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

Optimal. Leaf size=168 \[ -\frac{3 \sqrt{\pi } d^{3/2} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{32 b^{5/2}}-\frac{3 \sqrt{\pi } d^{3/2} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{32 b^{5/2}}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{16 b^2}-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b} \]

[Out]

-((c + d*x)^(3/2)*Cos[2*a + 2*b*x])/(4*b) - (3*d^(3/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[
c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(32*b^(5/2)) - (3*d^(3/2)*Sqrt[Pi]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*
Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(32*b^(5/2)) + (3*d*Sqrt[c + d*x]*Sin[2*a + 2*b*x])/(16*b^2)

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Rubi [A]  time = 0.29454, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {4406, 12, 3296, 3306, 3305, 3351, 3304, 3352} \[ -\frac{3 \sqrt{\pi } d^{3/2} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{32 b^{5/2}}-\frac{3 \sqrt{\pi } d^{3/2} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{32 b^{5/2}}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{16 b^2}-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c + d*x)^(3/2)*Cos[2*a + 2*b*x])/(4*b) - (3*d^(3/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[
c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(32*b^(5/2)) - (3*d^(3/2)*Sqrt[Pi]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*
Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(32*b^(5/2)) + (3*d*Sqrt[c + d*x]*Sin[2*a + 2*b*x])/(16*b^2)

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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} \cos (a+b x) \sin (a+b x) \, dx &=\int \frac{1}{2} (c+d x)^{3/2} \sin (2 a+2 b x) \, dx\\ &=\frac{1}{2} \int (c+d x)^{3/2} \sin (2 a+2 b x) \, dx\\ &=-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b}+\frac{(3 d) \int \sqrt{c+d x} \cos (2 a+2 b x) \, dx}{8 b}\\ &=-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{16 b^2}-\frac{\left (3 d^2\right ) \int \frac{\sin (2 a+2 b x)}{\sqrt{c+d x}} \, dx}{32 b^2}\\ &=-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{16 b^2}-\frac{\left (3 d^2 \cos \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{2 b c}{d}+2 b x\right )}{\sqrt{c+d x}} \, dx}{32 b^2}-\frac{\left (3 d^2 \sin \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{2 b c}{d}+2 b x\right )}{\sqrt{c+d x}} \, dx}{32 b^2}\\ &=-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{16 b^2}-\frac{\left (3 d \cos \left (2 a-\frac{2 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{16 b^2}-\frac{\left (3 d \sin \left (2 a-\frac{2 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{16 b^2}\\ &=-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b}-\frac{3 d^{3/2} \sqrt{\pi } \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{32 b^{5/2}}-\frac{3 d^{3/2} \sqrt{\pi } C\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right ) \sin \left (2 a-\frac{2 b c}{d}\right )}{32 b^{5/2}}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{16 b^2}\\ \end{align*}

Mathematica [A]  time = 0.886386, size = 157, normalized size = 0.93 \[ \frac{-3 \sqrt{\pi } d \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-3 \sqrt{\pi } d \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-2 \sqrt{\frac{b}{d}} \sqrt{c+d x} (4 b (c+d x) \cos (2 (a+b x))-3 d \sin (2 (a+b x)))}{32 d^2 \left (\frac{b}{d}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] - 3*d*Sqrt[Pi]*FresnelC[(2*
Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]]*Sin[2*a - (2*b*c)/d] - 2*Sqrt[b/d]*Sqrt[c + d*x]*(4*b*(c + d*x)*Cos[2*(a +
b*x)] - 3*d*Sin[2*(a + b*x)]))/(32*(b/d)^(5/2)*d^2)

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Maple [A]  time = 0.025, size = 187, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( -1/8\,{\frac{d \left ( dx+c \right ) ^{3/2}}{b}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{ad-bc}{d}} \right ) }+3/8\,{\frac{d}{b} \left ( 1/4\,{\frac{d\sqrt{dx+c}}{b}\sin \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{ad-bc}{d}} \right ) }-1/8\,{\frac{d\sqrt{\pi }}{b} \left ( \cos \left ( 2\,{\frac{ad-bc}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) +\sin \left ( 2\,{\frac{ad-bc}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\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)*cos(b*x+a)*sin(b*x+a),x)

