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

Optimal. Leaf size=351 \[ -\frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \sin \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{512 b^{5/2}}-\frac{3 \sqrt{\pi } d^{3/2} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{64 b^{5/2}}-\frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \cos \left (4 a-\frac{4 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{512 b^{5/2}}-\frac{3 \sqrt{\pi } d^{3/2} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{64 b^{5/2}}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{32 b^2}+\frac{3 d \sqrt{c+d x} \sin (4 a+4 b x)}{256 b^2}-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}-\frac{(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b} \]

[Out]

-((c + d*x)^(3/2)*Cos[2*a + 2*b*x])/(8*b) - ((c + d*x)^(3/2)*Cos[4*a + 4*b*x])/(32*b) - (3*d^(3/2)*Sqrt[Pi/2]*
Cos[4*a - (4*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(512*b^(5/2)) - (3*d^(3/2)*Sqrt[P
i]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(64*b^(5/2)) - (3*d^(3/2)*Sqrt
[Pi/2]*FresnelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(512*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])/(64*b^(5/2)) + (3*d*Sqr
t[c + d*x]*Sin[2*a + 2*b*x])/(32*b^2) + (3*d*Sqrt[c + d*x]*Sin[4*a + 4*b*x])/(256*b^2)

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Rubi [A]  time = 0.572442, antiderivative size = 351, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {4406, 3296, 3306, 3305, 3351, 3304, 3352} \[ -\frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \sin \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{512 b^{5/2}}-\frac{3 \sqrt{\pi } d^{3/2} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{64 b^{5/2}}-\frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \cos \left (4 a-\frac{4 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{512 b^{5/2}}-\frac{3 \sqrt{\pi } d^{3/2} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{64 b^{5/2}}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{32 b^2}+\frac{3 d \sqrt{c+d x} \sin (4 a+4 b x)}{256 b^2}-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}-\frac{(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c + d*x)^(3/2)*Cos[2*a + 2*b*x])/(8*b) - ((c + d*x)^(3/2)*Cos[4*a + 4*b*x])/(32*b) - (3*d^(3/2)*Sqrt[Pi/2]*
Cos[4*a - (4*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(512*b^(5/2)) - (3*d^(3/2)*Sqrt[P
i]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(64*b^(5/2)) - (3*d^(3/2)*Sqrt
[Pi/2]*FresnelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(512*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])/(64*b^(5/2)) + (3*d*Sqr
t[c + d*x]*Sin[2*a + 2*b*x])/(32*b^2) + (3*d*Sqrt[c + d*x]*Sin[4*a + 4*b*x])/(256*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 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 ^3(a+b x) \sin (a+b x) \, dx &=\int \left (\frac{1}{4} (c+d x)^{3/2} \sin (2 a+2 b x)+\frac{1}{8} (c+d x)^{3/2} \sin (4 a+4 b x)\right ) \, dx\\ &=\frac{1}{8} \int (c+d x)^{3/2} \sin (4 a+4 b x) \, dx+\frac{1}{4} \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)}{8 b}-\frac{(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b}+\frac{(3 d) \int \sqrt{c+d x} \cos (4 a+4 b x) \, dx}{64 b}+\frac{(3 d) \int \sqrt{c+d x} \cos (2 a+2 b x) \, dx}{16 b}\\ &=-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}-\frac{(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{32 b^2}+\frac{3 d \sqrt{c+d x} \sin (4 a+4 b x)}{256 b^2}-\frac{\left (3 d^2\right ) \int \frac{\sin (4 a+4 b x)}{\sqrt{c+d x}} \, dx}{512 b^2}-\frac{\left (3 d^2\right ) \int \frac{\sin (2 a+2 b x)}{\sqrt{c+d x}} \, dx}{64 b^2}\\ &=-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}-\frac{(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{32 b^2}+\frac{3 d \sqrt{c+d x} \sin (4 a+4 b x)}{256 b^2}-\frac{\left (3 d^2 \cos \left (4 a-\frac{4 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{4 b c}{d}+4 b x\right )}{\sqrt{c+d x}} \, dx}{512 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}{64 b^2}-\frac{\left (3 d^2 \sin \left (4 a-\frac{4 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{4 b c}{d}+4 b x\right )}{\sqrt{c+d x}} \, dx}{512 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}{64 b^2}\\ &=-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}-\frac{(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{32 b^2}+\frac{3 d \sqrt{c+d x} \sin (4 a+4 b x)}{256 b^2}-\frac{\left (3 d \cos \left (4 a-\frac{4 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{4 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{256 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 )}{32 b^2}-\frac{\left (3 d \sin \left (4 a-\frac{4 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{4 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{256 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 )}{32 b^2}\\ &=-\frac{(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}-\frac{(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b}-\frac{3 d^{3/2} \sqrt{\frac{\pi }{2}} \cos \left (4 a-\frac{4 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{512 b^{5/2}}-\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 )}{64 b^{5/2}}-\frac{3 d^{3/2} \sqrt{\frac{\pi }{2}} C\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (4 a-\frac{4 b c}{d}\right )}{512 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 )}{64 b^{5/2}}+\frac{3 d \sqrt{c+d x} \sin (2 a+2 b x)}{32 b^2}+\frac{3 d \sqrt{c+d x} \sin (4 a+4 b x)}{256 b^2}\\ \end{align*}

