3.58 \(\int \frac{\sin ^3(a+b x)}{(c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=292 \[ \frac{\sqrt{6 \pi } b^{3/2} \sin \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}-\frac{\sqrt{2 \pi } b^{3/2} \sin \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}-\frac{\sqrt{2 \pi } b^{3/2} \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}+\frac{\sqrt{6 \pi } b^{3/2} \cos \left (3 a-\frac{3 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}-\frac{4 b \sin ^2(a+b x) \cos (a+b x)}{d^2 \sqrt{c+d x}}-\frac{2 \sin ^3(a+b x)}{3 d (c+d x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.710384, antiderivative size = 292, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {3314, 3306, 3305, 3351, 3304, 3352, 3312} \[ \frac{\sqrt{6 \pi } b^{3/2} \sin \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}-\frac{\sqrt{2 \pi } b^{3/2} \sin \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}-\frac{\sqrt{2 \pi } b^{3/2} \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}+\frac{\sqrt{6 \pi } b^{3/2} \cos \left (3 a-\frac{3 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}-\frac{4 b \sin ^2(a+b x) \cos (a+b x)}{d^2 \sqrt{c+d x}}-\frac{2 \sin ^3(a+b x)}{3 d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3/(c + d*x)^(5/2),x]

[Out]

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

Rubi steps

\begin{align*} \int \frac{\sin ^3(a+b x)}{(c+d x)^{5/2}} \, dx &=-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{d^2 \sqrt{c+d x}}-\frac{2 \sin ^3(a+b x)}{3 d (c+d x)^{3/2}}+\frac{\left (8 b^2\right ) \int \frac{\sin (a+b x)}{\sqrt{c+d x}} \, dx}{d^2}-\frac{\left (12 b^2\right ) \int \frac{\sin ^3(a+b x)}{\sqrt{c+d x}} \, dx}{d^2}\\ &=-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{d^2 \sqrt{c+d x}}-\frac{2 \sin ^3(a+b x)}{3 d (c+d x)^{3/2}}-\frac{\left (12 b^2\right ) \int \left (\frac{3 \sin (a+b x)}{4 \sqrt{c+d x}}-\frac{\sin (3 a+3 b x)}{4 \sqrt{c+d x}}\right ) \, dx}{d^2}+\frac{\left (8 b^2 \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{d^2}+\frac{\left (8 b^2 \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{d^2}\\ &=-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{d^2 \sqrt{c+d x}}-\frac{2 \sin ^3(a+b x)}{3 d (c+d x)^{3/2}}+\frac{\left (3 b^2\right ) \int \frac{\sin (3 a+3 b x)}{\sqrt{c+d x}} \, dx}{d^2}-\frac{\left (9 b^2\right ) \int \frac{\sin (a+b x)}{\sqrt{c+d x}} \, dx}{d^2}+\frac{\left (16 b^2 \cos \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 )}{d^3}+\frac{\left (16 b^2 \sin \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 )}{d^3}\\ &=\frac{8 b^{3/2} \sqrt{2 \pi } \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}+\frac{8 b^{3/2} \sqrt{2 \pi } C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (a-\frac{b c}{d}\right )}{d^{5/2}}-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{d^2 \sqrt{c+d x}}-\frac{2 \sin ^3(a+b x)}{3 d (c+d x)^{3/2}}+\frac{\left (3 b^2 \cos \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}{d^2}-\frac{\left (9 b^2 \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{d^2}+\frac{\left (3 b^2 \sin \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}{d^2}-\frac{\left (9 b^2 \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{\sqrt{c+d x}} \, dx}{d^2}\\ &=\frac{8 b^{3/2} \sqrt{2 \pi } \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}+\frac{8 b^{3/2} \sqrt{2 \pi } C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (a-\frac{b c}{d}\right )}{d^{5/2}}-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{d^2 \sqrt{c+d x}}-\frac{2 \sin ^3(a+b x)}{3 d (c+d x)^{3/2}}+\frac{\left (6 b^2 \cos \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 )}{d^3}-\frac{\left (18 b^2 \cos \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 )}{d^3}+\frac{\left (6 b^2 \sin \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 )}{d^3}-\frac{\left (18 b^2 \sin \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 )}{d^3}\\ &=-\frac{b^{3/2} \sqrt{2 \pi } \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}+\frac{b^{3/2} \sqrt{6 \pi } \cos \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 )}{d^{5/2}}+\frac{b^{3/2} \sqrt{6 \pi } C\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 )}{d^{5/2}}-\frac{b^{3/2} \sqrt{2 \pi } C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (a-\frac{b c}{d}\right )}{d^{5/2}}-\frac{4 b \cos (a+b x) \sin ^2(a+b x)}{d^2 \sqrt{c+d x}}-\frac{2 \sin ^3(a+b x)}{3 d (c+d x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 2.42588, size = 496, normalized size = 1.7 \[ \frac{6 \sqrt{6 \pi } b d x \sqrt{\frac{b}{d}} \sqrt{c+d x} \sin \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )+6 \sqrt{6 \pi } b c \sqrt{\frac{b}{d}} \sqrt{c+d x} \sin \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )-6 \sqrt{2 \pi } b d x \sqrt{\frac{b}{d}} \sqrt{c+d x} \sin \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )-6 \sqrt{2 \pi } b c \sqrt{\frac{b}{d}} \sqrt{c+d x} \sin \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )-6 \sqrt{2 \pi } b \sqrt{\frac{b}{d}} (c+d x)^{3/2} \cos \left (a-\frac{b c}{d}\right ) S\left (\sqrt{\frac{b}{d}} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}\right )+6 \sqrt{6 \pi } b \sqrt{\frac{b}{d}} (c+d x)^{3/2} \cos \left (3 a-\frac{3 b c}{d}\right ) S\left (\sqrt{\frac{b}{d}} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}\right )-6 b c \cos (a+b x)+6 b c \cos (3 (a+b x))-3 d \sin (a+b x)+d \sin (3 (a+b x))-6 b d x \cos (a+b x)+6 b d x \cos (3 (a+b x))}{6 d^2 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3/(c + d*x)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.011, size = 368, normalized size = 1.3 \begin{align*} 2\,{\frac{1}{d} \left ( -1/4\,{\frac{1}{ \left ( dx+c \right ) ^{3/2}}\sin \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }+1/2\,{\frac{b}{d} \left ( -{\frac{1}{\sqrt{dx+c}}\cos \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 FresnelS} \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 FresnelC} \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 ) }+1/12\,{\frac{1}{ \left ( dx+c \right ) ^{3/2}}\sin \left ( 3\,{\frac{ \left ( dx+c \right ) b}{d}}+3\,{\frac{da-cb}{d}} \right ) }-1/2\,{\frac{b}{d} \left ( -{\frac{1}{\sqrt{dx+c}}\cos \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 FresnelS} \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 FresnelC} \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3/(d*x+c)^(5/2),x)

