3.124 \(\int (c+d x)^{5/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx\)

Optimal. Leaf size=228 \[ -\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \sin \left (4 a-\frac {4 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}-\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \cos \left (4 a-\frac {4 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac {5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}+\frac {(c+d x)^{7/2}}{28 d} \]

[Out]

1/28*(d*x+c)^(7/2)/d-5/256*d*(d*x+c)^(3/2)*cos(4*b*x+4*a)/b^2-1/32*(d*x+c)^(5/2)*sin(4*b*x+4*a)/b-15/8192*d^(5
/2)*cos(4*a-4*b*c/d)*FresnelS(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^(7/2)-15/81
92*d^(5/2)*FresnelC(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^(7/2
)+15/2048*d^2*sin(4*b*x+4*a)*(d*x+c)^(1/2)/b^3

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Rubi [A]  time = 0.40, antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ -\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \sin \left (4 a-\frac {4 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}-\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \cos \left (4 a-\frac {4 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac {5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}+\frac {(c+d x)^{7/2}}{28 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)*Cos[a + b*x]^2*Sin[a + b*x]^2,x]

[Out]

(c + d*x)^(7/2)/(28*d) - (5*d*(c + d*x)^(3/2)*Cos[4*a + 4*b*x])/(256*b^2) - (15*d^(5/2)*Sqrt[Pi/2]*Cos[4*a - (
4*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(4096*b^(7/2)) - (15*d^(5/2)*Sqrt[Pi/2]*Fres
nelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(4096*b^(7/2)) + (15*d^2*Sqrt[c + d*x
]*Sin[4*a + 4*b*x])/(2048*b^3) - ((c + d*x)^(5/2)*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)^{5/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx &=\int \left (\frac {1}{8} (c+d x)^{5/2}-\frac {1}{8} (c+d x)^{5/2} \cos (4 a+4 b x)\right ) \, dx\\ &=\frac {(c+d x)^{7/2}}{28 d}-\frac {1}{8} \int (c+d x)^{5/2} \cos (4 a+4 b x) \, dx\\ &=\frac {(c+d x)^{7/2}}{28 d}-\frac {(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}+\frac {(5 d) \int (c+d x)^{3/2} \sin (4 a+4 b x) \, dx}{64 b}\\ &=\frac {(c+d x)^{7/2}}{28 d}-\frac {5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}+\frac {\left (15 d^2\right ) \int \sqrt {c+d x} \cos (4 a+4 b x) \, dx}{512 b^2}\\ &=\frac {(c+d x)^{7/2}}{28 d}-\frac {5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}+\frac {15 d^2 \sqrt {c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac {(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}-\frac {\left (15 d^3\right ) \int \frac {\sin (4 a+4 b x)}{\sqrt {c+d x}} \, dx}{4096 b^3}\\ &=\frac {(c+d x)^{7/2}}{28 d}-\frac {5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}+\frac {15 d^2 \sqrt {c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac {(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}-\frac {\left (15 d^3 \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}{4096 b^3}-\frac {\left (15 d^3 \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}{4096 b^3}\\ &=\frac {(c+d x)^{7/2}}{28 d}-\frac {5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}+\frac {15 d^2 \sqrt {c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac {(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}-\frac {\left (15 d^2 \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 )}{2048 b^3}-\frac {\left (15 d^2 \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 )}{2048 b^3}\\ &=\frac {(c+d x)^{7/2}}{28 d}-\frac {5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}-\frac {15 d^{5/2} \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 )}{4096 b^{7/2}}-\frac {15 d^{5/2} \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 )}{4096 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac {(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}\\ \end {align*}

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Mathematica [A]  time = 3.42, size = 206, normalized size = 0.90 \[ \frac {\sqrt {\frac {b}{d}} \left (4 \sqrt {\frac {b}{d}} \sqrt {c+d x} \left (-7 d \sin (4 (a+b x)) \left (64 b^2 (c+d x)^2-15 d^2\right )-280 b d^2 (c+d x) \cos (4 (a+b x))+512 b^3 (c+d x)^3\right )-105 \sqrt {2 \pi } d^3 \sin \left (4 a-\frac {4 b c}{d}\right ) C\left (2 \sqrt {\frac {b}{d}} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}\right )-105 \sqrt {2 \pi } d^3 \cos \left (4 a-\frac {4 b c}{d}\right ) S\left (2 \sqrt {\frac {b}{d}} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}\right )\right )}{57344 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/2)*Cos[a + b*x]^2*Sin[a + b*x]^2,x]

