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

Optimal. Leaf size=231 \[ -\frac {15 \sqrt {\pi } d^{5/2} \sin \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{128 b^{7/2}}-\frac {15 \sqrt {\pi } d^{5/2} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{128 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \sin (2 a+2 b x)}{64 b^3}+\frac {5 d (c+d x)^{3/2} \sin ^2(a+b x)}{8 b^2}-\frac {(c+d x)^{5/2} \sin (a+b x) \cos (a+b x)}{2 b}-\frac {5 d (c+d x)^{3/2}}{16 b^2}+\frac {(c+d x)^{7/2}}{7 d} \]

[Out]

-5/16*d*(d*x+c)^(3/2)/b^2+1/7*(d*x+c)^(7/2)/d-1/2*(d*x+c)^(5/2)*cos(b*x+a)*sin(b*x+a)/b+5/8*d*(d*x+c)^(3/2)*si
n(b*x+a)^2/b^2-15/128*d^(5/2)*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^(
7/2)-15/128*d^(5/2)*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^(7/2)+15/64
*d^2*sin(2*b*x+2*a)*(d*x+c)^(1/2)/b^3

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Rubi [A]  time = 0.44, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3311, 32, 3312, 3296, 3306, 3305, 3351, 3304, 3352} \[ -\frac {15 \sqrt {\pi } d^{5/2} \sin \left (2 a-\frac {2 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {\pi } \sqrt {d}}\right )}{128 b^{7/2}}-\frac {15 \sqrt {\pi } d^{5/2} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{128 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \sin (2 a+2 b x)}{64 b^3}+\frac {5 d (c+d x)^{3/2} \sin ^2(a+b x)}{8 b^2}-\frac {(c+d x)^{5/2} \sin (a+b x) \cos (a+b x)}{2 b}-\frac {5 d (c+d x)^{3/2}}{16 b^2}+\frac {(c+d x)^{7/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)*Sin[a + b*x]^2,x]

[Out]

(-5*d*(c + d*x)^(3/2))/(16*b^2) + (c + d*x)^(7/2)/(7*d) - (15*d^(5/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelS[(
2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(128*b^(7/2)) - (15*d^(5/2)*Sqrt[Pi]*FresnelC[(2*Sqrt[b]*Sqrt[c
+ d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(128*b^(7/2)) - ((c + d*x)^(5/2)*Cos[a + b*x]*Sin[a + b*x])/
(2*b) + (5*d*(c + d*x)^(3/2)*Sin[a + b*x]^2)/(8*b^2) + (15*d^2*Sqrt[c + d*x]*Sin[2*a + 2*b*x])/(64*b^3)

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

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Mathematica [A]  time = 2.32, size = 194, normalized size = 0.84 \[ \frac {\sqrt {\frac {b}{d}} \left (2 \sqrt {\frac {b}{d}} \sqrt {c+d x} \left (-7 d \sin (2 (a+b x)) \left (16 b^2 (c+d x)^2-15 d^2\right )-140 b d^2 (c+d x) \cos (2 (a+b x))+64 b^3 (c+d x)^3\right )-105 \sqrt {\pi } d^3 \sin \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )-105 \sqrt {\pi } d^3 \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )\right )}{896 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b/d]*(-105*d^3*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] - 105*d^3*Sq
rt[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]*(64*b^3
*(c + d*x)^3 - 140*b*d^2*(c + d*x)*Cos[2*(a + b*x)] - 7*d*(-15*d^2 + 16*b^2*(c + d*x)^2)*Sin[2*(a + b*x)])))/(
896*b^4)

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fricas [A]  time = 0.73, size = 258, normalized size = 1.12 \[ -\frac {105 \, \pi d^{4} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 105 \, \pi d^{4} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 4 \, {\left (32 \, b^{4} d^{3} x^{3} + 96 \, b^{4} c d^{2} x^{2} + 32 \, b^{4} c^{3} + 70 \, b^{2} c d^{2} - 140 \, {\left (b^{2} d^{3} x + b^{2} c d^{2}\right )} \cos \left (b x + a\right )^{2} - 7 \, {\left (16 \, b^{3} d^{3} x^{2} + 32 \, b^{3} c d^{2} x + 16 \, b^{3} c^{2} d - 15 \, b d^{3}\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, {\left (48 \, b^{4} c^{2} d + 35 \, b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{896 \, b^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/896*(105*pi*d^4*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d))) + 105*pi*d
^4*sqrt(b/(pi*d))*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - 4*(32*b^4*d^3*x^3 + 96*b
^4*c*d^2*x^2 + 32*b^4*c^3 + 70*b^2*c*d^2 - 140*(b^2*d^3*x + b^2*c*d^2)*cos(b*x + a)^2 - 7*(16*b^3*d^3*x^2 + 32
*b^3*c*d^2*x + 16*b^3*c^2*d - 15*b*d^3)*cos(b*x + a)*sin(b*x + a) + 2*(48*b^4*c^2*d + 35*b^2*d^3)*x)*sqrt(d*x
+ c))/(b^4*d)

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giac [C]  time = 1.16, size = 1310, normalized size = 5.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8960*(2240*(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)) + 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^3 + 28*c*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) + d^3*(256*(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(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*
(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 - 35*(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*(-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) - 560*(
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)*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) - 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^2)/d

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maple [A]  time = 0.04, size = 242, normalized size = 1.05 \[ \frac {\frac {\left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {d \left (d x +c \right )^{\frac {5}{2}} \sin \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}+\frac {5 d \left (-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}-\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}\right )}{4 b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(1/14*(d*x+c)^(7/2)-1/8/b*d*(d*x+c)^(5/2)*sin(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)
*cos(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)*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)*Fre
snelC(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)))))

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maxima [C]  time = 1.48, size = 295, normalized size = 1.28 \[ \frac {\sqrt {2} {\left (\frac {512 \, \sqrt {2} {\left (d x + c\right )}^{\frac {7}{2}} b^{4}}{d} - 1120 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - {\left (\left (105 i + 105\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (105 i - 105\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) - {\left (-\left (105 i - 105\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (105 i + 105\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right ) - 56 \, {\left (16 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3} - 15 \, \sqrt {2} \sqrt {d x + c} b d^{2}\right )} \sin \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right )\right )}}{7168 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7168*sqrt(2)*(512*sqrt(2)*(d*x + c)^(7/2)*b^4/d - 1120*sqrt(2)*(d*x + c)^(3/2)*b^2*d*cos(2*((d*x + c)*b - b*
c + a*d)/d) - ((105*I + 105)*4^(1/4)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) - (105*I - 105)*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)) - (-(105*I - 105)*4^(1/
4)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) + (105*I + 105)*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)) - 56*(16*sqrt(2)*(d*x + c)^(5/2)*b^3 - 15*sqrt(2)*sqrt(d
*x + c)*b*d^2)*sin(2*((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 \[ \int {\sin \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^2*(c + d*x)^(5/2),x)

[Out]

int(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)*sin(b*x+a)**2,x)

[Out]

Timed out

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