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

Optimal. Leaf size=193 \[ -\frac{8 \sqrt{2 \pi } b^{5/2} \cos \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}+\frac{8 \sqrt{2 \pi } b^{5/2} \sin \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}+\frac{8 b^2 \sin (a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{4 b \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \sin (a+b x)}{5 d (c+d x)^{5/2}} \]

[Out]

(-4*b*Cos[a + b*x])/(15*d^2*(c + d*x)^(3/2)) - (8*b^(5/2)*Sqrt[2*Pi]*Cos[a - (b*c)/d]*FresnelC[(Sqrt[b]*Sqrt[2
/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(15*d^(7/2)) + (8*b^(5/2)*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])
/Sqrt[d]]*Sin[a - (b*c)/d])/(15*d^(7/2)) - (2*Sin[a + b*x])/(5*d*(c + d*x)^(5/2)) + (8*b^2*Sin[a + b*x])/(15*d
^3*Sqrt[c + d*x])

________________________________________________________________________________________

Rubi [A]  time = 0.296832, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{8 \sqrt{2 \pi } b^{5/2} \cos \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}+\frac{8 \sqrt{2 \pi } b^{5/2} \sin \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}+\frac{8 b^2 \sin (a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{4 b \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \sin (a+b x)}{5 d (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]/(c + d*x)^(7/2),x]

[Out]

(-4*b*Cos[a + b*x])/(15*d^2*(c + d*x)^(3/2)) - (8*b^(5/2)*Sqrt[2*Pi]*Cos[a - (b*c)/d]*FresnelC[(Sqrt[b]*Sqrt[2
/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(15*d^(7/2)) + (8*b^(5/2)*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])
/Sqrt[d]]*Sin[a - (b*c)/d])/(15*d^(7/2)) - (2*Sin[a + b*x])/(5*d*(c + d*x)^(5/2)) + (8*b^2*Sin[a + b*x])/(15*d
^3*Sqrt[c + d*x])

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)^{7/2}} \, dx &=-\frac{2 \sin (a+b x)}{5 d (c+d x)^{5/2}}+\frac{(2 b) \int \frac{\cos (a+b x)}{(c+d x)^{5/2}} \, dx}{5 d}\\ &=-\frac{4 b \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \sin (a+b x)}{5 d (c+d x)^{5/2}}-\frac{\left (4 b^2\right ) \int \frac{\sin (a+b x)}{(c+d x)^{3/2}} \, dx}{15 d^2}\\ &=-\frac{4 b \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \sin (a+b x)}{5 d (c+d x)^{5/2}}+\frac{8 b^2 \sin (a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{\left (8 b^3\right ) \int \frac{\cos (a+b x)}{\sqrt{c+d x}} \, dx}{15 d^3}\\ &=-\frac{4 b \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \sin (a+b x)}{5 d (c+d x)^{5/2}}+\frac{8 b^2 \sin (a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{\left (8 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}{15 d^3}+\frac{\left (8 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}{15 d^3}\\ &=-\frac{4 b \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \sin (a+b x)}{5 d (c+d x)^{5/2}}+\frac{8 b^2 \sin (a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{\left (16 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 )}{15 d^4}+\frac{\left (16 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 )}{15 d^4}\\ &=-\frac{4 b \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{8 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 )}{15 d^{7/2}}+\frac{8 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 )}{15 d^{7/2}}-\frac{2 \sin (a+b x)}{5 d (c+d x)^{5/2}}+\frac{8 b^2 \sin (a+b x)}{15 d^3 \sqrt{c+d x}}\\ \end{align*}

Mathematica [C]  time = 0.458893, size = 208, normalized size = 1.08 \[ -\frac{i \left (b (c+d x) \left (2 e^{i \left (a-\frac{b c}{d}\right )} \left (e^{\frac{i b (c+d x)}{d}} (2 b (c+d x)-i d)-2 i d \left (-\frac{i b (c+d x)}{d}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{i b (c+d x)}{d}\right )\right )-i e^{-i (a+b x)} \left (4 d e^{\frac{i b (c+d x)}{d}} \left (\frac{i b (c+d x)}{d}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},\frac{i b (c+d x)}{d}\right )-4 i b (c+d x)+2 d\right )\right )-6 i d^2 \sin (a+b x)\right )}{15 d^3 (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/15)*(b*(c + d*x)*(2*E^(I*(a - (b*c)/d))*(E^((I*b*(c + d*x))/d)*((-I)*d + 2*b*(c + d*x)) - (2*I)*d*(((-I)*
b*(c + d*x))/d)^(3/2)*Gamma[1/2, ((-I)*b*(c + d*x))/d]) - (I*(2*d - (4*I)*b*(c + d*x) + 4*d*E^((I*b*(c + d*x))
/d)*((I*b*(c + d*x))/d)^(3/2)*Gamma[1/2, (I*b*(c + d*x))/d]))/E^(I*(a + b*x))) - (6*I)*d^2*Sin[a + b*x]))/(d^3
*(c + d*x)^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 220, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( -1/5\,{\frac{1}{ \left ( dx+c \right ) ^{5/2}}\sin \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }+2/5\,{\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 ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-1/5/(d*x+c)^(5/2)*sin(1/d*(d*x+c)*b+(a*d-b*c)/d)+2/5*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)))))

________________________________________________________________________________________

Maxima [C]  time = 1.31408, size = 632, normalized size = 3.27 \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)^(7/2),x, algorithm="maxima")

[Out]

-1/4*(((I*gamma(-5/2, I*(d*x + c)*b/d) - 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))) + (I*gamma(-5/2, I*(d*x + c)*b/d) - 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))) - (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))) + (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) + ((ga
mma(-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))) + (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))) + (I*gamma(-5/2, I*(d*x + c)*b/d) - 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))) + (-I*gamma(-5/2, I*(d*x + c)*b/d) + 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)

________________________________________________________________________________________

Fricas [A]  time = 2.36828, size = 682, normalized size = 3.53 \begin{align*} -\frac{2 \,{\left (4 \, \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 ) - 4 \, \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 ) + \sqrt{d x + c}{\left (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} - 3 \, d^{2}\right )} \sin \left (b x + a\right )\right )}\right )}}{15 \,{\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)/(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(4*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/(pi*d))) - 4*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) + sqrt(d*x + c)*(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 - 3*d^2)
*sin(b*x + a)))/(d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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