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

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

[Out]

-16/105*b^2/d^3/(d*x+c)^(3/2)-8/35*b*cos(b*x+a)*sin(b*x+a)/d^2/(d*x+c)^(5/2)-2/7*sin(b*x+a)^2/d/(d*x+c)^(7/2)+
32/105*b^2*sin(b*x+a)^2/d^3/(d*x+c)^(3/2)-128/105*b^(7/2)*cos(2*a-2*b*c/d)*FresnelC(2*b^(1/2)*(d*x+c)^(1/2)/d^
(1/2)/Pi^(1/2))*Pi^(1/2)/d^(9/2)+128/105*b^(7/2)*FresnelS(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^(1/2))*sin(2*a-2*
b*c/d)*Pi^(1/2)/d^(9/2)+128/105*b^3*cos(b*x+a)*sin(b*x+a)/d^4/(d*x+c)^(1/2)

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Rubi [A]  time = 0.42, antiderivative size = 247, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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 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 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*}

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Mathematica [B]  time = 4.84, 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} C\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 (15 d^3 \sin \left (\frac {2 b (c+d x)}{d}\right )+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 )\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 (15 d^3 \sin \left (\frac {2 b (c+d x)}{d}\right )+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 )\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} C\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))

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fricas [B]  time = 0.85, size = 422, normalized size = 1.71 \[ -\frac {2 \, {\left (64 \, {\left (\pi b^{3} d^{4} x^{4} + 4 \, \pi b^{3} c d^{3} x^{3} + 6 \, \pi b^{3} c^{2} d^{2} x^{2} + 4 \, \pi b^{3} c^{3} d x + \pi b^{3} c^{4}\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 64 \, {\left (\pi b^{3} d^{4} x^{4} + 4 \, \pi b^{3} c d^{3} x^{3} + 6 \, \pi b^{3} c^{2} d^{2} x^{2} + 4 \, \pi b^{3} c^{3} d x + \pi b^{3} c^{4}\right )} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - {\left (8 \, b^{2} d^{3} x^{2} + 16 \, b^{2} c d^{2} x + 8 \, b^{2} c^{2} d - 15 \, d^{3} - {\left (16 \, b^{2} d^{3} x^{2} + 32 \, b^{2} c d^{2} x + 16 \, b^{2} c^{2} d - 15 \, d^{3}\right )} \cos \left (b x + a\right )^{2} + 4 \, {\left (16 \, b^{3} d^{3} x^{3} + 48 \, b^{3} c d^{2} x^{2} + 16 \, b^{3} c^{3} - 3 \, b c d^{2} + 3 \, {\left (16 \, b^{3} c^{2} d - b d^{3}\right )} x\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {d x + c}\right )}}{105 \, {\left (d^{8} x^{4} + 4 \, c d^{7} x^{3} + 6 \, c^{2} d^{6} x^{2} + 4 \, c^{3} d^{5} x + c^{4} d^{4}\right )}} \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac {9}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.03, size = 273, normalized size = 1.11 \[ \frac {-\frac {1}{7 \left (d x +c \right )^{\frac {7}{2}}}+\frac {\cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{7 \left (d x +c \right )^{\frac {7}{2}}}+\frac {4 b \left (-\frac {\sin \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{5 \left (d x +c \right )^{\frac {5}{2}}}+\frac {4 b \left (-\frac {\cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}-\frac {4 b \left (-\frac {\sin \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{\sqrt {d x +c}}+\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}\right )}{5 d}\right )}{7 d}}{d} \]

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))))))

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maxima [C]  time = 1.05, size = 135, normalized size = 0.55 \[ -\frac {\sqrt {2} {\left ({\left (\left (7 i - 7\right ) \, \sqrt {2} \Gamma \left (-\frac {7}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) - \left (7 i + 7\right ) \, \sqrt {2} \Gamma \left (-\frac {7}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + {\left (\left (7 i + 7\right ) \, \sqrt {2} \Gamma \left (-\frac {7}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) - \left (7 i - 7\right ) \, \sqrt {2} \Gamma \left (-\frac {7}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {7}{2}} + 1}{7 \, {\left (d x + c\right )}^{\frac {7}{2}} d} \]

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*I - 7)*sqrt(2)*gamma(-7/2, 2*I*(d*x + c)*b/d) - (7*I + 7)*sqrt(2)*gamma(-7/2, -2*I*(d*x + c
)*b/d))*cos(-2*(b*c - a*d)/d) + ((7*I + 7)*sqrt(2)*gamma(-7/2, 2*I*(d*x + c)*b/d) - (7*I - 7)*sqrt(2)*gamma(-7
/2, -2*I*(d*x + c)*b/d))*sin(-2*(b*c - a*d)/d))*((d*x + c)*b/d)^(7/2) + 1)/((d*x + c)^(7/2)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^2/(c + d*x)^(9/2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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