[Out]

2/d*(-1/8/b*d*(d*x+c)^(3/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+3/8/b*d*(1/4/b*d*(d*x+c)^(1/2)*sin(2/d*(d*x+c)*b+
2*(a*d-b*c)/d)-1/8/b*d*Pi^(1/2)/(b/d)^(1/2)*(cos(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*
b/d)+sin(2*(a*d-b*c)/d)*FresnelC(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d))))

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Maxima [C]  time = 1.88766, size = 869, normalized size = 5.17 \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)*cos(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/256*sqrt(2)*(32*sqrt(2)*(d*x + c)^(3/2)*b*d*sqrt(abs(b)/abs(d))*cos(2*((d*x + c)*b - b*c + a*d)/d) - 24*sqr
t(2)*sqrt(d*x + c)*d^2*sqrt(abs(b)/abs(d))*sin(2*((d*x + c)*b - b*c + a*d)/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*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(-2*(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*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)*si
n(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^2*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqr
t(2*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*arctan2(0, d/sqrt(d^2))) - 3*sqrt(pi)*sin(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))))*d^2*cos(-
2*(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*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*s
in(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-2*I*b/d)))/(b^2*d*sqrt(abs(b)/abs(d)))

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Fricas [A]  time = 0.531847, size = 413, normalized size = 2.46 \begin{align*} -\frac{3 \, \pi d^{2} \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{S}\left (2 \, \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) + 3 \, \pi d^{2} \sqrt{\frac{b}{\pi d}} \operatorname{C}\left (2 \, \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) - 4 \,{\left (2 \, b^{2} d x + 3 \, b d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, b^{2} c - 4 \,{\left (b^{2} d x + b^{2} c\right )} \cos \left (b x + a\right )^{2}\right )} \sqrt{d x + c}}{32 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(3*pi*d^2*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d))) + 3*pi*d^2*sq
rt(b/(pi*d))*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - 4*(2*b^2*d*x + 3*b*d*cos(b*x
+ a)*sin(b*x + a) + 2*b^2*c - 4*(b^2*d*x + b^2*c)*cos(b*x + a)^2)*sqrt(d*x + c))/b^3

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Sympy [B]  time = 130.886, size = 665, normalized size = 3.96 \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)*cos(b*x+a)*sin(b*x+a),x)

[Out]

