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

Optimal. Leaf size=196 \[ -\frac {15 \sqrt {\pi } d^{5/2} \cos \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{128 b^{7/2}}+\frac {15 \sqrt {\pi } d^{5/2} \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{128 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{64 b^3}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b} \]

[Out]

-1/4*(d*x+c)^(5/2)*cos(2*b*x+2*a)/b+5/16*d*(d*x+c)^(3/2)*sin(2*b*x+2*a)/b^2-15/128*d^(5/2)*cos(2*a-2*b*c/d)*Fr
esnelC(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^(1/2))*Pi^(1/2)/b^(7/2)+15/128*d^(5/2)*FresnelS(2*b^(1/2)*(d*x+c)^(1
/2)/d^(1/2)/Pi^(1/2))*sin(2*a-2*b*c/d)*Pi^(1/2)/b^(7/2)+15/64*d^2*cos(2*b*x+2*a)*(d*x+c)^(1/2)/b^3

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Rubi [A]  time = 0.33, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ -\frac {15 \sqrt {\pi } d^{5/2} \cos \left (2 a-\frac {2 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {\pi } \sqrt {d}}\right )}{128 b^{7/2}}+\frac {15 \sqrt {\pi } d^{5/2} \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{128 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{64 b^3}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*d^2*Sqrt[c + d*x]*Cos[2*a + 2*b*x])/(64*b^3) - ((c + d*x)^(5/2)*Cos[2*a + 2*b*x])/(4*b) - (15*d^(5/2)*Sqrt
[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(128*b^(7/2)) + (15*d^(5/2)*
Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(128*b^(7/2)) + (5*d*(c
+ d*x)^(3/2)*Sin[2*a + 2*b*x])/(16*b^2)

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 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 (a+b x) \sin (a+b x) \, dx &=\int \frac {1}{2} (c+d x)^{5/2} \sin (2 a+2 b x) \, dx\\ &=\frac {1}{2} \int (c+d x)^{5/2} \sin (2 a+2 b x) \, dx\\ &=-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}+\frac {(5 d) \int (c+d x)^{3/2} \cos (2 a+2 b x) \, dx}{8 b}\\ &=-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac {\left (15 d^2\right ) \int \sqrt {c+d x} \sin (2 a+2 b x) \, dx}{32 b^2}\\ &=\frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{64 b^3}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac {\left (15 d^3\right ) \int \frac {\cos (2 a+2 b x)}{\sqrt {c+d x}} \, dx}{128 b^3}\\ &=\frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{64 b^3}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac {\left (15 d^3 \cos \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}{128 b^3}+\frac {\left (15 d^3 \sin \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}{128 b^3}\\ &=\frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{64 b^3}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac {\left (15 d^2 \cos \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 )}{64 b^3}+\frac {\left (15 d^2 \sin \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 )}{64 b^3}\\ &=\frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{64 b^3}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}-\frac {15 d^{5/2} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{128 b^{7/2}}+\frac {15 d^{5/2} \sqrt {\pi } S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{128 b^{7/2}}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}\\ \end {align*}

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Mathematica [A]  time = 1.06, size = 179, normalized size = 0.91 \[ \frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x} \left (20 b d (c+d x) \sin (2 (a+b x))-\cos (2 (a+b x)) \left (16 b^2 (c+d x)^2-15 d^2\right )\right )-15 \sqrt {\pi } d^2 \cos \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )+15 \sqrt {\pi } d^2 \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )}{128 b^3 \sqrt {\frac {b}{d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*d^2*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] + 15*d^2*Sqrt[Pi]*Fresne
lS[(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]*(-((-15*d^2 + 16*b^2
*(c + d*x)^2)*Cos[2*(a + b*x)]) + 20*b*d*(c + d*x)*Sin[2*(a + b*x)]))/(128*b^3*Sqrt[b/d])