[Out]

2/d*(-1/4/(d*x+c)^(3/2)*sin(1/d*(d*x+c)*b+(a*d-b*c)/d)+1/2*b/d*(-1/(d*x+c)^(1/2)*cos(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)*FresnelS(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)+
sin((a*d-b*c)/d)*FresnelC(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)))+1/12/(d*x+c)^(3/2)*sin(3/d*(d*x+c)
*b+3*(a*d-b*c)/d)-1/2*b/d*(-1/(d*x+c)^(1/2)*cos(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)*FresnelS(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
)*FresnelC(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 = 1.51922, size = 1265, normalized size = 4.33 \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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.66328, size = 963, normalized size = 3.3 \begin{align*} \frac{3 \, \sqrt{6}{\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{3 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{S}\left (\sqrt{6} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - 3 \, \sqrt{2}{\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{b c - a d}{d}\right ) \operatorname{S}\left (\sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - 3 \, \sqrt{2}{\left (\pi b d^{2} x^{2} + 2 \, \pi b c d x + \pi b c^{2}\right )} \sqrt{\frac{b}{\pi d}} \operatorname{C}\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 d^{2} x^{2} + 2 \, \pi b c d x + \pi b c^{2}\right )} \sqrt{\frac{b}{\pi d}} \operatorname{C}\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 ) + 2 \,{\left (6 \,{\left (b d x + b c\right )} \cos \left (b x + a\right )^{3} - 6 \,{\left (b d x + b c\right )} \cos \left (b x + a\right ) +{\left (d \cos \left (b x + a\right )^{2} - d\right )} \sin \left (b x + a\right )\right )} \sqrt{d x + c}}{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)^3/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

1/3*(3*sqrt(6)*(pi*b*d^2*x^2 + 2*pi*b*c*d*x + pi*b*c^2)*sqrt(b/(pi*d))*cos(-3*(b*c - a*d)/d)*fresnel_sin(sqrt(
6)*sqrt(d*x + c)*sqrt(b/(pi*d))) - 3*sqrt(2)*(pi*b*d^2*x^2 + 2*pi*b*c*d*x + pi*b*c^2)*sqrt(b/(pi*d))*cos(-(b*c
- a*d)/d)*fresnel_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) - 3*sqrt(2)*(pi*b*d^2*x^2 + 2*pi*b*c*d*x + pi*b*c
^2)*sqrt(b/(pi*d))*fresnel_cos(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c - a*d)/d) + 3*sqrt(6)*(pi*b*d^2
*x^2 + 2*pi*b*c*d*x + pi*b*c^2)*sqrt(b/(pi*d))*fresnel_cos(sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-3*(b*c -
a*d)/d) + 2*(6*(b*d*x + b*c)*cos(b*x + a)^3 - 6*(b*d*x + b*c)*cos(b*x + a) + (d*cos(b*x + a)^2 - d)*sin(b*x +
a))*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 ^{3}{\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)**3/(d*x+c)**(5/2),x)

[Out]

Integral(sin(a + b*x)**3/(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 )^{3}}{{\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)^3/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^3/(d*x + c)^(5/2), x)