3.59 \(\int \frac{\sin ^3(a+b x)}{(c+d x)^{7/2}} \, dx\)

Optimal. Leaf size=356 \[ -\frac{2 \sqrt{2 \pi } b^{5/2} \cos \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{5 d^{7/2}}+\frac{6 \sqrt{6 \pi } b^{5/2} \cos \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{6 \sqrt{6 \pi } b^{5/2} \sin \left (3 a-\frac{3 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{5 d^{7/2}}+\frac{2 \sqrt{2 \pi } b^{5/2} \sin \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{5 d^{7/2}}+\frac{24 b^2 \sin ^3(a+b x)}{5 d^3 \sqrt{c+d x}}-\frac{16 b^2 \sin (a+b x)}{5 d^3 \sqrt{c+d x}}-\frac{4 b \sin ^2(a+b x) \cos (a+b x)}{5 d^2 (c+d x)^{3/2}}-\frac{2 \sin ^3(a+b x)}{5 d (c+d x)^{5/2}} \]

[Out]

(-2*b^(5/2)*Sqrt[2*Pi]*Cos[a - (b*c)/d]*FresnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(5*d^(7/2)) + (6
*b^(5/2)*Sqrt[6*Pi]*Cos[3*a - (3*b*c)/d]*FresnelC[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(5*d^(7/2)) - (
6*b^(5/2)*Sqrt[6*Pi]*FresnelS[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[3*a - (3*b*c)/d])/(5*d^(7/2)) +
(2*b^(5/2)*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(5*d^(7/2)) - (16
*b^2*Sin[a + b*x])/(5*d^3*Sqrt[c + d*x]) - (4*b*Cos[a + b*x]*Sin[a + b*x]^2)/(5*d^2*(c + d*x)^(3/2)) - (2*Sin[
a + b*x]^3)/(5*d*(c + d*x)^(5/2)) + (24*b^2*Sin[a + b*x]^3)/(5*d^3*Sqrt[c + d*x])

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Rubi [A]  time = 0.796731, antiderivative size = 356, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3314, 3297, 3306, 3305, 3351, 3304, 3352, 3313} \[ -\frac{2 \sqrt{2 \pi } b^{5/2} \cos \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{5 d^{7/2}}+\frac{6 \sqrt{6 \pi } b^{5/2} \cos \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{6 \sqrt{6 \pi } b^{5/2} \sin \left (3 a-\frac{3 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{5 d^{7/2}}+\frac{2 \sqrt{2 \pi } b^{5/2} \sin \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{5 d^{7/2}}+\frac{24 b^2 \sin ^3(a+b x)}{5 d^3 \sqrt{c+d x}}-\frac{16 b^2 \sin (a+b x)}{5 d^3 \sqrt{c+d x}}-\frac{4 b \sin ^2(a+b x) \cos (a+b x)}{5 d^2 (c+d x)^{3/2}}-\frac{2 \sin ^3(a+b x)}{5 d (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3/(c + d*x)^(7/2),x]

[Out]

(-2*b^(5/2)*Sqrt[2*Pi]*Cos[a - (b*c)/d]*FresnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(5*d^(7/2)) + (6
*b^(5/2)*Sqrt[6*Pi]*Cos[3*a - (3*b*c)/d]*FresnelC[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(5*d^(7/2)) - (
6*b^(5/2)*Sqrt[6*Pi]*FresnelS[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[3*a - (3*b*c)/d])/(5*d^(7/2)) +
(2*b^(5/2)*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(5*d^(7/2)) - (16
*b^2*Sin[a + b*x])/(5*d^3*Sqrt[c + d*x]) - (4*b*Cos[a + b*x]*Sin[a + b*x]^2)/(5*d^2*(c + d*x)^(3/2)) - (2*Sin[
a + b*x]^3)/(5*d*(c + d*x)^(5/2)) + (24*b^2*Sin[a + b*x]^3)/(5*d^3*Sqrt[c + d*x])

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 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && 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]

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]

