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

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

[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)

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Rubi [A]  time = 0.442284, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, 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 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 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 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 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 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 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 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 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 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*}

Mathematica [A]  time = 2.14366, size = 194, normalized size = 0.84 \[ \frac{\sqrt{\frac{b}{d}} \left (2 \sqrt{\frac{b}{d}} \sqrt{c+d x} \left (-7 d \sin (2 (a+b x)) \left (16 b^2 (c+d x)^2-15 d^2\right )-140 b d^2 (c+d x) \cos (2 (a+b x))+64 b^3 (c+d x)^3\right )-105 \sqrt{\pi } d^3 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-105 \sqrt{\pi } d^3 \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )\right )}{896 b^4} \]

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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Maple [A]  time = 0.017, size = 242, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( 1/14\, \left ( dx+c \right ) ^{7/2}-1/8\,{\frac{d \left ( dx+c \right ) ^{5/2}}{b}\sin \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }+5/8\,{\frac{d}{b} \left ( -1/4\,{\frac{d \left ( dx+c \right ) ^{3/2}}{b}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }+3/4\,{\frac{d}{b} \left ( 1/4\,{\frac{d\sqrt{dx+c}}{b}\sin \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }-1/8\,{\frac{d\sqrt{\pi }}{b} \left ( \cos \left ( 2\,{\frac{da-cb}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) +\sin \left ( 2\,{\frac{da-cb}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) } \right ) } \right ) } \end{align*}

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 = 1.85487, size = 940, normalized size = 4.07 \begin{align*} \text{result too large to display} \end{align*}

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^3*sqrt(abs(b)/abs(d)) - 1120*sqrt(2)*(d*x + c)^(3/2)*b*d^2*sqrt(
abs(b)/abs(d))*cos(2*((d*x + c)*b - b*c + a*d)/d) - ((105*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arct
an2(0, d/sqrt(d^2))) + 105*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 105*sqr
t(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 105*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0
, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^3*cos(-2*(b*c - a*d)/d) + (105*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b)
+ 1/2*arctan2(0, d/sqrt(d^2))) + 105*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) -
105*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 105*I*sqrt(pi)*sin(-1/4*pi + 1
/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^3*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/d)) -
((-105*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 105*I*sqrt(pi)*cos(-1/4*pi
+ 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 105*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2
(0, d/sqrt(d^2))) - 105*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^3*cos(-2*(b
*c - a*d)/d) + (105*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 105*sqrt(pi)*cos(
-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 105*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1
/2*arctan2(0, d/sqrt(d^2))) - 105*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d
^3*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^2*d*sqrt(abs(b)
/abs(d)) - 15*sqrt(2)*sqrt(d*x + c)*d^3*sqrt(abs(b)/abs(d)))*sin(2*((d*x + c)*b - b*c + a*d)/d))/(b^3*d*sqrt(a
bs(b)/abs(d)))

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Fricas [A]  time = 2.29634, size = 614, normalized size = 2.66 \begin{align*} -\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} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [C]  time = 1.34768, size = 1419, normalized size = 6.14 \begin{align*} \text{result too large to display} \end{align*}

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]

-1/26880*(560*(3*I*sqrt(pi)*d^2*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*I*sqrt(pi)*d^2*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) - 16*(d*x + c)^(3/2) - 6*I*s
qrt(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^2*(256*(15*(d*x + c)^(7/2) - 42*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2)/d^2 +
105*(sqrt(pi)*(-16*I*b^2*c^2*d + 24*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 - 3
2*I*(d*x + c)^(3/2)*b^2*c*d + 16*I*sqrt(d*x + c)*b^2*c^2*d + 20*(d*x + c)^(3/2)*b*d^2 - 24*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^2 + 105*(sqrt(pi)*(16*I*b^2*c
^2*d + 24*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 + 32*I*(d*x + c)^(3/2)*b^2
*c*d - 16*I*sqrt(d*x + c)*b^2*c^2*d + 20*(d*x + c)^(3/2)*b*d^2 - 24*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^2) - 56*(192*(d*x + c)^(5/2) - 320*(d*x + c)^(3/2)*c
+ 15*sqrt(pi)*(4*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) + 15*sqrt(pi)*(-4*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) -
30*(-4*I*(d*x + c)^(3/2)*b*d + 4*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 - 30*(4*I*(d*x + c)^(3/2)*b*d - 4*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)*c)/d