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

Optimal. Leaf size=382 \[ \frac {2 f^2 \sin \left (a+b (c+d x)^{3/2}\right )}{3 b^2 d^3}-\frac {4 f \sqrt {c+d x} (d e-c f) \cos \left (a+b (c+d x)^{3/2}\right )}{3 b d^3}-\frac {2 e^{i a} f \sqrt {c+d x} (d e-c f) \Gamma \left (\frac {1}{3},-i b (c+d x)^{3/2}\right )}{9 b d^3 \sqrt [3]{-i b (c+d x)^{3/2}}}-\frac {2 e^{-i a} f \sqrt {c+d x} (d e-c f) \Gamma \left (\frac {1}{3},i b (c+d x)^{3/2}\right )}{9 b d^3 \sqrt [3]{i b (c+d x)^{3/2}}}+\frac {i e^{i a} (c+d x) (d e-c f)^2 \Gamma \left (\frac {2}{3},-i b (c+d x)^{3/2}\right )}{3 d^3 \left (-i b (c+d x)^{3/2}\right )^{2/3}}-\frac {i e^{-i a} (c+d x) (d e-c f)^2 \Gamma \left (\frac {2}{3},i b (c+d x)^{3/2}\right )}{3 d^3 \left (i b (c+d x)^{3/2}\right )^{2/3}}-\frac {2 f^2 (c+d x)^{3/2} \cos \left (a+b (c+d x)^{3/2}\right )}{3 b d^3} \]

[Out]

-2/3*f^2*(d*x+c)^(3/2)*cos(a+b*(d*x+c)^(3/2))/b/d^3+1/3*I*exp(I*a)*(-c*f+d*e)^2*(d*x+c)*GAMMA(2/3,-I*b*(d*x+c)
^(3/2))/d^3/(-I*b*(d*x+c)^(3/2))^(2/3)-1/3*I*(-c*f+d*e)^2*(d*x+c)*GAMMA(2/3,I*b*(d*x+c)^(3/2))/d^3/exp(I*a)/(I
*b*(d*x+c)^(3/2))^(2/3)+2/3*f^2*sin(a+b*(d*x+c)^(3/2))/b^2/d^3-4/3*f*(-c*f+d*e)*cos(a+b*(d*x+c)^(3/2))*(d*x+c)
^(1/2)/b/d^3-2/9*exp(I*a)*f*(-c*f+d*e)*GAMMA(1/3,-I*b*(d*x+c)^(3/2))*(d*x+c)^(1/2)/b/d^3/(-I*b*(d*x+c)^(3/2))^
(1/3)-2/9*f*(-c*f+d*e)*GAMMA(1/3,I*b*(d*x+c)^(3/2))*(d*x+c)^(1/2)/b/d^3/exp(I*a)/(I*b*(d*x+c)^(3/2))^(1/3)

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Rubi [A]  time = 0.31, antiderivative size = 382, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {3433, 3389, 2218, 3385, 3356, 2208, 3379, 3296, 2637} \[ -\frac {2 e^{i a} f \sqrt {c+d x} (d e-c f) \text {Gamma}\left (\frac {1}{3},-i b (c+d x)^{3/2}\right )}{9 b d^3 \sqrt [3]{-i b (c+d x)^{3/2}}}-\frac {2 e^{-i a} f \sqrt {c+d x} (d e-c f) \text {Gamma}\left (\frac {1}{3},i b (c+d x)^{3/2}\right )}{9 b d^3 \sqrt [3]{i b (c+d x)^{3/2}}}+\frac {i e^{i a} (c+d x) (d e-c f)^2 \text {Gamma}\left (\frac {2}{3},-i b (c+d x)^{3/2}\right )}{3 d^3 \left (-i b (c+d x)^{3/2}\right )^{2/3}}-\frac {i e^{-i a} (c+d x) (d e-c f)^2 \text {Gamma}\left (\frac {2}{3},i b (c+d x)^{3/2}\right )}{3 d^3 \left (i b (c+d x)^{3/2}\right )^{2/3}}+\frac {2 f^2 \sin \left (a+b (c+d x)^{3/2}\right )}{3 b^2 d^3}-\frac {4 f \sqrt {c+d x} (d e-c f) \cos \left (a+b (c+d x)^{3/2}\right )}{3 b d^3}-\frac {2 f^2 (c+d x)^{3/2} \cos \left (a+b (c+d x)^{3/2}\right )}{3 b d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*f*(d*e - c*f)*Sqrt[c + d*x]*Cos[a + b*(c + d*x)^(3/2)])/(3*b*d^3) - (2*f^2*(c + d*x)^(3/2)*Cos[a + b*(c +
d*x)^(3/2)])/(3*b*d^3) - (2*E^(I*a)*f*(d*e - c*f)*Sqrt[c + d*x]*Gamma[1/3, (-I)*b*(c + d*x)^(3/2)])/(9*b*d^3*(
(-I)*b*(c + d*x)^(3/2))^(1/3)) - (2*f*(d*e - c*f)*Sqrt[c + d*x]*Gamma[1/3, I*b*(c + d*x)^(3/2)])/(9*b*d^3*E^(I
*a)*(I*b*(c + d*x)^(3/2))^(1/3)) + ((I/3)*E^(I*a)*(d*e - c*f)^2*(c + d*x)*Gamma[2/3, (-I)*b*(c + d*x)^(3/2)])/
(d^3*((-I)*b*(c + d*x)^(3/2))^(2/3)) - ((I/3)*(d*e - c*f)^2*(c + d*x)*Gamma[2/3, I*b*(c + d*x)^(3/2)])/(d^3*E^
(I*a)*(I*b*(c + d*x)^(3/2))^(2/3)) + (2*f^2*Sin[a + b*(c + d*x)^(3/2)])/(3*b^2*d^3)

