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

Optimal. Leaf size=170 \[ -\frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \sin \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 b^{5/2}}-\frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 b^{5/2}}+\frac{3 d \sqrt{c+d x} \sin (a+b x)}{2 b^2}-\frac{(c+d x)^{3/2} \cos (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.242042, antiderivative size = 170, 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 = {3296, 3306, 3305, 3351, 3304, 3352} \[ -\frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \sin \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 b^{5/2}}-\frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \cos \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 b^{5/2}}+\frac{3 d \sqrt{c+d x} \sin (a+b x)}{2 b^2}-\frac{(c+d x)^{3/2} \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)*Sin[a + b*x],x]

[Out]

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

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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

Mathematica [C]  time = 0.100351, size = 125, normalized size = 0.74 \[ -\frac{i d \sqrt{c+d x} e^{-\frac{i (a d+b c)}{d}} \left (\frac{e^{2 i a} \text{Gamma}\left (\frac{5}{2},-\frac{i b (c+d x)}{d}\right )}{\sqrt{-\frac{i b (c+d x)}{d}}}-\frac{e^{\frac{2 i b c}{d}} \text{Gamma}\left (\frac{5}{2},\frac{i b (c+d x)}{d}\right )}{\sqrt{\frac{i b (c+d x)}{d}}}\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*d*Sqrt[c + d*x]*((E^((2*I)*a)*Gamma[5/2, ((-I)*b*(c + d*x))/d])/Sqrt[((-I)*b*(c + d*x))/d] - (E^(((2*I
)*b*c)/d)*Gamma[5/2, (I*b*(c + d*x))/d])/Sqrt[(I*b*(c + d*x))/d]))/(b^2*E^((I*(b*c + a*d))/d))

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Maple [A]  time = 0.007, size = 188, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( -1/2\,{\frac{d \left ( dx+c \right ) ^{3/2}}{b}\cos \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }+3/2\,{\frac{d}{b} \left ( 1/2\,{\frac{d\sqrt{dx+c}}{b}\sin \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }-1/4\,{\frac{d\sqrt{2}\sqrt{\pi }}{b} \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((d*x+c)^(3/2)*sin(b*x+a),x)

[Out]

2/d*(-1/2/b*d*(d*x+c)^(3/2)*cos(1/d*(d*x+c)*b+(a*d-b*c)/d)+3/2/b*d*(1/2/b*d*(d*x+c)^(1/2)*sin(1/d*(d*x+c)*b+(a
*d-b*c)/d)-1/4/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 = 1.82981, size = 855, normalized size = 5.03 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(16*(d*x + c)^(3/2)*b*d*sqrt(abs(b)/abs(d))*cos(((d*x + c)*b - b*c + a*d)/d) - 24*sqrt(d*x + c)*d^2*sqrt
(abs(b)/abs(d))*sin(((d*x + c)*b - b*c + a*d)/d) - ((-3*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan
2(0, d/sqrt(d^2))) - 3*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 3*sqrt(pi)*
sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 3*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1
/2*arctan2(0, d/sqrt(d^2))))*d^2*cos(-(b*c - a*d)/d) - (3*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan
2(0, d/sqrt(d^2))) + 3*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 3*I*sqrt(pi)*
sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 3*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) +
1/2*arctan2(0, d/sqrt(d^2))))*d^2*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(I*b/d)) - ((3*I*sqrt(pi)*cos(1/
4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 3*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*a
rctan2(0, d/sqrt(d^2))) - 3*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 3*sqrt(pi
)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^2*cos(-(b*c - a*d)/d) - (3*sqrt(pi)*cos(1/
4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 3*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arc
tan2(0, d/sqrt(d^2))) + 3*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 3*I*sqrt(
pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^2*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)
*sqrt(-I*b/d)))/(b^2*d*sqrt(abs(b)/abs(d)))