[Out]

(Sqrt[b/d]*(-105*d^3*Sqrt[2*Pi]*Cos[4*a - (4*b*c)/d]*FresnelS[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]] - 105*d^3*
Sqrt[2*Pi]*FresnelC[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]*(51
2*b^3*(c + d*x)^3 - 280*b*d^2*(c + d*x)*Cos[4*(a + b*x)] - 7*d*(-15*d^2 + 64*b^2*(c + d*x)^2)*Sin[4*(a + b*x)]
)))/(57344*b^4)

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fricas [A]  time = 0.51, size = 347, normalized size = 1.52 \[ -\frac {105 \, \sqrt {2} \pi d^{4} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 105 \, \sqrt {2} \pi d^{4} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - 16 \, {\left (128 \, b^{4} d^{3} x^{3} + 384 \, b^{4} c d^{2} x^{2} + 128 \, b^{4} c^{3} - 70 \, b^{2} c d^{2} - 560 \, {\left (b^{2} d^{3} x + b^{2} c d^{2}\right )} \cos \left (b x + a\right )^{4} + 560 \, {\left (b^{2} d^{3} x + b^{2} c d^{2}\right )} \cos \left (b x + a\right )^{2} + 2 \, {\left (192 \, b^{4} c^{2} d - 35 \, b^{2} d^{3}\right )} x - 7 \, {\left (2 \, {\left (64 \, b^{3} d^{3} x^{2} + 128 \, b^{3} c d^{2} x + 64 \, b^{3} c^{2} d - 15 \, b d^{3}\right )} \cos \left (b x + a\right )^{3} - {\left (64 \, b^{3} d^{3} x^{2} + 128 \, b^{3} c d^{2} x + 64 \, b^{3} c^{2} d - 15 \, b d^{3}\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{57344 \, b^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*cos(b*x+a)^2*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/57344*(105*sqrt(2)*pi*d^4*sqrt(b/(pi*d))*cos(-4*(b*c - a*d)/d)*fresnel_sin(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(
pi*d))) + 105*sqrt(2)*pi*d^4*sqrt(b/(pi*d))*fresnel_cos(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-4*(b*c -
a*d)/d) - 16*(128*b^4*d^3*x^3 + 384*b^4*c*d^2*x^2 + 128*b^4*c^3 - 70*b^2*c*d^2 - 560*(b^2*d^3*x + b^2*c*d^2)*c
os(b*x + a)^4 + 560*(b^2*d^3*x + b^2*c*d^2)*cos(b*x + a)^2 + 2*(192*b^4*c^2*d - 35*b^2*d^3)*x - 7*(2*(64*b^3*d
^3*x^2 + 128*b^3*c*d^2*x + 64*b^3*c^2*d - 15*b*d^3)*cos(b*x + a)^3 - (64*b^3*d^3*x^2 + 128*b^3*c*d^2*x + 64*b^
3*c^2*d - 15*b*d^3)*cos(b*x + a))*sin(b*x + a))*sqrt(d*x + c))/(b^4*d)

