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

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

[Out]

1/20*(d*x+c)^(5/2)/d-1/32*(d*x+c)^(3/2)*sin(4*b*x+4*a)/b+3/1024*d^(3/2)*cos(4*a-4*b*c/d)*FresnelC(2*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)-3/1024*d^(3/2)*FresnelS(2*b^(1/2)*2^(1/2)/Pi^(1
/2)*(d*x+c)^(1/2)/d^(1/2))*sin(4*a-4*b*c/d)*2^(1/2)*Pi^(1/2)/b^(5/2)-3/256*d*cos(4*b*x+4*a)*(d*x+c)^(1/2)/b^2

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Rubi [A]  time = 0.32, antiderivative size = 200, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 1.27, size = 187, normalized size = 0.94 \[ \frac {\sqrt {\frac {b}{d}} \left (15 \sqrt {2 \pi } d^2 \cos \left (4 a-\frac {4 b c}{d}\right ) C\left (2 \sqrt {\frac {b}{d}} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}\right )-15 \sqrt {2 \pi } d^2 \sin \left (4 a-\frac {4 b c}{d}\right ) S\left (2 \sqrt {\frac {b}{d}} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}\right )+4 \sqrt {\frac {b}{d}} \sqrt {c+d x} \left (8 b (c+d x) (8 b (c+d x)-5 d \sin (4 (a+b x)))-15 d^2 \cos (4 (a+b x))\right )\right )}{5120 b^3} \]

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)

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fricas [A]  time = 0.52, size = 249, normalized size = 1.24 \[ \frac {15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + 4 \, {\left (64 \, b^{3} d^{2} x^{2} - 120 \, b d^{2} \cos \left (b x + a\right )^{4} + 128 \, b^{3} c d x + 64 \, b^{3} c^{2} + 120 \, b d^{2} \cos \left (b x + a\right )^{2} - 15 \, b d^{2} - 160 \, {\left (2 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )^{3} - {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{5120 \, b^{3} d} \]

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)

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giac [C]  time = 4.12, size = 842, normalized size = 4.21 \[ \text {result too large to display} \]

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*(960*(sqrt(2)*sqrt(pi)*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)) + sqrt(2)*sqrt(pi)*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)) + 8*sqrt(d*x
+ c))*c^2 + d^2*(512*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)/d^2 + 15*(sqrt(2)*sqrt(
pi)*(64*b^2*c^2 + 16*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) - 4*(8*I*(d*x + c)^(3/2)*b*d - 16*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^2 + 15*(sqrt(2)*sqrt(pi)*
(64*b^2*c^2 - 16*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) - 4*(-8*I*(d*x + c)^(3/2)*b*d + 16*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^2) - 80*(3*sqrt(2)*sqrt(pi
)*(8*b*c + I*d)*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) + 3*sqrt(2)*sqrt(pi)*(8*b*c - I*d)*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) - 64*(d*
x + c)^(3/2) + 192*sqrt(d*x + c)*c - 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)/d

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maple [A]  time = 0.00, size = 206, normalized size = 1.03 \[ \frac {\frac {\left (d x +c \right )^{\frac {5}{2}}}{20}-\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {4 \left (d x +c \right ) b}{d}+\frac {4 d a -4 c b}{d}\right )}{32 b}+\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {4 \left (d x +c \right ) b}{d}+\frac {4 d a -4 c b}{d}\right )}{8 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 d a -4 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {4 d a -4 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{32 b \sqrt {\frac {b}{d}}}\right )}{32 b}}{d} \]

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))))

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maxima [C]  time = 0.78, size = 264, normalized size = 1.32 \[ \frac {\sqrt {2} {\left (\frac {512 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3}}{d} - 320 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} \sin \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 120 \, \sqrt {2} \sqrt {d x + c} b d \cos \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - {\left (\left (15 i - 15\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + \left (15 i + 15\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) - {\left (-\left (15 i + 15\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - \left (15 i - 15\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right )\right )}}{20480 \, b^{3}} \]

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*sqrt(2)*(512*sqrt(2)*(d*x + c)^(5/2)*b^3/d - 320*sqrt(2)*(d*x + c)^(3/2)*b^2*sin(4*((d*x + c)*b - b*c
+ a*d)/d) - 120*sqrt(2)*sqrt(d*x + c)*b*d*cos(4*((d*x + c)*b - b*c + a*d)/d) - ((15*I - 15)*sqrt(pi)*d^2*(b^2/
d^2)^(1/4)*cos(-4*(b*c - a*d)/d) + (15*I + 15)*sqrt(pi)*d^2*(b^2/d^2)^(1/4)*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(
d*x + c)*sqrt(I*b/d)) - (-(15*I + 15)*sqrt(pi)*d^2*(b^2/d^2)^(1/4)*cos(-4*(b*c - a*d)/d) - (15*I - 15)*sqrt(pi
)*d^2*(b^2/d^2)^(1/4)*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(-I*b/d)))/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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]

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

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