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

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

[Out]

-(d*x+c)^(3/2)*cos(b*x+a)/b-3/4*d^(3/2)*cos(a-b*c/d)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))*
2^(1/2)*Pi^(1/2)/b^(5/2)-3/4*d^(3/2)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))*sin(a-b*c/d)*2^(
1/2)*Pi^(1/2)/b^(5/2)+3/2*d*sin(b*x+a)*(d*x+c)^(1/2)/b^2

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Rubi [A]  time = 0.24, antiderivative size = 170, normalized size of antiderivative = 1.00, 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 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 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 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 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 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*}

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Mathematica [C]  time = 0.11, size = 125, normalized size = 0.74 \[ -\frac {i d \sqrt {c+d x} e^{-\frac {i (a d+b c)}{d}} \left (\frac {e^{2 i a} \Gamma \left (\frac {5}{2},-\frac {i b (c+d x)}{d}\right )}{\sqrt {-\frac {i b (c+d x)}{d}}}-\frac {e^{\frac {2 i b c}{d}} \Gamma \left (\frac {5}{2},\frac {i b (c+d x)}{d}\right )}{\sqrt {\frac {i b (c+d x)}{d}}}\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*I)*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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fricas [A]  time = 0.64, size = 156, normalized size = 0.92 \[ -\frac {3 \, \sqrt {2} \pi d^{2} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 3 \, \sqrt {2} \pi d^{2} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) - 2 \, {\left (3 \, b d \sin \left (b x + a\right ) - 2 \, {\left (b^{2} d x + b^{2} c\right )} \cos \left (b x + a\right )\right )} \sqrt {d x + c}}{4 \, b^{3}} \]

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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giac [C]  time = 2.18, size = 779, normalized size = 4.58 \[ -\frac {4 \, {\left (\frac {i \, \sqrt {2} \sqrt {\pi } 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 )}} - \frac {i \, \sqrt {2} \sqrt {\pi } 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 )}}\right )} c^{2} + d^{2} {\left (\frac {\frac {i \, \sqrt {2} \sqrt {\pi } {\left (4 \, b^{2} c^{2} + 4 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 - 4 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}}}{d^{2}} + \frac {-\frac {i \, \sqrt {2} \sqrt {\pi } {\left (4 \, b^{2} c^{2} - 4 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 - 4 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}}}{d^{2}}\right )} + 4 \, {\left (-\frac {i \, \sqrt {2} \sqrt {\pi } {\left (2 \, b c + i \, d\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} + \frac {i \, \sqrt {2} \sqrt {\pi } {\left (2 \, b c - i \, d\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} + \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}{8 \, d} \]

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*(4*(I*sqrt(2)*sqrt(pi)*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)) - I*sqrt(2)*sqrt(pi)*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)))*c^2 + d^2*((I
*sqrt(2)*sqrt(pi)*(4*b^2*c^2 + 4*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 - 4
*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^2 + (-I*sqrt(2)*sq
rt(pi)*(4*b^2*c^2 - 4*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 - 4*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^2) + 4*(-I*sqrt(2)*sqrt(pi)
*(2*b*c + I*d)*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) + I*sqrt(2)*sqrt(pi)*(2*b*c - I*d)*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*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)/d

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maple [A]  time = 0.01, size = 188, normalized size = 1.11 \[ \frac {-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{2 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {d a -c b}{d}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {d a -c b}{d}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{4 b \sqrt {\frac {b}{d}}}\right )}{b}}{d} \]

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 = 1.76, size = 242, normalized size = 1.42 \[ -\frac {\sqrt {2} {\left (8 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} \cos \left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right ) - 12 \, \sqrt {2} \sqrt {d x + c} b d \sin \left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right ) - {\left (-\left (3 i + 3\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) + \left (3 i - 3\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) - {\left (\left (3 i - 3\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) - \left (3 i + 3\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right )\right )}}{16 \, b^{3}} \]

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*sqrt(2)*(8*sqrt(2)*(d*x + c)^(3/2)*b^2*cos(((d*x + c)*b - b*c + a*d)/d) - 12*sqrt(2)*sqrt(d*x + c)*b*d*s
in(((d*x + c)*b - b*c + a*d)/d) - (-(3*I + 3)*sqrt(pi)*d^2*(b^2/d^2)^(1/4)*cos(-(b*c - a*d)/d) + (3*I - 3)*sqr
t(pi)*d^2*(b^2/d^2)^(1/4)*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(I*b/d)) - ((3*I - 3)*sqrt(pi)*d^2*(b^2/d
^2)^(1/4)*cos(-(b*c - a*d)/d) - (3*I + 3)*sqrt(pi)*d^2*(b^2/d^2)^(1/4)*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*
sqrt(-I*b/d)))/b^3

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

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

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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