}{105 d^5}+\frac{\left (256 b^4 \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 )}{105 d^5}\\ &=-\frac{16 b^2}{105 d^3 (c+d x)^{3/2}}-\frac{128 b^{7/2} \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 )}{105 d^{9/2}}+\frac{128 b^{7/2} \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 )}{105 d^{9/2}}-\frac{8 b \cos (a+b x) \sin (a+b x)}{35 d^2 (c+d x)^{5/2}}+\frac{128 b^3 \cos (a+b x) \sin (a+b x)}{105 d^4 \sqrt{c+d x}}-\frac{2 \sin ^2(a+b x)}{7 d (c+d x)^{7/2}}+\frac{32 b^2 \sin ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}\\ \end{align*}

Mathematica [B]  time = 4.63769, size = 661, normalized size = 2.68 \[ \frac{\cos (2 a) \left (2 \cos \left (\frac{2 b c}{d}\right ) \left (15 d^3 \cos \left (\frac{2 b (c+d x)}{d}\right )-4 b (c+d x) \left (3 d^2 \sin \left (\frac{2 b (c+d x)}{d}\right )+4 b (c+d x) \left (8 \sqrt{\pi } b \sqrt{\frac{b}{d}} (c+d x)^{3/2} \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-4 b (c+d x) \sin \left (\frac{2 b (c+d x)}{d}\right )+d \cos \left (\frac{2 b (c+d x)}{d}\right )\right )\right )\right )+4 \sin \left (\frac{b c}{d}\right ) \cos \left (\frac{b c}{d}\right ) \left (4 b (c+d x) \left (3 d^2 \cos \left (\frac{2 b (c+d x)}{d}\right )-4 b (c+d x) \left (8 \sqrt{\pi } b \sqrt{\frac{b}{d}} (c+d x)^{3/2} S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+d \sin \left (\frac{2 b (c+d x)}{d}\right )+4 b (c+d x) \cos \left (\frac{2 b (c+d x)}{d}\right )\right )\right )+15 d^3 \sin \left (\frac{2 b (c+d x)}{d}\right )\right )\right )-2 \sin (a) \cos (a) \left (2 \left (\cos \left (\frac{b c}{d}\right )-\sin \left (\frac{b c}{d}\right )\right ) \left (\sin \left (\frac{b c}{d}\right )+\cos \left (\frac{b c}{d}\right )\right ) \left (4 b (c+d x) \left (3 d^2 \cos \left (\frac{2 b (c+d x)}{d}\right )-4 b (c+d x) \left (8 \sqrt{\pi } b \sqrt{\frac{b}{d}} (c+d x)^{3/2} S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+d \sin \left (\frac{2 b (c+d x)}{d}\right )+4 b (c+d x) \cos \left (\frac{2 b (c+d x)}{d}\right )\right )\right )+15 d^3 \sin \left (\frac{2 b (c+d x)}{d}\right )\right )-2 \sin \left (\frac{2 b c}{d}\right ) \left (15 d^3 \cos \left (\frac{2 b (c+d x)}{d}\right )-4 b (c+d x) \left (3 d^2 \sin \left (\frac{2 b (c+d x)}{d}\right )+4 b (c+d x) \left (8 \sqrt{\pi } b \sqrt{\frac{b}{d}} (c+d x)^{3/2} \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-4 b (c+d x) \sin \left (\frac{2 b (c+d x)}{d}\right )+d \cos \left (\frac{2 b (c+d x)}{d}\right )\right )\right )\right )\right )-30 d^3}{210 d^4 (c+d x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2/(c + d*x)^(9/2),x]

[Out]

