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

Optimal. Leaf size=203 \[ \frac {3 \sqrt {\pi } d^{3/2} \cos \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{32 b^{5/2}}-\frac {3 \sqrt {\pi } d^{3/2} \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{32 b^{5/2}}+\frac {3 d \sqrt {c+d x} \sin ^2(a+b x)}{8 b^2}-\frac {(c+d x)^{3/2} \sin (a+b x) \cos (a+b x)}{2 b}-\frac {3 d \sqrt {c+d x}}{16 b^2}+\frac {(c+d x)^{5/2}}{5 d} \]

[Out]

1/5*(d*x+c)^(5/2)/d-1/2*(d*x+c)^(3/2)*cos(b*x+a)*sin(b*x+a)/b+3/32*d^(3/2)*cos(2*a-2*b*c/d)*FresnelC(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)*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^(5/2)-3/16*d*(d*x+c)^(1/2)/b^2+3/8*d*sin(b*x+a)^2*(d*x+c)^(1/2)/b^2

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Rubi [A]  time = 0.36, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3311, 32, 3312, 3306, 3305, 3351, 3304, 3352} \[ \frac {3 \sqrt {\pi } d^{3/2} \cos \left (2 a-\frac {2 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {\pi } \sqrt {d}}\right )}{32 b^{5/2}}-\frac {3 \sqrt {\pi } d^{3/2} \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{32 b^{5/2}}+\frac {3 d \sqrt {c+d x} \sin ^2(a+b x)}{8 b^2}-\frac {(c+d x)^{3/2} \sin (a+b x) \cos (a+b x)}{2 b}-\frac {3 d \sqrt {c+d x}}{16 b^2}+\frac {(c+d x)^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)*Sin[a + b*x]^2,x]

[Out]

(-3*d*Sqrt[c + d*x])/(16*b^2) + (c + d*x)^(5/2)/(5*d) + (3*d^(3/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*S
qrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(32*b^(5/2)) - (3*d^(3/2)*Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x
])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(32*b^(5/2)) - ((c + d*x)^(3/2)*Cos[a + b*x]*Sin[a + b*x])/(2*b)
+ (3*d*Sqrt[c + d*x]*Sin[a + b*x]^2)/(8*b^2)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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]

Rubi steps

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

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Mathematica [A]  time = 1.79, size = 175, normalized size = 0.86 \[ \frac {\sqrt {\frac {b}{d}} \left (15 \sqrt {\pi } d^2 \cos \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )-15 \sqrt {\pi } d^2 \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )+2 \sqrt {\frac {b}{d}} \sqrt {c+d x} \left (4 b (c+d x) (4 b (c+d x)-5 d \sin (2 (a+b x)))-15 d^2 \cos (2 (a+b x))\right )\right )}{160 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b/d]*(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]*FresnelS[(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*C
os[2*(a + b*x)] + 4*b*(c + d*x)*(4*b*(c + d*x) - 5*d*Sin[2*(a + b*x)]))))/(160*b^3)

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fricas [A]  time = 0.65, size = 195, normalized size = 0.96 \[ \frac {15 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 15 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, {\left (16 \, b^{3} d^{2} x^{2} + 32 \, b^{3} c d x + 16 \, b^{3} c^{2} - 30 \, b d^{2} \cos \left (b x + a\right )^{2} + 15 \, b d^{2} - 40 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{160 \, b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/160*(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 - 30*b*d^2*cos(b*x + a)^2 + 15*b*d^2 - 40*(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 = 0.92, size = 806, normalized size = 3.97 \[ \frac {240 \, {\left (\frac {\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 {\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 )}} + 4 \, \sqrt {d x + c}\right )} c^{2} + d^{2} {\left (\frac {64 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )}}{d^{2}} + \frac {15 \, {\left (\frac {\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 \, {\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}}\right )}}{d^{2}} + \frac {15 \, {\left (\frac {\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 \, {\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}}\right )}}{d^{2}}\right )} - 40 \, {\left (\frac {3 \, \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 {3 \, \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} - 16 \, {\left (d x + c\right )}^{\frac {3}{2}} + 48 \, \sqrt {d x + c} c - \frac {6 i \, \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 {6 i \, \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}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(240*(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)/(sq
rt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)) + 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)) + 4*sqrt(d*x + c))*c^2 + d^2*(64*(3*(d*x + c)^(
5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)/d^2 + 15*(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*(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 + 15*(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*(-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) - 40*(3*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) + 3*sqrt(pi)*(4*b*c - I*d)*d*erf(-sqrt(b*d)*sqr
t(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) -
16*(d*x + c)^(3/2) + 48*sqrt(d*x + c)*c - 6*I*sqrt(d*x + c)*d*e^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b +
6*I*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 = 0.03, size = 197, normalized size = 0.97 \[ \frac {\frac {\left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}+\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(1/10*(d*x+c)^(5/2)-1/8/b*d*(d*x+c)^(3/2)*sin(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)
*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 = 0.51, size = 274, normalized size = 1.35 \[ \frac {\sqrt {2} {\left (\frac {128 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3}}{d} - 160 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} \sin \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 120 \, \sqrt {2} \sqrt {d x + c} b d \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - {\left (\left (15 i - 15\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (15 i + 15\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) - {\left (-\left (15 i + 15\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (15 i - 15\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right )\right )}}{1280 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1280*sqrt(2)*(128*sqrt(2)*(d*x + c)^(5/2)*b^3/d - 160*sqrt(2)*(d*x + c)^(3/2)*b^2*sin(2*((d*x + c)*b - b*c +
a*d)/d) - 120*sqrt(2)*sqrt(d*x + c)*b*d*cos(2*((d*x + c)*b - b*c + a*d)/d) - ((15*I - 15)*4^(1/4)*sqrt(pi)*d^
2*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) + (15*I + 15)*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)) - (-(15*I + 15)*4^(1/4)*sqrt(pi)*d^2*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d
) - (15*I - 15)*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 = -0.00 \[ \int {\sin \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^2*(c + d*x)^(3/2),x)

[Out]

int(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 \[ \int \left (c + d x\right )^{\frac {3}{2}} \sin ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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