Rubi steps

\begin{align*} \int \frac{\sin ^3(a+b x)}{(c+d x)^{7/2}} \, dx &=-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{5 d^2 (c+d x)^{3/2}}-\frac{2 \sin ^3(a+b x)}{5 d (c+d x)^{5/2}}+\frac{\left (8 b^2\right ) \int \frac{\sin (a+b x)}{(c+d x)^{3/2}} \, dx}{5 d^2}-\frac{\left (12 b^2\right ) \int \frac{\sin ^3(a+b x)}{(c+d x)^{3/2}} \, dx}{5 d^2}\\ &=-\frac{16 b^2 \sin (a+b x)}{5 d^3 \sqrt{c+d x}}-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{5 d^2 (c+d x)^{3/2}}-\frac{2 \sin ^3(a+b x)}{5 d (c+d x)^{5/2}}+\frac{24 b^2 \sin ^3(a+b x)}{5 d^3 \sqrt{c+d x}}+\frac{\left (16 b^3\right ) \int \frac{\cos (a+b x)}{\sqrt{c+d x}} \, dx}{5 d^3}-\frac{\left (72 b^3\right ) \int \left (\frac{\cos (a+b x)}{4 \sqrt{c+d x}}-\frac{\cos (3 a+3 b x)}{4 \sqrt{c+d x}}\right ) \, dx}{5 d^3}\\ &=-\frac{16 b^2 \sin (a+b x)}{5 d^3 \sqrt{c+d x}}-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{5 d^2 (c+d x)^{3/2}}-\frac{2 \sin ^3(a+b x)}{5 d (c+d x)^{5/2}}+\frac{24 b^2 \sin ^3(a+b x)}{5 d^3 \sqrt{c+d x}}-\frac{\left (18 b^3\right ) \int \frac{\cos (a+b x)}{\sqrt{c+d x}} \, dx}{5 d^3}+\frac{\left (18 b^3\right ) \int \frac{\cos (3 a+3 b x)}{\sqrt{c+d x}} \, dx}{5 d^3}+\frac{\left (16 b^3 \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{5 d^3}-\frac{\left (16 b^3 \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{5 d^3}\\ &=-\frac{16 b^2 \sin (a+b x)}{5 d^3 \sqrt{c+d x}}-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{5 d^2 (c+d x)^{3/2}}-\frac{2 \sin ^3(a+b x)}{5 d (c+d x)^{5/2}}+\frac{24 b^2 \sin ^3(a+b x)}{5 d^3 \sqrt{c+d x}}+\frac{\left (18 b^3 \cos \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{3 b c}{d}+3 b x\right )}{\sqrt{c+d x}} \, dx}{5 d^3}+\frac{\left (32 b^3 \cos \left (a-\frac{b c}{d}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{5 d^4}-\frac{\left (18 b^3 \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{5 d^3}-\frac{\left (18 b^3 \sin \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{3 b c}{d}+3 b x\right )}{\sqrt{c+d x}} \, dx}{5 d^3}-\frac{\left (32 b^3 \sin \left (a-\frac{b c}{d}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{5 d^4}+\frac{\left (18 b^3 \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{5 d^3}\\ &=\frac{16 b^{5/2} \sqrt{2 \pi } \cos \left (a-\frac{b c}{d}\right ) C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{16 b^{5/2} \sqrt{2 \pi } S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (a-\frac{b c}{d}\right )}{5 d^{7/2}}-\frac{16 b^2 \sin (a+b x)}{5 d^3 \sqrt{c+d x}}-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{5 d^2 (c+d x)^{3/2}}-\frac{2 \sin ^3(a+b x)}{5 d (c+d x)^{5/2}}+\frac{24 b^2 \sin ^3(a+b x)}{5 d^3 \sqrt{c+d x}}+\frac{\left (36 b^3 \cos \left (3 a-\frac{3 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{3 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{5 d^4}-\frac{\left (36 b^3 \cos \left (a-\frac{b c}{d}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{5 d^4}-\frac{\left (36 b^3 \sin \left (3 a-\frac{3 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{3 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{5 d^4}+\frac{\left (36 b^3 \sin \left (a-\frac{b c}{d}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{5 d^4}\\ &=-\frac{2 b^{5/2} \sqrt{2 \pi } \cos \left (a-\frac{b c}{d}\right ) C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{5 d^{7/2}}+\frac{6 b^{5/2} \sqrt{6 \pi } \cos \left (3 a-\frac{3 b c}{d}\right ) C\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{6 b^{5/2} \sqrt{6 \pi } S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (3 a-\frac{3 b c}{d}\right )}{5 d^{7/2}}+\frac{2 b^{5/2} \sqrt{2 \pi } S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (a-\frac{b c}{d}\right )}{5 d^{7/2}}-\frac{16 b^2 \sin (a+b x)}{5 d^3 \sqrt{c+d x}}-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{5 d^2 (c+d x)^{3/2}}-\frac{2 \sin ^3(a+b x)}{5 d (c+d x)^{5/2}}+\frac{24 b^2 \sin ^3(a+b x)}{5 d^3 \sqrt{c+d x}}\\ \end{align*}