Mathematica [A]  time = 1.61926, size = 393, normalized size = 1.12 \[ \frac{-3 \sqrt{2 \pi } d \sin \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )-48 \sqrt{\pi } d \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-3 \sqrt{2 \pi } d \cos \left (4 a-\frac{4 b c}{d}\right ) S\left (2 \sqrt{\frac{b}{d}} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}\right )-48 \sqrt{\pi } d \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+96 d \sqrt{\frac{b}{d}} \sqrt{c+d x} \sin (2 (a+b x))+12 d \sqrt{\frac{b}{d}} \sqrt{c+d x} \sin (4 (a+b x))-128 b d x \sqrt{\frac{b}{d}} \sqrt{c+d x} \cos (2 (a+b x))-128 b c \sqrt{\frac{b}{d}} \sqrt{c+d x} \cos (2 (a+b x))-32 b d x \sqrt{\frac{b}{d}} \sqrt{c+d x} \cos (4 (a+b x))-32 b c \sqrt{\frac{b}{d}} \sqrt{c+d x} \cos (4 (a+b x))}{1024 b^2 \sqrt{\frac{b}{d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-128*b*c*Sqrt[b/d]*Sqrt[c + d*x]*Cos[2*(a + b*x)] - 128*b*Sqrt[b/d]*d*x*Sqrt[c + d*x]*Cos[2*(a + b*x)] - 32*b
*c*Sqrt[b/d]*Sqrt[c + d*x]*Cos[4*(a + b*x)] - 32*b*Sqrt[b/d]*d*x*Sqrt[c + d*x]*Cos[4*(a + b*x)] - 3*d*Sqrt[2*P
i]*Cos[4*a - (4*b*c)/d]*FresnelS[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]] - 48*d*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*Fr
esnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] - 3*d*Sqrt[2*Pi]*FresnelC[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]]*S
in[4*a - (4*b*c)/d] - 48*d*Sqrt[Pi]*FresnelC[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]]*Sin[2*a - (2*b*c)/d] + 96*S
qrt[b/d]*d*Sqrt[c + d*x]*Sin[2*(a + b*x)] + 12*Sqrt[b/d]*d*Sqrt[c + d*x]*Sin[4*(a + b*x)])/(1024*b^2*Sqrt[b/d]
)

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Maple [A]  time = 0.033, size = 376, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( -1/16\,{\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 ) }+3/16\,{\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 ) }-{\frac{d \left ( dx+c \right ) ^{3/2}}{64\,b}\cos \left ( 4\,{\frac{ \left ( dx+c \right ) b}{d}}+4\,{\frac{ad-bc}{d}} \right ) }+{\frac{3\,d}{64\,b} \left ( 1/8\,{\frac{d\sqrt{dx+c}}{b}\sin \left ( 4\,{\frac{ \left ( dx+c \right ) b}{d}}+4\,{\frac{ad-bc}{d}} \right ) }-1/32\,{\frac{d\sqrt{2}\sqrt{\pi }}{b} \left ( \cos \left ( 4\,{\frac{ad-bc}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{2}\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) +\sin \left ( 4\,{\frac{ad-bc}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{2}\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)^3*sin(b*x+a),x)

[Out]

2/d*(-1/16/b*d*(d*x+c)^(3/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+3/16/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)))-1/64/b*d*(d*x+c)^(3/2)*cos(4/d*
(d*x+c)*b+4*(a*d-b*c)/d)+3/64/b*d*(1/8/b*d*(d*x+c)^(1/2)*sin(4/d*(d*x+c)*b+4*(a*d-b*c)/d)-1/32/b*d*2^(1/2)*Pi^
(1/2)/(b/d)^(1/2)*(cos(4*(a*d-b*c)/d)*FresnelS(2*2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)+sin(4*(a*d-b*
c)/d)*FresnelC(2*2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d))))

