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

Optimal. Leaf size=168 \[ -\frac{4 \sqrt{2 \pi } b^{3/2} \sin \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{4 \sqrt{2 \pi } b^{3/2} \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{4 b \cos (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sin (a+b x)}{3 d (c+d x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.238313, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3297, 3306, 3305, 3351, 3304, 3352} \[ -\frac{4 \sqrt{2 \pi } b^{3/2} \sin \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{4 \sqrt{2 \pi } b^{3/2} \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{4 b \cos (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sin (a+b x)}{3 d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]/(c + d*x)^(5/2),x]

[Out]

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

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]

Rubi steps

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

Mathematica [C]  time = 0.629594, size = 162, normalized size = 0.96 \[ \frac{2 \left (-d \sin (a+b x)-b (c+d x) \left (e^{-i (a+b x)} \left (-e^{\frac{i b (c+d x)}{d}} \sqrt{\frac{i b (c+d x)}{d}} \text{Gamma}\left (\frac{1}{2},\frac{i b (c+d x)}{d}\right )+e^{2 i (a+b x)}+1\right )-e^{i \left (a-\frac{b c}{d}\right )} \sqrt{-\frac{i b (c+d x)}{d}} \text{Gamma}\left (\frac{1}{2},-\frac{i b (c+d x)}{d}\right )\right )\right )}{3 d^2 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-(b*(c + d*x)*(-(E^(I*(a - (b*c)/d))*Sqrt[((-I)*b*(c + d*x))/d]*Gamma[1/2, ((-I)*b*(c + d*x))/d]) + (1 + E
^((2*I)*(a + b*x)) - E^((I*b*(c + d*x))/d)*Sqrt[(I*b*(c + d*x))/d]*Gamma[1/2, (I*b*(c + d*x))/d])/E^(I*(a + b*
x)))) - d*Sin[a + b*x]))/(3*d^2*(c + d*x)^(3/2))

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Maple [A]  time = 0.009, size = 180, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( -1/3\,{\frac{1}{ \left ( dx+c \right ) ^{3/2}}\sin \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }+2/3\,{\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 ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-1/3/(d*x+c)^(3/2)*sin(1/d*(d*x+c)*b+(a*d-b*c)/d)+2/3*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))))

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Maxima [C]  time = 1.31056, size = 632, normalized size = 3.76 \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)/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

-1/4*(((I*gamma(-3/2, I*(d*x + c)*b/d) - 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))) + (I*gamma(-3/2, I*(d*x + c)*b/d) - 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))) - (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))) + (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) + ((ga
mma(-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))) + (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))) + (I*gamma(-3/2, I*(d*x + c)*b/d) - 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))) + (-I*gamma(-3/2, I*(d*x + c)*b/d) + 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 = 2.26133, size = 510, normalized size = 3.04 \begin{align*} -\frac{2 \,{\left (2 \, \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 ) + 2 \, \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 ) + \sqrt{d x + c}{\left (2 \,{\left (b d x + b c\right )} \cos \left (b x + a\right ) + d \sin \left (b x + a\right )\right )}\right )}}{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)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(2*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))) + 2*sqrt(2)*(pi*b*d^2*x^2 + 2*pi*b*c*d*x + pi*b*c^2)*sqrt(b/(pi*d))*fresnel_co
s(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c - a*d)/d) + sqrt(d*x + c)*(2*(b*d*x + b*c)*cos(b*x + a) + d*
sin(b*x + a)))/(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{\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)/(d*x+c)**(5/2),x)

[Out]

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

[Out]

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