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

Optimal. Leaf size=168 \[ -\frac {3 \sqrt {\pi } d^{3/2} \sin \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{32 b^{5/2}}-\frac {3 \sqrt {\pi } d^{3/2} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{32 b^{5/2}}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{16 b^2}-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b} \]

[Out]

-1/4*(d*x+c)^(3/2)*cos(2*b*x+2*a)/b-3/32*d^(3/2)*cos(2*a-2*b*c/d)*FresnelS(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^
(1/2))*Pi^(1/2)/b^(5/2)-3/32*d^(3/2)*FresnelC(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^(5/2)+3/16*d*sin(2*b*x+2*a)*(d*x+c)^(1/2)/b^2

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Rubi [A]  time = 0.28, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ -\frac {3 \sqrt {\pi } d^{3/2} \sin \left (2 a-\frac {2 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {\pi } \sqrt {d}}\right )}{32 b^{5/2}}-\frac {3 \sqrt {\pi } d^{3/2} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{32 b^{5/2}}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{16 b^2}-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c + d*x)^(3/2)*Cos[2*a + 2*b*x])/(4*b) - (3*d^(3/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[
c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(32*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])/(32*b^(5/2)) + (3*d*Sqrt[c + d*x]*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)^{3/2} \cos (a+b x) \sin (a+b x) \, dx &=\int \frac {1}{2} (c+d x)^{3/2} \sin (2 a+2 b x) \, dx\\ &=\frac {1}{2} \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)}{4 b}+\frac {(3 d) \int \sqrt {c+d x} \cos (2 a+2 b x) \, dx}{8 b}\\ &=-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{16 b^2}-\frac {\left (3 d^2\right ) \int \frac {\sin (2 a+2 b x)}{\sqrt {c+d x}} \, dx}{32 b^2}\\ &=-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{16 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}{32 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}{32 b^2}\\ &=-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{16 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 )}{16 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 )}{16 b^2}\\ &=-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{4 b}-\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 )}{32 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 )}{32 b^{5/2}}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{16 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 157, normalized size = 0.93 \[ \frac {-3 \sqrt {\pi } d \sin \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )-3 \sqrt {\pi } d \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )-2 \sqrt {\frac {b}{d}} \sqrt {c+d x} (4 b (c+d x) \cos (2 (a+b x))-3 d \sin (2 (a+b x)))}{32 d^2 \left (\frac {b}{d}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 167, normalized size = 0.99 \[ -\frac {3 \, \pi d^{2} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 3 \, \pi d^{2} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 4 \, {\left (2 \, b^{2} d x + 3 \, b d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, b^{2} c - 4 \, {\left (b^{2} d x + b^{2} c\right )} \cos \left (b x + a\right )^{2}\right )} \sqrt {d x + c}}{32 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(3*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))) + 3*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) - 4*(2*b^2*d*x + 3*b*d*cos(b*x
+ a)*sin(b*x + a) + 2*b^2*c - 4*(b^2*d*x + b^2*c)*cos(b*x + a)^2)*sqrt(d*x + c))/b^3

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giac [C]  time = 1.02, size = 743, normalized size = 4.42 \[ -\frac {16 \, {\left (\frac {i \, \sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {2 i \, b c - 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}} - \frac {i \, \sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {-2 i \, b c + 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}\right )} c^{2} + d^{2} {\left (\frac {\frac {i \, \sqrt {\pi } {\left (16 \, b^{2} c^{2} + 8 i \, b c d - 3 \, d^{2}\right )} d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {2 i \, b c - 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b^{2}} - \frac {2 i \, {\left (4 i \, {\left (d x + c\right )}^{\frac {3}{2}} b d - 8 i \, \sqrt {d x + c} b c d + 3 \, \sqrt {d x + c} d^{2}\right )} e^{\left (\frac {-2 i \, {\left (d x + c\right )} b + 2 i \, b c - 2 i \, a d}{d}\right )}}{b^{2}}}{d^{2}} + \frac {-\frac {i \, \sqrt {\pi } {\left (16 \, b^{2} c^{2} - 8 i \, b c d - 3 \, d^{2}\right )} d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {-2 i \, b c + 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b^{2}} - \frac {2 i \, {\left (4 i \, {\left (d x + c\right )}^{\frac {3}{2}} b d - 8 i \, \sqrt {d x + c} b c d - 3 \, \sqrt {d x + c} d^{2}\right )} e^{\left (\frac {2 i \, {\left (d x + c\right )} b - 2 i \, b c + 2 i \, a d}{d}\right )}}{b^{2}}}{d^{2}}\right )} + 8 \, {\left (-\frac {i \, \sqrt {\pi } {\left (4 \, b c + i \, d\right )} d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {2 i \, b c - 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} + \frac {i \, \sqrt {\pi } {\left (4 \, b c - i \, d\right )} d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {-2 i \, b c + 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} + \frac {2 \, \sqrt {d x + c} d e^{\left (\frac {2 i \, {\left (d x + c\right )} b - 2 i \, b c + 2 i \, a d}{d}\right )}}{b} + \frac {2 \, \sqrt {d x + c} d e^{\left (\frac {-2 i \, {\left (d x + c\right )} b + 2 i \, b c - 2 i \, a d}{d}\right )}}{b}\right )} c}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(16*(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)/(s
qrt(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^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 + (-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) + 8*(-I*sqrt(pi)*(4*b*c + I*d)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqr
t(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(-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/s
qrt(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)/d