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giac [C]  time = 3.30, size = 1358, normalized size = 5.96 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*cos(b*x+a)^2*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/573440*(17920*(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^3 + 56*c*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*sq
rt(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)*s
qrt(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*sq
rt(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) + d^3*(4096*(5*
(d*x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2 - 35*sqrt(d*x + c)*c^3)/d^3 - 35*(sqrt(2)*sqrt
(pi)*(512*b^3*c^3 + 192*I*b^2*c^2*d - 72*b*c*d^2 - 15*I*d^3)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqr
t(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) + 4*(64*I*(d*x + c)^(5/
2)*b^2*d - 192*I*(d*x + c)^(3/2)*b^2*c*d + 192*I*sqrt(d*x + c)*b^2*c^2*d + 40*(d*x + c)^(3/2)*b*d^2 - 72*sqrt(
d*x + c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)*e^((-4*I*(d*x + c)*b + 4*I*b*c - 4*I*a*d)/d)/b^3)/d^3 - 35*(sqrt(2)
*sqrt(pi)*(512*b^3*c^3 - 192*I*b^2*c^2*d - 72*b*c*d^2 + 15*I*d^3)*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) + 4*(-64*I*(d*x
+ c)^(5/2)*b^2*d + 192*I*(d*x + c)^(3/2)*b^2*c*d - 192*I*sqrt(d*x + c)*b^2*c^2*d + 40*(d*x + c)^(3/2)*b*d^2 -
72*sqrt(d*x + c)*b*c*d^2 + 15*I*sqrt(d*x + c)*d^3)*e^((4*I*(d*x + c)*b - 4*I*b*c + 4*I*a*d)/d)/b^3)/d^3) - 22
40*(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^2)/d

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maple [A]  time = 0.05, size = 251, normalized size = 1.10 \[ \frac {\frac {\left (d x +c \right )^{\frac {7}{2}}}{28}-\frac {d \left (d x +c \right )^{\frac {5}{2}} \sin \left (\frac {4 \left (d x +c \right ) b}{d}+\frac {4 d a -4 c b}{d}\right )}{32 b}+\frac {5 d \left (-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {4 \left (d x +c \right ) b}{d}+\frac {4 d a -4 c b}{d}\right )}{8 b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {4 \left (d x +c \right ) b}{d}+\frac {4 d a -4 c b}{d}\right )}{8 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 d a -4 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {4 d a -4 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{32 b \sqrt {\frac {b}{d}}}\right )}{8 b}\right )}{32 b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)*cos(b*x+a)^2*sin(b*x+a)^2,x)

[Out]

2/d*(1/56*(d*x+c)^(7/2)-1/64/b*d*(d*x+c)^(5/2)*sin(4/d*(d*x+c)*b+4*(a*d-b*c)/d)+5/64/b*d*(-1/8/b*d*(d*x+c)^(3/
2)*cos(4/d*(d*x+c)*b+4*(a*d-b*c)/d)+3/8/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 = 0.47, size = 285, normalized size = 1.25 \[ \frac {\sqrt {2} {\left (\frac {4096 \, \sqrt {2} {\left (d x + c\right )}^{\frac {7}{2}} b^{4}}{d} - 2240 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d \cos \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - {\left (\left (105 i + 105\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - \left (105 i - 105\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) - {\left (-\left (105 i - 105\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + \left (105 i + 105\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right ) - 56 \, {\left (64 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3} - 15 \, \sqrt {2} \sqrt {d x + c} b d^{2}\right )} \sin \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right )\right )}}{229376 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*cos(b*x+a)^2*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/229376*sqrt(2)*(4096*sqrt(2)*(d*x + c)^(7/2)*b^4/d - 2240*sqrt(2)*(d*x + c)^(3/2)*b^2*d*cos(4*((d*x + c)*b -
b*c + a*d)/d) - ((105*I + 105)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*cos(-4*(b*c - a*d)/d) - (105*I - 105)*sqrt(pi)*d^
3*(b^2/d^2)^(1/4)*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(I*b/d)) - (-(105*I - 105)*sqrt(pi)*d^3*(b^2/
d^2)^(1/4)*cos(-4*(b*c - a*d)/d) + (105*I + 105)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*sin(-4*(b*c - a*d)/d))*erf(2*sqr
t(d*x + c)*sqrt(-I*b/d)) - 56*(64*sqrt(2)*(d*x + c)^(5/2)*b^3 - 15*sqrt(2)*sqrt(d*x + c)*b*d^2)*sin(4*((d*x +
c)*b - b*c + a*d)/d))/b^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^{5/2} \,d x \]

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)^(5/2),x)

[Out]

int(cos(a + b*x)^2*sin(a + b*x)^2*(c + d*x)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/2)*cos(b*x+a)**2*sin(b*x+a)**2,x)

[Out]

Timed out

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