Rule 2208

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> -Simp[(F^a*(c + d*x)*Gamma[1/n, -(b*(c + d*x)
^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^(1/n)), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, 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 3356

Int[Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Dist[1/2, Int[E^(-(c*I) - d*I*(e + f*x)^n), x],
x] + Dist[1/2, Int[E^(c*I + d*I*(e + f*x)^n), x], x] /; FreeQ[{c, d, e, f}, x] && IGtQ[n, 2]

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3389

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[I/2, Int[(e*x)^m*E^(-(c*I) - d*I*x^n),
x], x] - Dist[I/2, Int[(e*x)^m*E^(c*I + d*I*x^n), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]

Rule 3433

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[k/f^(m + 1), Subst[Int[ExpandIntegrand[(a + b*Sin[c +
d*x^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f,
g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int (e+f x)^2 \sin \left (a+b (c+d x)^{3/2}\right ) \, dx &=\frac {2 \operatorname {Subst}\left (\int \left ((d e-c f)^2 x \sin \left (a+b x^3\right )-2 f (-d e+c f) x^3 \sin \left (a+b x^3\right )+f^2 x^5 \sin \left (a+b x^3\right )\right ) \, dx,x,\sqrt {c+d x}\right )}{d^3}\\ &=\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int x^5 \sin \left (a+b x^3\right ) \, dx,x,\sqrt {c+d x}\right )}{d^3}+\frac {(4 f (d e-c f)) \operatorname {Subst}\left (\int x^3 \sin \left (a+b x^3\right ) \, dx,x,\sqrt {c+d x}\right )}{d^3}+\frac {\left (2 (d e-c f)^2\right ) \operatorname {Subst}\left (\int x \sin \left (a+b x^3\right ) \, dx,x,\sqrt {c+d x}\right )}{d^3}\\ &=-\frac {4 f (d e-c f) \sqrt {c+d x} \cos \left (a+b (c+d x)^{3/2}\right )}{3 b d^3}+\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int x \sin (a+b x) \, dx,x,(c+d x)^{3/2}\right )}{3 d^3}+\frac {(4 f (d e-c f)) \operatorname {Subst}\left (\int \cos \left (a+b x^3\right ) \, dx,x,\sqrt {c+d x}\right )}{3 b d^3}+\frac {\left (i (d e-c f)^2\right ) \operatorname {Subst}\left (\int e^{-i a-i b x^3} x \, dx,x,\sqrt {c+d x}\right )}{d^3}-\frac {\left (i (d e-c f)^2\right ) \operatorname {Subst}\left (\int e^{i a+i b x^3} x \, dx,x,\sqrt {c+d x}\right )}{d^3}\\ &=-\frac {4 f (d e-c f) \sqrt {c+d x} \cos \left (a+b (c+d x)^{3/2}\right )}{3 b d^3}-\frac {2 f^2 (c+d x)^{3/2} \cos \left (a+b (c+d x)^{3/2}\right )}{3 b d^3}+\frac {i e^{i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {2}{3},-i b (c+d x)^{3/2}\right )}{3 d^3 \left (-i b (c+d x)^{3/2}\right )^{2/3}}-\frac {i e^{-i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {2}{3},i b (c+d x)^{3/2}\right )}{3 d^3 \left (i b (c+d x)^{3/2}\right )^{2/3}}+\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int \cos (a+b x) \, dx,x,(c+d x)^{3/2}\right )}{3 b d^3}+\frac {(2 f (d e-c f)) \operatorname {Subst}\left (\int e^{-i a-i b x^3} \, dx,x,\sqrt {c+d x}\right )}{3 b d^3}+\frac {(2 f (d e-c f)) \operatorname {Subst}\left (\int e^{i a+i b x^3} \, dx,x,\sqrt {c+d x}\right )}{3 b d^3}\\ &=-\frac {4 f (d e-c f) \sqrt {c+d x} \cos \left (a+b (c+d x)^{3/2}\right )}{3 b d^3}-\frac {2 f^2 (c+d x)^{3/2} \cos \left (a+b (c+d x)^{3/2}\right )}{3 b d^3}-\frac {2 e^{i a} f (d e-c f) \sqrt {c+d x} \Gamma \left (\frac {1}{3},-i b (c+d x)^{3/2}\right )}{9 b d^3 \sqrt [3]{-i b (c+d x)^{3/2}}}-\frac {2 e^{-i a} f (d e-c f) \sqrt {c+d x} \Gamma \left (\frac {1}{3},i b (c+d x)^{3/2}\right )}{9 b d^3 \sqrt [3]{i b (c+d x)^{3/2}}}+\frac {i e^{i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {2}{3},-i b (c+d x)^{3/2}\right )}{3 d^3 \left (-i b (c+d x)^{3/2}\right )^{2/3}}-\frac {i e^{-i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {2}{3},i b (c+d x)^{3/2}\right )}{3 d^3 \left (i b (c+d x)^{3/2}\right )^{2/3}}+\frac {2 f^2 \sin \left (a+b (c+d x)^{3/2}\right )}{3 b^2 d^3}\\ \end {align*}