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maple [A]  time = 0.00, size = 187, normalized size = 1.11 \[ \frac {-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}-\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-1/8/b*d*(d*x+c)^(3/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+3/8/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))))

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maxima [C]  time = 0.55, size = 256, normalized size = 1.52 \[ -\frac {\sqrt {2} {\left (32 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 24 \, \sqrt {2} \sqrt {d x + c} b d \sin \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - {\left (-\left (3 i + 3\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (3 i - 3\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) - {\left (\left (3 i - 3\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (3 i + 3\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right )\right )}}{256 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/256*sqrt(2)*(32*sqrt(2)*(d*x + c)^(3/2)*b^2*cos(2*((d*x + c)*b - b*c + a*d)/d) - 24*sqrt(2)*sqrt(d*x + c)*b
*d*sin(2*((d*x + c)*b - b*c + a*d)/d) - (-(3*I + 3)*4^(1/4)*sqrt(pi)*d^2*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d)
+ (3*I - 3)*4^(1/4)*sqrt(pi)*d^2*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/d)) - ((
3*I - 3)*4^(1/4)*sqrt(pi)*d^2*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) - (3*I + 3)*4^(1/4)*sqrt(pi)*d^2*(b^2/d^2)
^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-2*I*b/d)))/b^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 42.22, size = 665, normalized size = 3.96 \[ - \frac {5 \sqrt {\pi } \sqrt {\frac {d}{b}} \left (c + d x\right )^{2} \sin {\left (2 a - \frac {2 b c}{d} \right )} C\left (\frac {2 \sqrt {b} \sqrt {c + d x}}{\sqrt {\pi } \sqrt {d}}\right ) \Gamma \left (\frac {1}{4}\right )}{32 d \Gamma \left (\frac {9}{4}\right )} + \frac {\sqrt {\pi } \sqrt {\frac {d}{b}} \left (c + d x\right )^{2} \sin {\left (2 a - \frac {2 b c}{d} \right )} C\left (\frac {2 b \sqrt {c + d x}}{\sqrt {\pi } d \sqrt {\frac {b}{d}}}\right )}{2 d} - \frac {21 \sqrt {\pi } \sqrt {\frac {d}{b}} \left (c + d x\right )^{2} \cos {\left (2 a - \frac {2 b c}{d} \right )} S\left (\frac {2 \sqrt {b} \sqrt {c + d x}}{\sqrt {\pi } \sqrt {d}}\right ) \Gamma \left (\frac {3}{4}\right )}{32 d \Gamma \left (\frac {11}{4}\right )} + \frac {\sqrt {\pi } \sqrt {\frac {d}{b}} \left (c + d x\right )^{2} \cos {\left (2 a - \frac {2 b c}{d} \right )} S\left (\frac {2 b \sqrt {c + d x}}{\sqrt {\pi } d \sqrt {\frac {b}{d}}}\right )}{2 d} - \frac {15 \sqrt {\pi } d \sqrt {\frac {d}{b}} \sin {\left (2 a - \frac {2 b c}{d} \right )} C\left (\frac {2 \sqrt {b} \sqrt {c + d x}}{\sqrt {\pi } \sqrt {d}}\right ) \Gamma \left (\frac {1}{4}\right )}{512 b^{2} \Gamma \left (\frac {9}{4}\right )} - \frac {63 \sqrt {\pi } d \sqrt {\frac {d}{b}} \cos {\left (2 a - \frac {2 b c}{d} \right )} S\left (\frac {2 \sqrt {b} \sqrt {c + d x}}{\sqrt {\pi } \sqrt {d}}\right ) \Gamma \left (\frac {3}{4}\right )}{512 b^{2} \Gamma \left (\frac {11}{4}\right )} + \frac {5 \sqrt {\frac {d}{b}} \left (c + d x\right )^{\frac {3}{2}} \sin {\left (2 a - \frac {2 b c}{d} \right )} \sin {\left (\frac {2 b c}{d} + 2 b x \right )} \Gamma \left (\frac {1}{4}\right )}{64 \sqrt {b} \sqrt {d} \Gamma \left (\frac {9}{4}\right )} - \frac {21 \sqrt {\frac {d}{b}} \left (c + d x\right )^{\frac {3}{2}} \cos {\left (2 a - \frac {2 b c}{d} \right )} \cos {\left (\frac {2 b c}{d} + 2 b x \right )} \Gamma \left (\frac {3}{4}\right )}{64 \sqrt {b} \sqrt {d} \Gamma \left (\frac {11}{4}\right )} + \frac {15 \sqrt {d} \sqrt {\frac {d}{b}} \sqrt {c + d x} \sin {\left (2 a - \frac {2 b c}{d} \right )} \cos {\left (\frac {2 b c}{d} + 2 b x \right )} \Gamma \left (\frac {1}{4}\right )}{256 b^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {63 \sqrt {d} \sqrt {\frac {d}{b}} \sqrt {c + d x} \sin {\left (\frac {2 b c}{d} + 2 b x \right )} \cos {\left (2 a - \frac {2 b c}{d} \right )} \Gamma \left (\frac {3}{4}\right )}{256 b^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*sqrt(pi)*sqrt(d/b)*(c + d*x)**2*sin(2*a - 2*b*c/d)*fresnelc(2*sqrt(b)*sqrt(c + d*x)/(sqrt(pi)*sqrt(d)))*gam
ma(1/4)/(32*d*gamma(9/4)) + sqrt(pi)*sqrt(d/b)*(c + d*x)**2*sin(2*a - 2*b*c/d)*fresnelc(2*b*sqrt(c + d*x)/(sqr
t(pi)*d*sqrt(b/d)))/(2*d) - 21*sqrt(pi)*sqrt(d/b)*(c + d*x)**2*cos(2*a - 2*b*c/d)*fresnels(2*sqrt(b)*sqrt(c +
d*x)/(sqrt(pi)*sqrt(d)))*gamma(3/4)/(32*d*gamma(11/4)) + sqrt(pi)*sqrt(d/b)*(c + d*x)**2*cos(2*a - 2*b*c/d)*fr
esnels(2*b*sqrt(c + d*x)/(sqrt(pi)*d*sqrt(b/d)))/(2*d) - 15*sqrt(pi)*d*sqrt(d/b)*sin(2*a - 2*b*c/d)*fresnelc(2
*sqrt(b)*sqrt(c + d*x)/(sqrt(pi)*sqrt(d)))*gamma(1/4)/(512*b**2*gamma(9/4)) - 63*sqrt(pi)*d*sqrt(d/b)*cos(2*a
- 2*b*c/d)*fresnels(2*sqrt(b)*sqrt(c + d*x)/(sqrt(pi)*sqrt(d)))*gamma(3/4)/(512*b**2*gamma(11/4)) + 5*sqrt(d/b
)*(c + d*x)**(3/2)*sin(2*a - 2*b*c/d)*sin(2*b*c/d + 2*b*x)*gamma(1/4)/(64*sqrt(b)*sqrt(d)*gamma(9/4)) - 21*sqr
t(d/b)*(c + d*x)**(3/2)*cos(2*a - 2*b*c/d)*cos(2*b*c/d + 2*b*x)*gamma(3/4)/(64*sqrt(b)*sqrt(d)*gamma(11/4)) +
15*sqrt(d)*sqrt(d/b)*sqrt(c + d*x)*sin(2*a - 2*b*c/d)*cos(2*b*c/d + 2*b*x)*gamma(1/4)/(256*b**(3/2)*gamma(9/4)
) + 63*sqrt(d)*sqrt(d/b)*sqrt(c + d*x)*sin(2*b*c/d + 2*b*x)*cos(2*a - 2*b*c/d)*gamma(3/4)/(256*b**(3/2)*gamma(
11/4))

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