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

Optimal. Leaf size=200 \[ \frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \cos \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{512 b^{5/2}}-\frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \sin \left (4 a-\frac{4 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{512 b^{5/2}}-\frac{3 d \sqrt{c+d x} \cos (4 a+4 b x)}{256 b^2}-\frac{(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}+\frac{(c+d x)^{5/2}}{20 d} \]

[Out]

(c + d*x)^(5/2)/(20*d) - (3*d*Sqrt[c + d*x]*Cos[4*a + 4*b*x])/(256*b^2) + (3*d^(3/2)*Sqrt[Pi/2]*Cos[4*a - (4*b
*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(512*b^(5/2)) - (3*d^(3/2)*Sqrt[Pi/2]*FresnelS[
(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(512*b^(5/2)) - ((c + d*x)^(3/2)*Sin[4*a +
4*b*x])/(32*b)

________________________________________________________________________________________

Rubi [A]  time = 0.328737, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {4406, 3296, 3306, 3305, 3351, 3304, 3352} \[ \frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \cos \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{512 b^{5/2}}-\frac{3 \sqrt{\frac{\pi }{2}} d^{3/2} \sin \left (4 a-\frac{4 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{512 b^{5/2}}-\frac{3 d \sqrt{c+d x} \cos (4 a+4 b x)}{256 b^2}-\frac{(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}+\frac{(c+d x)^{5/2}}{20 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^(5/2)/(20*d) - (3*d*Sqrt[c + d*x]*Cos[4*a + 4*b*x])/(256*b^2) + (3*d^(3/2)*Sqrt[Pi/2]*Cos[4*a - (4*b
*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(512*b^(5/2)) - (3*d^(3/2)*Sqrt[Pi/2]*FresnelS[
(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(512*b^(5/2)) - ((c + d*x)^(3/2)*Sin[4*a +
4*b*x])/(32*b)

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 ^2(a+b x) \sin ^2(a+b x) \, dx &=\int \left (\frac{1}{8} (c+d x)^{3/2}-\frac{1}{8} (c+d x)^{3/2} \cos (4 a+4 b x)\right ) \, dx\\ &=\frac{(c+d x)^{5/2}}{20 d}-\frac{1}{8} \int (c+d x)^{3/2} \cos (4 a+4 b x) \, dx\\ &=\frac{(c+d x)^{5/2}}{20 d}-\frac{(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}+\frac{(3 d) \int \sqrt{c+d x} \sin (4 a+4 b x) \, dx}{64 b}\\ &=\frac{(c+d x)^{5/2}}{20 d}-\frac{3 d \sqrt{c+d x} \cos (4 a+4 b x)}{256 b^2}-\frac{(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}+\frac{\left (3 d^2\right ) \int \frac{\cos (4 a+4 b x)}{\sqrt{c+d x}} \, dx}{512 b^2}\\ &=\frac{(c+d x)^{5/2}}{20 d}-\frac{3 d \sqrt{c+d x} \cos (4 a+4 b x)}{256 b^2}-\frac{(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}+\frac{\left (3 d^2 \cos \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 (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{(c+d x)^{5/2}}{20 d}-\frac{3 d \sqrt{c+d x} \cos (4 a+4 b x)}{256 b^2}-\frac{(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}+\frac{\left (3 d \cos \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 (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{(c+d x)^{5/2}}{20 d}-\frac{3 d \sqrt{c+d x} \cos (4 a+4 b x)}{256 b^2}+\frac{3 d^{3/2} \sqrt{\frac{\pi }{2}} \cos \left (4 a-\frac{4 b c}{d}\right ) C\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{\frac{\pi }{2}} S\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{(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}\\ \end{align*}

Mathematica [A]  time = 3.15495, size = 187, normalized size = 0.94 \[ \frac{\sqrt{\frac{b}{d}} \left (15 \sqrt{2 \pi } d^2 \cos \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )-15 \sqrt{2 \pi } d^2 \sin \left (4 a-\frac{4 b c}{d}\right ) S\left (2 \sqrt{\frac{b}{d}} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}\right )+4 \sqrt{\frac{b}{d}} \sqrt{c+d x} \left (8 b (c+d x) (8 b (c+d x)-5 d \sin (4 (a+b x)))-15 d^2 \cos (4 (a+b x))\right )\right )}{5120 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b/d]*(15*d^2*Sqrt[2*Pi]*Cos[4*a - (4*b*c)/d]*FresnelC[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]] - 15*d^2*Sqr
t[2*Pi]*FresnelS[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]]*Sin[4*a - (4*b*c)/d] + 4*Sqrt[b/d]*Sqrt[c + d*x]*(-15*d
^2*Cos[4*(a + b*x)] + 8*b*(c + d*x)*(8*b*(c + d*x) - 5*d*Sin[4*(a + b*x)]))))/(5120*b^3)