Mathematica [B]  time = 6.39826, size = 1429, normalized size = 4.01 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3/(c + d*x)^(7/2),x]

[Out]

(3*(Cos[a]*((2*(b/d)^(5/2)*Sin[(b*c)/d]*(Cos[(b*(c + d*x))/d]/((b/d)^(5/2)*(c + d*x)^(5/2)) - (2*(2*(Cos[(b*(c
+ d*x))/d]/(Sqrt[b/d]*Sqrt[c + d*x]) + Sqrt[2*Pi]*FresnelS[Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]]) + Sin[(b*(c +
d*x))/d]/((b/d)^(3/2)*(c + d*x)^(3/2))))/3))/(5*d) - (2*(b/d)^(5/2)*Cos[(b*c)/d]*(Sin[(b*(c + d*x))/d]/((b/d)
^(5/2)*(c + d*x)^(5/2)) + (2*(Cos[(b*(c + d*x))/d]/((b/d)^(3/2)*(c + d*x)^(3/2)) - 2*(-(Sqrt[2*Pi]*FresnelC[Sq
rt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]]) + Sin[(b*(c + d*x))/d]/(Sqrt[b/d]*Sqrt[c + d*x]))))/3))/(5*d)) + Sin[a]*((-
2*(b/d)^(5/2)*Cos[(b*c)/d]*(Cos[(b*(c + d*x))/d]/((b/d)^(5/2)*(c + d*x)^(5/2)) - (2*(2*(Cos[(b*(c + d*x))/d]/(
Sqrt[b/d]*Sqrt[c + d*x]) + Sqrt[2*Pi]*FresnelS[Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]]) + Sin[(b*(c + d*x))/d]/((b
/d)^(3/2)*(c + d*x)^(3/2))))/3))/(5*d) - (2*(b/d)^(5/2)*Sin[(b*c)/d]*(Sin[(b*(c + d*x))/d]/((b/d)^(5/2)*(c + d
*x)^(5/2)) + (2*(Cos[(b*(c + d*x))/d]/((b/d)^(3/2)*(c + d*x)^(3/2)) - 2*(-(Sqrt[2*Pi]*FresnelC[Sqrt[b/d]*Sqrt[
2/Pi]*Sqrt[c + d*x]]) + Sin[(b*(c + d*x))/d]/(Sqrt[b/d]*Sqrt[c + d*x]))))/3))/(5*d))))/4 + (-(Cos[3*a]*((18*Sq
rt[3]*(b/d)^(5/2)*Sin[(3*b*c)/d]*(Cos[(3*b*(c + d*x))/d]/(9*Sqrt[3]*(b/d)^(5/2)*(c + d*x)^(5/2)) - (2*(2*(Cos[
(3*b*(c + d*x))/d]/(Sqrt[3]*Sqrt[b/d]*Sqrt[c + d*x]) + Sqrt[2*Pi]*FresnelS[Sqrt[b/d]*Sqrt[6/Pi]*Sqrt[c + d*x]]
) + Sin[(3*b*(c + d*x))/d]/(3*Sqrt[3]*(b/d)^(3/2)*(c + d*x)^(3/2))))/3))/(5*d) - (18*Sqrt[3]*(b/d)^(5/2)*Cos[(
3*b*c)/d]*(Sin[(3*b*(c + d*x))/d]/(9*Sqrt[3]*(b/d)^(5/2)*(c + d*x)^(5/2)) + (2*(Cos[(3*b*(c + d*x))/d]/(3*Sqrt
[3]*(b/d)^(3/2)*(c + d*x)^(3/2)) - 2*(-(Sqrt[2*Pi]*FresnelC[Sqrt[b/d]*Sqrt[6/Pi]*Sqrt[c + d*x]]) + Sin[(3*b*(c
+ d*x))/d]/(Sqrt[3]*Sqrt[b/d]*Sqrt[c + d*x]))))/3))/(5*d))) - Sin[3*a]*((-18*Sqrt[3]*(b/d)^(5/2)*Cos[(3*b*c)/
d]*(Cos[(3*b*(c + d*x))/d]/(9*Sqrt[3]*(b/d)^(5/2)*(c + d*x)^(5/2)) - (2*(2*(Cos[(3*b*(c + d*x))/d]/(Sqrt[3]*Sq
rt[b/d]*Sqrt[c + d*x]) + Sqrt[2*Pi]*FresnelS[Sqrt[b/d]*Sqrt[6/Pi]*Sqrt[c + d*x]]) + Sin[(3*b*(c + d*x))/d]/(3*
Sqrt[3]*(b/d)^(3/2)*(c + d*x)^(3/2))))/3))/(5*d) - (18*Sqrt[3]*(b/d)^(5/2)*Sin[(3*b*c)/d]*(Sin[(3*b*(c + d*x))
/d]/(9*Sqrt[3]*(b/d)^(5/2)*(c + d*x)^(5/2)) + (2*(Cos[(3*b*(c + d*x))/d]/(3*Sqrt[3]*(b/d)^(3/2)*(c + d*x)^(3/2
)) - 2*(-(Sqrt[2*Pi]*FresnelC[Sqrt[b/d]*Sqrt[6/Pi]*Sqrt[c + d*x]]) + Sin[(3*b*(c + d*x))/d]/(Sqrt[3]*Sqrt[b/d]
*Sqrt[c + d*x]))))/3))/(5*d)))/4

