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

Optimal. Leaf size=270 \[ \frac{3 \sqrt{\frac{\pi }{2}} \sqrt{b} \cos \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}-\frac{\sqrt{\frac{3 \pi }{2}} \sqrt{b} \cos \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{\sqrt{\frac{3 \pi }{2}} \sqrt{b} \sin \left (3 a-\frac{3 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}-\frac{3 \sqrt{\frac{\pi }{2}} \sqrt{b} \sin \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}-\frac{2 \sin ^3(a+b x)}{d \sqrt{c+d x}} \]

[Out]

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

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Rubi [A]  time = 0.563834, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3313, 3306, 3305, 3351, 3304, 3352} \[ \frac{3 \sqrt{\frac{\pi }{2}} \sqrt{b} \cos \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}-\frac{\sqrt{\frac{3 \pi }{2}} \sqrt{b} \cos \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{\sqrt{\frac{3 \pi }{2}} \sqrt{b} \sin \left (3 a-\frac{3 b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}-\frac{3 \sqrt{\frac{\pi }{2}} \sqrt{b} \sin \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}-\frac{2 \sin ^3(a+b x)}{d \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3/(c + d*x)^(3/2),x]

[Out]

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

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

Mathematica [A]  time = 1.00836, size = 300, normalized size = 1.11 \[ \frac{3 \sqrt{2 \pi } \sqrt{\frac{b}{d}} \sqrt{c+d x} \cos \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )-\sqrt{6 \pi } \sqrt{\frac{b}{d}} \sqrt{c+d x} \cos \left (3 a-\frac{3 b c}{d}\right ) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )+\sqrt{6 \pi } \sqrt{\frac{b}{d}} \sqrt{c+d x} \sin \left (3 a-\frac{3 b c}{d}\right ) S\left (\sqrt{\frac{b}{d}} \sqrt{\frac{6}{\pi }} \sqrt{c+d x}\right )-3 \sqrt{2 \pi } \sqrt{\frac{b}{d}} \sqrt{c+d x} \sin \left (a-\frac{b c}{d}\right ) S\left (\sqrt{\frac{b}{d}} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}\right )-3 \sin (a+b x)+\sin (3 (a+b x))}{2 d \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 288, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( -3/4\,{\frac{1}{\sqrt{dx+c}}\sin \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }+3/4\,{\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}}}}}}+1/4\,{\frac{1}{\sqrt{dx+c}}\sin \left ( 3\,{\frac{ \left ( dx+c \right ) b}{d}}+3\,{\frac{da-cb}{d}} \right ) }-1/4\,{\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-3/4/(d*x+c)^(1/2)*sin(1/d*(d*x+c)*b+(a*d-b*c)/d)+3/4*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))+1/4/(d*x+c)^(1/2)*sin(3/d*(d*x+c)*b+3*(a*d-b*c)/d)-1/4*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 = 1.49037, size = 1264, normalized size = 4.68 \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)^(3/2),x, algorithm="maxima")

[Out]

1/16*(sqrt(3)*(((I*gamma(-1/2, 3*I*(d*x + c)*b/d) - I*gamma(-1/2, -3*I*(d*x + c)*b/d))*cos(1/4*pi + 1/2*arctan
2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + (I*gamma(-1/2, 3*I*(d*x + c)*b/d) - I*gamma(-1/2, -3*I*(d*x + c)*b/d)
)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - (gamma(-1/2, 3*I*(d*x + c)*b/d) + gamma(-1/
2, -3*I*(d*x + c)*b/d))*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + (gamma(-1/2, 3*I*(d*x
+ c)*b/d) + gamma(-1/2, -3*I*(d*x + c)*b/d))*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*c
os(-3*(b*c - a*d)/d) + ((gamma(-1/2, 3*I*(d*x + c)*b/d) + gamma(-1/2, -3*I*(d*x + c)*b/d))*cos(1/4*pi + 1/2*ar
ctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + (gamma(-1/2, 3*I*(d*x + c)*b/d) + gamma(-1/2, -3*I*(d*x + c)*b/d)
)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + (I*gamma(-1/2, 3*I*(d*x + c)*b/d) - I*gamma
(-1/2, -3*I*(d*x + c)*b/d))*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + (-I*gamma(-1/2, 3*
I*(d*x + c)*b/d) + I*gamma(-1/2, -3*I*(d*x + c)*b/d))*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(
d^2))))*sin(-3*(b*c - a*d)/d))*sqrt((d*x + c)*abs(b)/abs(d)) + (((-3*I*gamma(-1/2, I*(d*x + c)*b/d) + 3*I*gamm
a(-1/2, -I*(d*x + c)*b/d))*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + (-3*I*gamma(-1/2, I
*(d*x + c)*b/d) + 3*I*gamma(-1/2, -I*(d*x + c)*b/d))*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d
^2))) + 3*(gamma(-1/2, I*(d*x + c)*b/d) + gamma(-1/2, -I*(d*x + c)*b/d))*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*
arctan2(0, d/sqrt(d^2))) - 3*(gamma(-1/2, I*(d*x + c)*b/d) + gamma(-1/2, -I*(d*x + c)*b/d))*sin(-1/4*pi + 1/2*
arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*cos(-(b*c - a*d)/d) - (3*(gamma(-1/2, I*(d*x + c)*b/d) + gamma(-
1/2, -I*(d*x + c)*b/d))*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 3*(gamma(-1/2, I*(d*x
+ c)*b/d) + gamma(-1/2, -I*(d*x + c)*b/d))*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - (-
3*I*gamma(-1/2, I*(d*x + c)*b/d) + 3*I*gamma(-1/2, -I*(d*x + c)*b/d))*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arc
tan2(0, d/sqrt(d^2))) - (3*I*gamma(-1/2, I*(d*x + c)*b/d) - 3*I*gamma(-1/2, -I*(d*x + c)*b/d))*sin(-1/4*pi + 1
/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*sin(-(b*c - a*d)/d))*sqrt((d*x + c)*abs(b)/abs(d)))/(sqrt(d*x
+ c)*d)

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Fricas [A]  time = 2.45337, size = 707, normalized size = 2.62 \begin{align*} -\frac{\sqrt{6}{\left (\pi d x + \pi c\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 ) - 3 \, \sqrt{2}{\left (\pi d x + \pi c\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 ) + 3 \, \sqrt{2}{\left (\pi d x + \pi c\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 ) - \sqrt{6}{\left (\pi d x + \pi c\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 ) - 4 \, \sqrt{d x + c}{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )}{2 \,{\left (d^{2} x + c d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(6)*(pi*d*x + pi*c)*sqrt(b/(pi*d))*cos(-3*(b*c - a*d)/d)*fresnel_cos(sqrt(6)*sqrt(d*x + c)*sqrt(b/(p
i*d))) - 3*sqrt(2)*(pi*d*x + pi*c)*sqrt(b/(pi*d))*cos(-(b*c - a*d)/d)*fresnel_cos(sqrt(2)*sqrt(d*x + c)*sqrt(b
/(pi*d))) + 3*sqrt(2)*(pi*d*x + pi*c)*sqrt(b/(pi*d))*fresnel_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b
*c - a*d)/d) - sqrt(6)*(pi*d*x + pi*c)*sqrt(b/(pi*d))*fresnel_sin(sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-3
*(b*c - a*d)/d) - 4*sqrt(d*x + c)*(cos(b*x + a)^2 - 1)*sin(b*x + a))/(d^2*x + c*d)

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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{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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