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fricas [A]  time = 0.68, size = 222, normalized size = 1.13 \[ -\frac {15 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 15 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left (16 \, b^{3} d^{2} x^{2} + 32 \, b^{3} c d x + 16 \, b^{3} c^{2} - 15 \, b d^{2} - 2 \, {\left (16 \, b^{3} d^{2} x^{2} + 32 \, b^{3} c d x + 16 \, b^{3} c^{2} - 15 \, b d^{2}\right )} \cos \left (b x + a\right )^{2} + 40 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{128 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(15*pi*d^3*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d))) - 15*pi*d^3
*sqrt(b/(pi*d))*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - 2*(16*b^3*d^2*x^2 + 32*b^3
*c*d*x + 16*b^3*c^2 - 15*b*d^2 - 2*(16*b^3*d^2*x^2 + 32*b^3*c*d*x + 16*b^3*c^2 - 15*b*d^2)*cos(b*x + a)^2 + 40
*(b^2*d^2*x + b^2*c*d)*cos(b*x + a)*sin(b*x + a))*sqrt(d*x + c))/b^4

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giac [C]  time = 0.68, size = 1198, normalized size = 6.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/256*(64*(I*sqrt(pi)*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)) - I*sqrt(pi)*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)))*c^3 + 12*c*d^2*((I*sqrt(pi)*(16*b^2*c^2 +
8*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) - 2*I*(4*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^((-2*I*(d*x + c)*b + 2*I*b*c - 2*I*a*d)/d)/b^2)/d^2 + (-I*sqrt(pi)*(16*b^2*c^2 - 8*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) - 2*I*(4*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^((2*I*(d
*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b^2)/d^2) + d^3*((-I*sqrt(pi)*(64*b^3*c^3 + 48*I*b^2*c^2*d - 36*b*c*d^2 - 15
*I*d^3)*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^3) - 2*I*(16*I*(d*x + c)^(5/2)*b^2*d - 48*I*(d*x + c)^(3/2)*b^2*c*d + 48*I*sqrt(d*x + c
)*b^2*c^2*d + 20*(d*x + c)^(3/2)*b*d^2 - 36*sqrt(d*x + c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)*e^((-2*I*(d*x + c)
*b + 2*I*b*c - 2*I*a*d)/d)/b^3)/d^3 + (I*sqrt(pi)*(64*b^3*c^3 - 48*I*b^2*c^2*d - 36*b*c*d^2 + 15*I*d^3)*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^3) - 2*I*(16*I*(d*x + c)^(5/2)*b^2*d - 48*I*(d*x + c)^(3/2)*b^2*c*d + 48*I*sqrt(d*x + c)*b^2*c^2*d
- 20*(d*x + c)^(3/2)*b*d^2 + 36*sqrt(d*x + c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)*e^((2*I*(d*x + c)*b - 2*I*b*c
+ 2*I*a*d)/d)/b^3)/d^3) + 48*(-I*sqrt(pi)*(4*b*c + I*d)*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) + I*sqrt(pi)*(4*b*c - I*d)*d*erf(-sqr
t(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^2)/d

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maple [A]  time = 0.00, size = 234, normalized size = 1.19 \[ \frac {-\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}+\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}\right )}{4 b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-1/8/b*d*(d*x+c)^(5/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+5/8/b*d*(1/4/b*d*(d*x+c)^(3/2)*sin(2/d*(d*x+c)*b+
2*(a*d-b*c)/d)-3/4/b*d*(-1/4/b*d*(d*x+c)^(1/2)*cos(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)*FresnelC(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)-sin(2*(a*d-b*c)/d)*FresnelS(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 = 275, normalized size = 1.40 \[ \frac {\sqrt {2} {\left (160 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d \sin \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 8 \, {\left (16 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3} - 15 \, \sqrt {2} \sqrt {d x + c} b d^{2}\right )} \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) + {\left (\left (15 i - 15\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (15 i + 15\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) + {\left (-\left (15 i + 15\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (15 i - 15\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right )\right )}}{1024 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1024*sqrt(2)*(160*sqrt(2)*(d*x + c)^(3/2)*b^2*d*sin(2*((d*x + c)*b - b*c + a*d)/d) - 8*(16*sqrt(2)*(d*x + c)
^(5/2)*b^3 - 15*sqrt(2)*sqrt(d*x + c)*b*d^2)*cos(2*((d*x + c)*b - b*c + a*d)/d) + ((15*I - 15)*4^(1/4)*sqrt(pi
)*d^3*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) + (15*I + 15)*4^(1/4)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*sin(-2*(b*c - a
*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/d)) + (-(15*I + 15)*4^(1/4)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*
d)/d) - (15*I - 15)*4^(1/4)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-2*I*b/
d)))/b^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(a + b*x)*sin(a + b*x)*(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)*sin(b*x+a),x)

[Out]

Timed out

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