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Maple [A]  time = 0.012, size = 450, normalized size = 1.3 \begin{align*} 2\,{\frac{1}{d} \left ( -{\frac{3}{20\, \left ( dx+c \right ) ^{5/2}}\sin \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }+3/10\,{\frac{b}{d} \left ( -1/3\,{\frac{1}{ \left ( dx+c \right ) ^{3/2}}\cos \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }-2/3\,{\frac{b}{d} \left ( -{\frac{1}{\sqrt{dx+c}}\sin \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }+{\frac{b\sqrt{2}\sqrt{\pi }}{d} \left ( \cos \left ({\frac{da-cb}{d}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) -\sin \left ({\frac{da-cb}{d}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) } \right ) }+1/20\,{\frac{1}{ \left ( dx+c \right ) ^{5/2}}\sin \left ( 3\,{\frac{ \left ( dx+c \right ) b}{d}}+3\,{\frac{da-cb}{d}} \right ) }-3/10\,{\frac{b}{d} \left ( -1/3\,{\frac{1}{ \left ( dx+c \right ) ^{3/2}}\cos \left ( 3\,{\frac{ \left ( dx+c \right ) b}{d}}+3\,{\frac{da-cb}{d}} \right ) }-2\,{\frac{b}{d} \left ( -{\frac{1}{\sqrt{dx+c}}\sin \left ( 3\,{\frac{ \left ( dx+c \right ) b}{d}}+3\,{\frac{da-cb}{d}} \right ) }+{\frac{b\sqrt{2}\sqrt{\pi }\sqrt{3}}{d} \left ( \cos \left ( 3\,{\frac{da-cb}{d}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{3}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) -\sin \left ( 3\,{\frac{da-cb}{d}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{3}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3/(d*x+c)^(7/2),x)