-5*sqrt(pi)*sqrt(d/b)*(c + d*x)**2*sin(2*a - 2*b*c/d)*fresnelc(2*sqrt(b)*sqrt(c + d*x)/(sqrt(pi)*sqrt(d)))*gam
ma(1/4)/(32*d*gamma(9/4)) + sqrt(pi)*sqrt(d/b)*(c + d*x)**2*sin(2*a - 2*b*c/d)*fresnelc(2*b*sqrt(c + d*x)/(sqr
t(pi)*d*sqrt(b/d)))/(2*d) - 21*sqrt(pi)*sqrt(d/b)*(c + d*x)**2*cos(2*a - 2*b*c/d)*fresnels(2*sqrt(b)*sqrt(c +
d*x)/(sqrt(pi)*sqrt(d)))*gamma(3/4)/(32*d*gamma(11/4)) + sqrt(pi)*sqrt(d/b)*(c + d*x)**2*cos(2*a - 2*b*c/d)*fr
esnels(2*b*sqrt(c + d*x)/(sqrt(pi)*d*sqrt(b/d)))/(2*d) - 15*sqrt(pi)*d*sqrt(d/b)*sin(2*a - 2*b*c/d)*fresnelc(2
*sqrt(b)*sqrt(c + d*x)/(sqrt(pi)*sqrt(d)))*gamma(1/4)/(512*b**2*gamma(9/4)) - 63*sqrt(pi)*d*sqrt(d/b)*cos(2*a
- 2*b*c/d)*fresnels(2*sqrt(b)*sqrt(c + d*x)/(sqrt(pi)*sqrt(d)))*gamma(3/4)/(512*b**2*gamma(11/4)) + 5*sqrt(d/b
)*(c + d*x)**(3/2)*sin(2*a - 2*b*c/d)*sin(2*b*c/d + 2*b*x)*gamma(1/4)/(64*sqrt(b)*sqrt(d)*gamma(9/4)) - 21*sqr
t(d/b)*(c + d*x)**(3/2)*cos(2*a - 2*b*c/d)*cos(2*b*c/d + 2*b*x)*gamma(3/4)/(64*sqrt(b)*sqrt(d)*gamma(11/4)) +
15*sqrt(d)*sqrt(d/b)*sqrt(c + d*x)*sin(2*a - 2*b*c/d)*cos(2*b*c/d + 2*b*x)*gamma(1/4)/(256*b**(3/2)*gamma(9/4)
) + 63*sqrt(d)*sqrt(d/b)*sqrt(c + d*x)*sin(2*b*c/d + 2*b*x)*cos(2*a - 2*b*c/d)*gamma(3/4)/(256*b**(3/2)*gamma(
11/4))

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Giac [C]  time = 1.22975, size = 732, normalized size = 4.36 \begin{align*} -\frac{4 \,{\left (\frac{\sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{b d} \sqrt{d x + c}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac{2 i \, b c - 2 i \, a d}{d}\right )}}{\sqrt{b d}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} + \frac{\sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{b d} \sqrt{d x + c}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac{-2 i \, b c + 2 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{2 i \,{\left (d x + c\right )} b - 2 i \, b c + 2 i \, a d}{d}\right )}}{b} + \frac{2 \, \sqrt{d x + c} d e^{\left (\frac{-2 i \,{\left (d x + c\right )} b + 2 i \, b c - 2 i \, a d}{d}\right )}}{b}\right )} c + \frac{i \, \sqrt{\pi }{\left (4 i \, b c d - 3 \, d^{2}\right )} d \operatorname{erf}\left (-\frac{\sqrt{b d} \sqrt{d x + c}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac{2 i \, b c - 2 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{\pi }{\left (4 i \, b c d + 3 \, d^{2}\right )} d \operatorname{erf}\left (-\frac{\sqrt{b d} \sqrt{d x + c}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac{-2 i \, b c + 2 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 (4 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{2 i \,{\left (d x + c\right )} b - 2 i \, b c + 2 i \, a d}{d}\right )}}{b^{2}} - \frac{2 i \,{\left (4 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{-2 i \,{\left (d x + c\right )} b + 2 i \, b c - 2 i \, a d}{d}\right )}}{b^{2}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(4*(sqrt(pi)*d^2*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*I*a*d)/d)/(sq
rt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + sqrt(pi)*d^2*erf(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)
*e^((-2*I*b*c + 2*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) + 2*sqrt(d*x + c)*d*e^((2*I*(d*x + c)*b -
2*I*b*c + 2*I*a*d)/d)/b + 2*sqrt(d*x + c)*d*e^((-2*I*(d*x + c)*b + 2*I*b*c - 2*I*a*d)/d)/b)*c + I*sqrt(pi)*(4
*I*b*c*d - 3*d^2)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*I*a*d)/d)/(sqrt(
b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b^2) + I*sqrt(pi)*(4*I*b*c*d + 3*d^2)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sq
rt(b^2*d^2) + 1)/d)*e^((-2*I*b*c + 2*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b^2) - 2*I*(4*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^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b^2 -
2*I*(4*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^((-2*I*(d*x + c)*b + 2*I*b*c -
2*I*a*d)/d)/b^2)/d