(-30*d^3 + Cos[2*a]*(4*Cos[(b*c)/d]*Sin[(b*c)/d]*(15*d^3*Sin[(2*b*(c + d*x))/d] + 4*b*(c + d*x)*(3*d^2*Cos[(2*
b*(c + d*x))/d] - 4*b*(c + d*x)*(4*b*(c + d*x)*Cos[(2*b*(c + d*x))/d] + 8*b*Sqrt[b/d]*Sqrt[Pi]*(c + d*x)^(3/2)
*FresnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] + d*Sin[(2*b*(c + d*x))/d]))) + 2*Cos[(2*b*c)/d]*(15*d^3*Cos[(
2*b*(c + d*x))/d] - 4*b*(c + d*x)*(3*d^2*Sin[(2*b*(c + d*x))/d] + 4*b*(c + d*x)*(d*Cos[(2*b*(c + d*x))/d] + 8*
b*Sqrt[b/d]*Sqrt[Pi]*(c + d*x)^(3/2)*FresnelC[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] - 4*b*(c + d*x)*Sin[(2*b*(
c + d*x))/d])))) - 2*Cos[a]*Sin[a]*(2*(Cos[(b*c)/d] - Sin[(b*c)/d])*(Cos[(b*c)/d] + Sin[(b*c)/d])*(15*d^3*Sin[
(2*b*(c + d*x))/d] + 4*b*(c + d*x)*(3*d^2*Cos[(2*b*(c + d*x))/d] - 4*b*(c + d*x)*(4*b*(c + d*x)*Cos[(2*b*(c +
d*x))/d] + 8*b*Sqrt[b/d]*Sqrt[Pi]*(c + d*x)^(3/2)*FresnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] + d*Sin[(2*b*
(c + d*x))/d]))) - 2*Sin[(2*b*c)/d]*(15*d^3*Cos[(2*b*(c + d*x))/d] - 4*b*(c + d*x)*(3*d^2*Sin[(2*b*(c + d*x))/
d] + 4*b*(c + d*x)*(d*Cos[(2*b*(c + d*x))/d] + 8*b*Sqrt[b/d]*Sqrt[Pi]*(c + d*x)^(3/2)*FresnelC[(2*Sqrt[b/d]*Sq
rt[c + d*x])/Sqrt[Pi]] - 4*b*(c + d*x)*Sin[(2*b*(c + d*x))/d])))))/(210*d^4*(c + d*x)^(7/2))

________________________________________________________________________________________

Maple [A]  time = 0.015, size = 273, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( -1/14\, \left ( dx+c \right ) ^{-7/2}+1/14\,{\frac{1}{ \left ( dx+c \right ) ^{7/2}}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }+2/7\,{\frac{b}{d} \left ( -1/5\,{\frac{1}{ \left ( dx+c \right ) ^{5/2}}\sin \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }+4/5\,{\frac{b}{d} \left ( -1/3\,{\frac{1}{ \left ( dx+c \right ) ^{3/2}}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }-4/3\,{\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 ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^2/(d*x+c)^(9/2),x)

[Out]

2/d*(-1/14/(d*x+c)^(7/2)+1/14/(d*x+c)^(7/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+2/7*b/d*(-1/5/(d*x+c)^(5/2)*sin(2
/d*(d*x+c)*b+2*(a*d-b*c)/d)+4/5*b/d*(-1/3/(d*x+c)^(3/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)-4/3*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))))))

________________________________________________________________________________________

Maxima [C]  time = 1.29426, size = 644, normalized size = 2.61 \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)^(9/2),x, algorithm="maxima")

[Out]

1/7*(sqrt(2)*((7*(gamma(-7/2, 2*I*(d*x + c)*b/d) + gamma(-7/2, -2*I*(d*x + c)*b/d))*cos(7/4*pi + 7/2*arctan2(0
, b) + 7/2*arctan2(0, d/sqrt(d^2))) + 7*(gamma(-7/2, 2*I*(d*x + c)*b/d) + gamma(-7/2, -2*I*(d*x + c)*b/d))*cos
(-7/4*pi + 7/2*arctan2(0, b) + 7/2*arctan2(0, d/sqrt(d^2))) + (7*I*gamma(-7/2, 2*I*(d*x + c)*b/d) - 7*I*gamma(
-7/2, -2*I*(d*x + c)*b/d))*sin(7/4*pi + 7/2*arctan2(0, b) + 7/2*arctan2(0, d/sqrt(d^2))) + (-7*I*gamma(-7/2, 2
*I*(d*x + c)*b/d) + 7*I*gamma(-7/2, -2*I*(d*x + c)*b/d))*sin(-7/4*pi + 7/2*arctan2(0, b) + 7/2*arctan2(0, d/sq
rt(d^2))))*cos(-2*(b*c - a*d)/d) + ((-7*I*gamma(-7/2, 2*I*(d*x + c)*b/d) + 7*I*gamma(-7/2, -2*I*(d*x + c)*b/d)
)*cos(7/4*pi + 7/2*arctan2(0, b) + 7/2*arctan2(0, d/sqrt(d^2))) + (-7*I*gamma(-7/2, 2*I*(d*x + c)*b/d) + 7*I*g
amma(-7/2, -2*I*(d*x + c)*b/d))*cos(-7/4*pi + 7/2*arctan2(0, b) + 7/2*arctan2(0, d/sqrt(d^2))) + 7*(gamma(-7/2
, 2*I*(d*x + c)*b/d) + gamma(-7/2, -2*I*(d*x + c)*b/d))*sin(7/4*pi + 7/2*arctan2(0, b) + 7/2*arctan2(0, d/sqrt
(d^2))) - 7*(gamma(-7/2, 2*I*(d*x + c)*b/d) + gamma(-7/2, -2*I*(d*x + c)*b/d))*sin(-7/4*pi + 7/2*arctan2(0, b)
+ 7/2*arctan2(0, d/sqrt(d^2))))*sin(-2*(b*c - a*d)/d))*((d*x + c)*abs(b)/abs(d))^(7/2) - 1)/((d*x + c)^(7/2)*
d)