[Out]

2/d*(-3/20/(d*x+c)^(5/2)*sin(1/d*(d*x+c)*b+(a*d-b*c)/d)+3/10*b/d*(-1/3/(d*x+c)^(3/2)*cos(1/d*(d*x+c)*b+(a*d-b*
c)/d)-2/3*b/d*(-1/(d*x+c)^(1/2)*sin(1/d*(d*x+c)*b+(a*d-b*c)/d)+b/d*2^(1/2)*Pi^(1/2)/(b/d)^(1/2)*(cos((a*d-b*c)
/d)*FresnelC(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)-sin((a*d-b*c)/d)*FresnelS(2^(1/2)/Pi^(1/2)/(b/d)^
(1/2)*(d*x+c)^(1/2)*b/d))))+1/20/(d*x+c)^(5/2)*sin(3/d*(d*x+c)*b+3*(a*d-b*c)/d)-3/10*b/d*(-1/3/(d*x+c)^(3/2)*c
os(3/d*(d*x+c)*b+3*(a*d-b*c)/d)-2*b/d*(-1/(d*x+c)^(1/2)*sin(3/d*(d*x+c)*b+3*(a*d-b*c)/d)+b/d*2^(1/2)*Pi^(1/2)*
3^(1/2)/(b/d)^(1/2)*(cos(3*(a*d-b*c)/d)*FresnelC(2^(1/2)/Pi^(1/2)*3^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)-sin(3
*(a*d-b*c)/d)*FresnelS(2^(1/2)/Pi^(1/2)*3^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)))))

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Maxima [C]  time = 1.48198, size = 1265, normalized size = 3.55 \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)^3/(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/16*(9*sqrt(3)*(((I*gamma(-5/2, 3*I*(d*x + c)*b/d) - I*gamma(-5/2, -3*I*(d*x + c)*b/d))*cos(5/4*pi + 5/2*arct
an2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + (I*gamma(-5/2, 3*I*(d*x + c)*b/d) - I*gamma(-5/2, -3*I*(d*x + c)*b/
d))*cos(-5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) - (gamma(-5/2, 3*I*(d*x + c)*b/d) + gamma(-
5/2, -3*I*(d*x + c)*b/d))*sin(5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + (gamma(-5/2, 3*I*(d*
x + c)*b/d) + gamma(-5/2, -3*I*(d*x + c)*b/d))*sin(-5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))))
*cos(-3*(b*c - a*d)/d) + ((gamma(-5/2, 3*I*(d*x + c)*b/d) + gamma(-5/2, -3*I*(d*x + c)*b/d))*cos(5/4*pi + 5/2*
arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + (gamma(-5/2, 3*I*(d*x + c)*b/d) + gamma(-5/2, -3*I*(d*x + c)*b/
d))*cos(-5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + (I*gamma(-5/2, 3*I*(d*x + c)*b/d) - I*gam
ma(-5/2, -3*I*(d*x + c)*b/d))*sin(5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + (-I*gamma(-5/2,
3*I*(d*x + c)*b/d) + I*gamma(-5/2, -3*I*(d*x + c)*b/d))*sin(-5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqr
t(d^2))))*sin(-3*(b*c - a*d)/d))*((d*x + c)*abs(b)/abs(d))^(5/2) + (((-3*I*gamma(-5/2, I*(d*x + c)*b/d) + 3*I*
gamma(-5/2, -I*(d*x + c)*b/d))*cos(5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + (-3*I*gamma(-5/
2, I*(d*x + c)*b/d) + 3*I*gamma(-5/2, -I*(d*x + c)*b/d))*cos(-5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sq
rt(d^2))) + 3*(gamma(-5/2, I*(d*x + c)*b/d) + gamma(-5/2, -I*(d*x + c)*b/d))*sin(5/4*pi + 5/2*arctan2(0, b) +
5/2*arctan2(0, d/sqrt(d^2))) - 3*(gamma(-5/2, I*(d*x + c)*b/d) + gamma(-5/2, -I*(d*x + c)*b/d))*sin(-5/4*pi +
5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))))*cos(-(b*c - a*d)/d) - (3*(gamma(-5/2, I*(d*x + c)*b/d) + gam
ma(-5/2, -I*(d*x + c)*b/d))*cos(5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + 3*(gamma(-5/2, I*(
d*x + c)*b/d) + gamma(-5/2, -I*(d*x + c)*b/d))*cos(-5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2)))
- (-3*I*gamma(-5/2, I*(d*x + c)*b/d) + 3*I*gamma(-5/2, -I*(d*x + c)*b/d))*sin(5/4*pi + 5/2*arctan2(0, b) + 5/2
*arctan2(0, d/sqrt(d^2))) - (3*I*gamma(-5/2, I*(d*x + c)*b/d) - 3*I*gamma(-5/2, -I*(d*x + c)*b/d))*sin(-5/4*pi
+ 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))))*sin(-(b*c - a*d)/d))*((d*x + c)*abs(b)/abs(d))^(5/2))/((d
*x + c)^(5/2)*d)