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Mathematica [A]  time = 3.25, size = 419, normalized size = 1.10 \[ -\frac {i \left ((\cos (a)+i \sin (a)) \left (\frac {i f^2 \sin \left (b (c+d x)^{3/2}\right )}{b^2}+\frac {f^2 \cos \left (b (c+d x)^{3/2}\right )}{b^2}-\frac {2 f (c+d x)^2 (d e-c f) \Gamma \left (\frac {4}{3},-i b (c+d x)^{3/2}\right )}{\left (-i b (c+d x)^{3/2}\right )^{4/3}}-\frac {(c+d x) (d e-c f)^2 \Gamma \left (\frac {2}{3},-i b (c+d x)^{3/2}\right )}{\left (-i b (c+d x)^{3/2}\right )^{2/3}}+\frac {f^2 (c+d x)^{3/2} \left (\sin \left (b (c+d x)^{3/2}\right )-i \cos \left (b (c+d x)^{3/2}\right )\right )}{b}\right )-(\cos (a)-i \sin (a)) \left (-\frac {i f^2 \sin \left (b (c+d x)^{3/2}\right )}{b^2}+\frac {f^2 \cos \left (b (c+d x)^{3/2}\right )}{b^2}-\frac {2 f (c+d x)^2 (d e-c f) \Gamma \left (\frac {4}{3},i b (c+d x)^{3/2}\right )}{\left (i b (c+d x)^{3/2}\right )^{4/3}}-\frac {(c+d x) (d e-c f)^2 \Gamma \left (\frac {2}{3},i b (c+d x)^{3/2}\right )}{\left (i b (c+d x)^{3/2}\right )^{2/3}}+\frac {f^2 (c+d x)^{3/2} \left (\sin \left (b (c+d x)^{3/2}\right )+i \cos \left (b (c+d x)^{3/2}\right )\right )}{b}\right )\right )}{3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/3*I)*((Cos[a] + I*Sin[a])*((f^2*Cos[b*(c + d*x)^(3/2)])/b^2 - ((d*e - c*f)^2*(c + d*x)*Gamma[2/3, (-I)*b*
(c + d*x)^(3/2)])/((-I)*b*(c + d*x)^(3/2))^(2/3) - (2*f*(d*e - c*f)*(c + d*x)^2*Gamma[4/3, (-I)*b*(c + d*x)^(3
/2)])/((-I)*b*(c + d*x)^(3/2))^(4/3) + (I*f^2*Sin[b*(c + d*x)^(3/2)])/b^2 + (f^2*(c + d*x)^(3/2)*((-I)*Cos[b*(
c + d*x)^(3/2)] + Sin[b*(c + d*x)^(3/2)]))/b) - (Cos[a] - I*Sin[a])*((f^2*Cos[b*(c + d*x)^(3/2)])/b^2 - ((d*e
- c*f)^2*(c + d*x)*Gamma[2/3, I*b*(c + d*x)^(3/2)])/(I*b*(c + d*x)^(3/2))^(2/3) - (2*f*(d*e - c*f)*(c + d*x)^2
*Gamma[4/3, I*b*(c + d*x)^(3/2)])/(I*b*(c + d*x)^(3/2))^(4/3) - (I*f^2*Sin[b*(c + d*x)^(3/2)])/b^2 + (f^2*(c +
d*x)^(3/2)*(I*Cos[b*(c + d*x)^(3/2)] + Sin[b*(c + d*x)^(3/2)]))/b)))/d^3