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Maxima [C]  time = 2.2755, size = 1806, normalized size = 5.15 \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)^3*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/8192*sqrt(2)*(128*sqrt(2)*(d*x + c)^(3/2)*b*d*abs(b)*cos(4*((d*x + c)*b - b*c + a*d)/d)/abs(d) + 512*sqrt(2
)*(d*x + c)^(3/2)*b*d*abs(b)*cos(2*((d*x + c)*b - b*c + a*d)/d)/abs(d) - 48*sqrt(2)*sqrt(d*x + c)*d^2*abs(b)*s
in(4*((d*x + c)*b - b*c + a*d)/d)/abs(d) - 384*sqrt(2)*sqrt(d*x + c)*d^2*abs(b)*sin(2*((d*x + c)*b - b*c + a*d
)/d)/abs(d) - ((-48*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 48*I*sqrt(pi)*c
os(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 48*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1
/2*arctan2(0, d/sqrt(d^2))) + 48*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^2*
sqrt(abs(b)/abs(d))*cos(-2*(b*c - a*d)/d) - (48*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqr
t(d^2))) + 48*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 48*I*sqrt(pi)*sin(1/4*
pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 48*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*ar
ctan2(0, d/sqrt(d^2))))*d^2*sqrt(abs(b)/abs(d))*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/d)) - (sqr
t(2)*(-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*sqrt(abs(b)/ab
s(d))*cos(-4*(b*c - a*d)/d) - sqrt(2)*(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*sqrt(abs(b)/abs(d))*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(I*b/d)) - (sqrt(2)*(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*arctan
2(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*sqrt(abs(b)/abs(d))*cos(-
4*(b*c - a*d)/d) - sqrt(2)*(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*sqrt(abs(b)/abs(d))*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(-I*b/d)) - ((48*I*sqrt(pi)*cos(1/4*
pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 48*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*ar
ctan2(0, d/sqrt(d^2))) - 48*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 48*sqrt(p
i)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^2*sqrt(abs(b)/abs(d))*cos(-2*(b*c - a*d)/
d) - (48*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 48*sqrt(pi)*cos(-1/4*pi + 1/
2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 48*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0,
d/sqrt(d^2))) - 48*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^2*sqrt(abs(b)
/abs(d))*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-2*I*b/d)))*abs(d)/(b^2*d*abs(b))

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Fricas [A]  time = 0.63998, size = 749, normalized size = 2.13 \begin{align*} -\frac{3 \, \sqrt{2} \pi d^{2} \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{S}\left (2 \, \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 (2 \, \sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right ) + 48 \, \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 ) + 48 \, \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 ) + 16 \,{\left (16 \,{\left (b^{2} d x + b^{2} c\right )} \cos \left (b x + a\right )^{4} - 6 \, b^{2} d x - 6 \, b^{2} c - 3 \,{\left (2 \, b d \cos \left (b x + a\right )^{3} + 3 \, b d \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )\right )} \sqrt{d x + c}}{1024 \, 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)^3*sin(b*x+a),x, algorithm="fricas")

[Out]

-1/1024*(3*sqrt(2)*pi*d^2*sqrt(b/(pi*d))*cos(-4*(b*c - a*d)/d)*fresnel_sin(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*
d))) + 3*sqrt(2)*pi*d^2*sqrt(b/(pi*d))*fresnel_cos(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-4*(b*c - a*d)/
d) + 48*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))) + 48*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) + 16*(16*(b^2*d*x + b^2*c)*cos(
b*x + a)^4 - 6*b^2*d*x - 6*b^2*c - 3*(2*b*d*cos(b*x + a)^3 + 3*b*d*cos(b*x + a))*sin(b*x + a))*sqrt(d*x + c))/
b^3

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

[Out]

Timed out

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Giac [C]  time = 1.53742, size = 1485, normalized size = 4.23 \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)^3*sin(b*x+a),x, algorithm="giac")

[Out]

-1/2048*(8*(sqrt(2)*sqrt(pi)*d^2*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((4*I*b*c
- 4*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + sqrt(2)*sqrt(pi)*d^2*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x
+ c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-4*I*b*c + 4*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) + 8*sqr
t(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) + 8*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) + 4*sqrt(d*x + c)*d*e^((4*I*(d*x + c)*b - 4*I*b*c +
4*I*a*d)/d)/b + 16*sqrt(d*x + c)*d*e^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b + 16*sqrt(d*x + c)*d*e^((-2*I
*(d*x + c)*b + 2*I*b*c - 2*I*a*d)/d)/b + 4*sqrt(d*x + c)*d*e^((-4*I*(d*x + c)*b + 4*I*b*c - 4*I*a*d)/d)/b)*c +
I*sqrt(2)*sqrt(pi)*(8*I*b*c*d - 3*d^2)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^
((4*I*b*c - 4*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b^2) + I*sqrt(2)*sqrt(pi)*(8*I*b*c*d + 3*d^2)*d*e
rf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-4*I*b*c + 4*I*a*d)/d)/(sqrt(b*d)*(-I*b*
d/sqrt(b^2*d^2) + 1)*b^2) + 16*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) + 16*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) - 4*I*(8*I*(d*x + c)^(3/2)*b*d - 8*I*sqrt(d*x + c)*b*c*d - 3*sqrt(d*x + c)*d^2)*e^(
(4*I*(d*x + c)*b - 4*I*b*c + 4*I*a*d)/d)/b^2 - 32*I*(4*I*(d*x + c)^(3/2)*b*d - 4*I*sqrt(d*x + c)*b*c*d - 3*sqr
t(d*x + c)*d^2)*e^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b^2 - 32*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 - 4*I*(8*I*(d*x + c)^(3/2)
*b*d - 8*I*sqrt(d*x + c)*b*c*d + 3*sqrt(d*x + c)*d^2)*e^((-4*I*(d*x + c)*b + 4*I*b*c - 4*I*a*d)/d)/b^2)/d