3.1.42 \(\int \frac {\sin (a+b x)}{(c+d x)^{3/2}} \, dx\) [42]

Optimal. Leaf size=139 \[ \frac {2 \sqrt {b} \sqrt {2 \pi } \cos \left (a-\frac {b c}{d}\right ) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {2 \sqrt {b} \sqrt {2 \pi } S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (a-\frac {b c}{d}\right )}{d^{3/2}}-\frac {2 \sin (a+b x)}{d \sqrt {c+d x}} \]

[Out]

2*cos(a-b*c/d)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))*b^(1/2)*2^(1/2)*Pi^(1/2)/d^(3/2)-2*Fre
snelS(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))*sin(a-b*c/d)*b^(1/2)*2^(1/2)*Pi^(1/2)/d^(3/2)-2*sin(b*x+
a)/d/(d*x+c)^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3378, 3387, 3386, 3432, 3385, 3433} \begin {gather*} \frac {2 \sqrt {2 \pi } \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 {2 \sqrt {2 \pi } \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 (a+b x)}{d \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]/(c + d*x)^(3/2),x]

[Out]

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

Rule 3378

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 3385

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 3386

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 3387

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 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.21, size = 148, normalized size = 1.06 \begin {gather*} \frac {i e^{-\frac {i (b c+a d)}{d}} \left (-e^{2 i a} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {i b (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {i b (c+d x)}{d}\right )+2 i e^{\frac {i (b c+a d)}{d}} \sin (a+b x)\right )}{d \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 140, normalized size = 1.01

method	result	size
derivativedivides	\(\frac {-\frac {2 \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {d a -c b}{d}\right )}{\sqrt {d x +c}}+\frac {2 b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {d a -c b}{d}\right ) \FresnelC \left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {d a -c b}{d}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}}{d}\)	\(140\)
default	\(\frac {-\frac {2 \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {d a -c b}{d}\right )}{\sqrt {d x +c}}+\frac {2 b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {d a -c b}{d}\right ) \FresnelC \left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {d a -c b}{d}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}}{d}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-1/(d*x+c)^(1/2)*sin(1/d*b*(d*x+c)+(a*d-b*c)/d)+b/d*2^(1/2)*Pi^(1/2)/(b/d)^(1/2)*(cos((a*d-b*c)/d)*Fresne
lC(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)-sin((a*d-b*c)/d)*FresnelS(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*b*(d
*x+c)^(1/2)/d)))

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Maxima [C] Result contains complex when optimal does not.
time = 0.60, size = 129, normalized size = 0.93 \begin {gather*} -\frac {{\left ({\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )} \sqrt {\frac {{\left (d x + c\right )} b}{d}}}{4 \, \sqrt {d x + c} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(((I - 1)*sqrt(2)*gamma(-1/2, I*(d*x + c)*b/d) - (I + 1)*sqrt(2)*gamma(-1/2, -I*(d*x + c)*b/d))*cos(-(b*c
- a*d)/d) + ((I + 1)*sqrt(2)*gamma(-1/2, I*(d*x + c)*b/d) - (I - 1)*sqrt(2)*gamma(-1/2, -I*(d*x + c)*b/d))*si
n(-(b*c - a*d)/d))*sqrt((d*x + c)*b/d)/(sqrt(d*x + c)*d)

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Fricas [A]
time = 0.35, size = 146, normalized size = 1.05 \begin {gather*} \frac {2 \, {\left (\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 ) - \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 {d x + c} \sin \left (b x + a\right )\right )}}{d^{2} x + c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (a + b x \right )}}{\left (c + d x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sin \left (a+b\,x\right )}{{\left (c+d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)/(c + d*x)^(3/2),x)

[Out]

int(sin(a + b*x)/(c + d*x)^(3/2), x)

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