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Fricas [A]  time = 3.42978, size = 1269, normalized size = 3.56 \begin{align*} \frac{2 \,{\left (3 \, \sqrt{6}{\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{C}\left (\sqrt{6} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - \sqrt{2}{\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{b c - a d}{d}\right ) \operatorname{C}\left (\sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) + \sqrt{2}{\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt{\frac{b}{\pi d}} \operatorname{S}\left (\sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{b c - a d}{d}\right ) - 3 \, \sqrt{6}{\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt{\frac{b}{\pi d}} \operatorname{S}\left (\sqrt{6} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) +{\left (2 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{3} - 2 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) +{\left (4 \, b^{2} d^{2} x^{2} + 8 \, b^{2} c d x + 4 \, b^{2} c^{2} -{\left (12 \, b^{2} d^{2} x^{2} + 24 \, b^{2} c d x + 12 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{2} - d^{2}\right )} \sin \left (b x + a\right )\right )} \sqrt{d x + c}\right )}}{5 \,{\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/5*(3*sqrt(6)*(pi*b^2*d^3*x^3 + 3*pi*b^2*c*d^2*x^2 + 3*pi*b^2*c^2*d*x + pi*b^2*c^3)*sqrt(b/(pi*d))*cos(-3*(b*
c - a*d)/d)*fresnel_cos(sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d))) - sqrt(2)*(pi*b^2*d^3*x^3 + 3*pi*b^2*c*d^2*x^2 +
3*pi*b^2*c^2*d*x + pi*b^2*c^3)*sqrt(b/(pi*d))*cos(-(b*c - a*d)/d)*fresnel_cos(sqrt(2)*sqrt(d*x + c)*sqrt(b/(p
i*d))) + sqrt(2)*(pi*b^2*d^3*x^3 + 3*pi*b^2*c*d^2*x^2 + 3*pi*b^2*c^2*d*x + pi*b^2*c^3)*sqrt(b/(pi*d))*fresnel_
sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c - a*d)/d) - 3*sqrt(6)*(pi*b^2*d^3*x^3 + 3*pi*b^2*c*d^2*x^2
+ 3*pi*b^2*c^2*d*x + pi*b^2*c^3)*sqrt(b/(pi*d))*fresnel_sin(sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-3*(b*c
- a*d)/d) + (2*(b*d^2*x + b*c*d)*cos(b*x + a)^3 - 2*(b*d^2*x + b*c*d)*cos(b*x + a) + (4*b^2*d^2*x^2 + 8*b^2*c
*d*x + 4*b^2*c^2 - (12*b^2*d^2*x^2 + 24*b^2*c*d*x + 12*b^2*c^2 - d^2)*cos(b*x + a)^2 - d^2)*sin(b*x + a))*sqrt
(d*x + c))/(d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)

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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)**3/(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )^{3}}{{\left (d x + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^3/(d*x + c)^(7/2), x)