________________________________________________________________________________________

Maple [A]  time = 0.036, size = 206, normalized size = 1. \begin{align*} 2\,{\frac{1}{d} \left ( 1/40\, \left ( dx+c \right ) ^{5/2}-{\frac{d \left ( dx+c \right ) ^{3/2}}{64\,b}\sin \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}\cos \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 FresnelC} \left ( 2\,{\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) -\sin \left ( 4\,{\frac{ad-bc}{d}} \right ){\it FresnelS} \left ( 2\,{\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)*cos(b*x+a)^2*sin(b*x+a)^2,x)

[Out]

2/d*(1/40*(d*x+c)^(5/2)-1/64/b*d*(d*x+c)^(3/2)*sin(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)*cos(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)*FresnelC(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)*FresnelS(2*2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^
(1/2)*b/d))))

________________________________________________________________________________________

Maxima [C]  time = 1.94746, size = 886, normalized size = 4.43 \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)^2*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/20480*(1024*(d*x + c)^(5/2)*b^2*sqrt(abs(b)/abs(d)) - 640*(d*x + c)^(3/2)*b*d*sqrt(abs(b)/abs(d))*sin(4*((d*
x + c)*b - b*c + a*d)/d) - 240*sqrt(d*x + c)*d^2*sqrt(abs(b)/abs(d))*cos(4*((d*x + c)*b - b*c + a*d)/d) + ((15
*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*sqrt(pi)*cos(-1/4*pi + 1/2*arctan
2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(
d^2))) + 15*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^2*cos(-4*(b*c - a*d)/
d) - (15*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*I*sqrt(pi)*cos(-1/4*pi
+ 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(
0, d/sqrt(d^2))) - 15*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^2*sin(-4*(b*c
- a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(I*b/d)) + ((15*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d
/sqrt(d^2))) + 15*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*I*sqrt(pi)*sin(
1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/
2*arctan2(0, d/sqrt(d^2))))*d^2*cos(-4*(b*c - a*d)/d) - (-15*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*a
rctan2(0, d/sqrt(d^2))) - 15*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*sq
rt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0
, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^2*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(-I*b/d)))/(b^2*d*sqrt
(abs(b)/abs(d)))

________________________________________________________________________________________

Fricas [A]  time = 0.597456, size = 621, normalized size = 3.1 \begin{align*} \frac{15 \, \sqrt{2} \pi d^{3} \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{C}\left (2 \, \sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - 15 \, \sqrt{2} \pi d^{3} \sqrt{\frac{b}{\pi d}} \operatorname{S}\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 ) + 4 \,{\left (64 \, b^{3} d^{2} x^{2} - 120 \, b d^{2} \cos \left (b x + a\right )^{4} + 128 \, b^{3} c d x + 64 \, b^{3} c^{2} + 120 \, b d^{2} \cos \left (b x + a\right )^{2} - 15 \, b d^{2} - 160 \,{\left (2 \,{\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )^{3} -{\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )\right )} \sqrt{d x + c}}{5120 \, b^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5120*(15*sqrt(2)*pi*d^3*sqrt(b/(pi*d))*cos(-4*(b*c - a*d)/d)*fresnel_cos(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*
d))) - 15*sqrt(2)*pi*d^3*sqrt(b/(pi*d))*fresnel_sin(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-4*(b*c - a*d)
/d) + 4*(64*b^3*d^2*x^2 - 120*b*d^2*cos(b*x + a)^4 + 128*b^3*c*d*x + 64*b^3*c^2 + 120*b*d^2*cos(b*x + a)^2 - 1
5*b*d^2 - 160*(2*(b^2*d^2*x + b^2*c*d)*cos(b*x + a)^3 - (b^2*d^2*x + b^2*c*d)*cos(b*x + a))*sin(b*x + a))*sqrt
(d*x + c))/(b^3*d)

________________________________________________________________________________________

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)**2*sin(b*x+a)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [C]  time = 1.45801, size = 805, normalized size = 4.02 \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)^2*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/30720*(1536*(d*x + c)^(5/2) - 2560*(d*x + c)^(3/2)*c - 40*(3*I*sqrt(2)*sqrt(pi)*d^2*erf(-sqrt(2)*sqrt(b*d)*s
qrt(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) -
3*I*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) - 64*(d*x + c)^(3/2) - 12*I*sqrt(d*x + c)*d*e^((4*I*(d*x + c
)*b - 4*I*b*c + 4*I*a*d)/d)/b + 12*I*sqrt(d*x + c)*d*e^((-4*I*(d*x + c)*b + 4*I*b*c - 4*I*a*d)/d)/b)*c + 15*sq
rt(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) + 15*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/s
qrt(b^2*d^2) + 1)*b^2) - 60*(-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 - 60*(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