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Fricas [A]  time = 1.79597, size = 394, normalized size = 2.32 \begin{align*} -\frac{3 \, \sqrt{2} \pi d^{2} \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} \pi d^{2} \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 ) - 2 \,{\left (3 \, b d \sin \left (b x + a\right ) - 2 \,{\left (b^{2} d x + b^{2} c\right )} \cos \left (b x + a\right )\right )} \sqrt{d x + c}}{4 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.25184, size = 764, normalized size = 4.49 \begin{align*} -\frac{2 \,{\left (\frac{\sqrt{2} \sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{2} \sqrt{b d} \sqrt{d x + c}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac{i \, b c - i \, a d}{d}\right )}}{\sqrt{b d}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} + \frac{\sqrt{2} \sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{2} \sqrt{b d} \sqrt{d x + c}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac{-i \, b c + i \, a d}{d}\right )}}{\sqrt{b d}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} + \frac{2 \, \sqrt{d x + c} d e^{\left (\frac{i \,{\left (d x + c\right )} b - i \, b c + i \, a d}{d}\right )}}{b} + \frac{2 \, \sqrt{d x + c} d e^{\left (\frac{-i \,{\left (d x + c\right )} b + i \, b c - i \, a d}{d}\right )}}{b}\right )} c + \frac{i \, \sqrt{2} \sqrt{\pi }{\left (2 i \, b c d - 3 \, d^{2}\right )} d \operatorname{erf}\left (-\frac{\sqrt{2} \sqrt{b d} \sqrt{d x + c}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac{i \, b c - i \, a d}{d}\right )}}{\sqrt{b d}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b^{2}} + \frac{i \, \sqrt{2} \sqrt{\pi }{\left (2 i \, b c d + 3 \, d^{2}\right )} d \operatorname{erf}\left (-\frac{\sqrt{2} \sqrt{b d} \sqrt{d x + c}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac{-i \, b c + i \, a d}{d}\right )}}{\sqrt{b d}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b^{2}} - \frac{2 i \,{\left (2 i \,{\left (d x + c\right )}^{\frac{3}{2}} b d - 2 i \, \sqrt{d x + c} b c d - 3 \, \sqrt{d x + c} d^{2}\right )} e^{\left (\frac{i \,{\left (d x + c\right )} b - i \, b c + i \, a d}{d}\right )}}{b^{2}} - \frac{2 i \,{\left (2 i \,{\left (d x + c\right )}^{\frac{3}{2}} b d - 2 i \, \sqrt{d x + c} b c d + 3 \, \sqrt{d x + c} d^{2}\right )} e^{\left (\frac{-i \,{\left (d x + c\right )} b + i \, b c - i \, a d}{d}\right )}}{b^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*(sqrt(2)*sqrt(pi)*d^2*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((I*b*c
- I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + sqrt(2)*sqrt(pi)*d^2*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x
+ c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) + 2*sqrt(d
*x + c)*d*e^((I*(d*x + c)*b - I*b*c + I*a*d)/d)/b + 2*sqrt(d*x + c)*d*e^((-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b
)*c + I*sqrt(2)*sqrt(pi)*(2*I*b*c*d - 3*d^2)*d*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) +
1)/d)*e^((I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b^2) + I*sqrt(2)*sqrt(pi)*(2*I*b*c*d + 3*d^2
)*d*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-I*b*c + I*a*d)/d)/(sqrt(b*d)*(
-I*b*d/sqrt(b^2*d^2) + 1)*b^2) - 2*I*(2*I*(d*x + c)^(3/2)*b*d - 2*I*sqrt(d*x + c)*b*c*d - 3*sqrt(d*x + c)*d^2)
*e^((I*(d*x + c)*b - I*b*c + I*a*d)/d)/b^2 - 2*I*(2*I*(d*x + c)^(3/2)*b*d - 2*I*sqrt(d*x + c)*b*c*d + 3*sqrt(d
*x + c)*d^2)*e^((-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b^2)/d