________________________________________________________________________________________

Fricas [B]  time = 2.89485, size = 941, normalized size = 3.81 \begin{align*} -\frac{2 \,{\left (64 \,{\left (\pi b^{3} d^{4} x^{4} + 4 \, \pi b^{3} c d^{3} x^{3} + 6 \, \pi b^{3} c^{2} d^{2} x^{2} + 4 \, \pi b^{3} c^{3} d x + \pi b^{3} c^{4}\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 ) - 64 \,{\left (\pi b^{3} d^{4} x^{4} + 4 \, \pi b^{3} c d^{3} x^{3} + 6 \, \pi b^{3} c^{2} d^{2} x^{2} + 4 \, \pi b^{3} c^{3} d x + \pi b^{3} c^{4}\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 (8 \, b^{2} d^{3} x^{2} + 16 \, b^{2} c d^{2} x + 8 \, b^{2} c^{2} d - 15 \, d^{3} -{\left (16 \, b^{2} d^{3} x^{2} + 32 \, b^{2} c d^{2} x + 16 \, b^{2} c^{2} d - 15 \, d^{3}\right )} \cos \left (b x + a\right )^{2} + 4 \,{\left (16 \, b^{3} d^{3} x^{3} + 48 \, b^{3} c d^{2} x^{2} + 16 \, b^{3} c^{3} - 3 \, b c d^{2} + 3 \,{\left (16 \, b^{3} c^{2} d - b d^{3}\right )} x\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt{d x + c}\right )}}{105 \,{\left (d^{8} x^{4} + 4 \, c d^{7} x^{3} + 6 \, c^{2} d^{6} x^{2} + 4 \, c^{3} d^{5} x + c^{4} d^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^2/(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

-2/105*(64*(pi*b^3*d^4*x^4 + 4*pi*b^3*c*d^3*x^3 + 6*pi*b^3*c^2*d^2*x^2 + 4*pi*b^3*c^3*d*x + pi*b^3*c^4)*sqrt(b
/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d))) - 64*(pi*b^3*d^4*x^4 + 4*pi*b^3*c*d
^3*x^3 + 6*pi*b^3*c^2*d^2*x^2 + 4*pi*b^3*c^3*d*x + pi*b^3*c^4)*sqrt(b/(pi*d))*fresnel_sin(2*sqrt(d*x + c)*sqrt
(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - (8*b^2*d^3*x^2 + 16*b^2*c*d^2*x + 8*b^2*c^2*d - 15*d^3 - (16*b^2*d^3*x^2 +
32*b^2*c*d^2*x + 16*b^2*c^2*d - 15*d^3)*cos(b*x + a)^2 + 4*(16*b^3*d^3*x^3 + 48*b^3*c*d^2*x^2 + 16*b^3*c^3 -
3*b*c*d^2 + 3*(16*b^3*c^2*d - b*d^3)*x)*cos(b*x + a)*sin(b*x + a))*sqrt(d*x + c))/(d^8*x^4 + 4*c*d^7*x^3 + 6*c
^2*d^6*x^2 + 4*c^3*d^5*x + c^4*d^4)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**2/(d*x+c)**(9/2),x)

[Out]

Timed out

________________________________________________________________________________________

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{9}{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)^(9/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^2/(d*x + c)^(9/2), x)