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fricas [A]  time = 0.69, size = 276, normalized size = 0.72 \[ \frac {{\left (2 i \, d e f - 2 i \, c f^{2}\right )} \left (i \, b\right )^{\frac {2}{3}} e^{\left (-i \, a\right )} \Gamma \left (\frac {1}{3}, {\left (i \, b d x + i \, b c\right )} \sqrt {d x + c}\right ) + {\left (-2 i \, d e f + 2 i \, c f^{2}\right )} \left (-i \, b\right )^{\frac {2}{3}} e^{\left (i \, a\right )} \Gamma \left (\frac {1}{3}, {\left (-i \, b d x - i \, b c\right )} \sqrt {d x + c}\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \left (i \, b\right )^{\frac {1}{3}} e^{\left (-i \, a\right )} \Gamma \left (\frac {2}{3}, {\left (i \, b d x + i \, b c\right )} \sqrt {d x + c}\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \left (-i \, b\right )^{\frac {1}{3}} e^{\left (i \, a\right )} \Gamma \left (\frac {2}{3}, {\left (-i \, b d x - i \, b c\right )} \sqrt {d x + c}\right ) + 6 \, f^{2} \sin \left ({\left (b d x + b c\right )} \sqrt {d x + c} + a\right ) - 6 \, {\left (b d f^{2} x + 2 \, b d e f - b c f^{2}\right )} \sqrt {d x + c} \cos \left ({\left (b d x + b c\right )} \sqrt {d x + c} + a\right )}{9 \, b^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*((2*I*d*e*f - 2*I*c*f^2)*(I*b)^(2/3)*e^(-I*a)*gamma(1/3, (I*b*d*x + I*b*c)*sqrt(d*x + c)) + (-2*I*d*e*f +
2*I*c*f^2)*(-I*b)^(2/3)*e^(I*a)*gamma(1/3, (-I*b*d*x - I*b*c)*sqrt(d*x + c)) - 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*
c^2*f^2)*(I*b)^(1/3)*e^(-I*a)*gamma(2/3, (I*b*d*x + I*b*c)*sqrt(d*x + c)) - 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2
*f^2)*(-I*b)^(1/3)*e^(I*a)*gamma(2/3, (-I*b*d*x - I*b*c)*sqrt(d*x + c)) + 6*f^2*sin((b*d*x + b*c)*sqrt(d*x + c
) + a) - 6*(b*d*f^2*x + 2*b*d*e*f - b*c*f^2)*sqrt(d*x + c)*cos((b*d*x + b*c)*sqrt(d*x + c) + a))/(b^2*d^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (f x + e\right )}^{2} \sin \left ({\left (d x + c\right )}^{\frac {3}{2}} b + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \left (f x +e \right )^{2} \sin \left (a +b \left (d x +c \right )^{\frac {3}{2}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.20, size = 694, normalized size = 1.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(3*((d*x + c)^(3/2)*b)^(1/3)*(((sqrt(3) + I)*gamma(2/3, I*(d*x + c)^(3/2)*b) + (sqrt(3) - I)*gamma(2/3,
-I*(d*x + c)^(3/2)*b))*cos(a) - ((I*sqrt(3) - 1)*gamma(2/3, I*(d*x + c)^(3/2)*b) + (-I*sqrt(3) - 1)*gamma(2/3,
-I*(d*x + c)^(3/2)*b))*sin(a))*e^2/(sqrt(d*x + c)*b) - 6*((d*x + c)^(3/2)*b)^(1/3)*(((sqrt(3) + I)*gamma(2/3,
I*(d*x + c)^(3/2)*b) + (sqrt(3) - I)*gamma(2/3, -I*(d*x + c)^(3/2)*b))*cos(a) - ((I*sqrt(3) - 1)*gamma(2/3, I
*(d*x + c)^(3/2)*b) + (-I*sqrt(3) - 1)*gamma(2/3, -I*(d*x + c)^(3/2)*b))*sin(a))*c*e*f/(sqrt(d*x + c)*b*d) + 3
*((d*x + c)^(3/2)*b)^(1/3)*(((sqrt(3) + I)*gamma(2/3, I*(d*x + c)^(3/2)*b) + (sqrt(3) - I)*gamma(2/3, -I*(d*x
+ c)^(3/2)*b))*cos(a) - ((I*sqrt(3) - 1)*gamma(2/3, I*(d*x + c)^(3/2)*b) + (-I*sqrt(3) - 1)*gamma(2/3, -I*(d*x
+ c)^(3/2)*b))*sin(a))*c^2*f^2/(sqrt(d*x + c)*b*d^2) + 2*(12*((d*x + c)^(3/2)*b)^(1/3)*sqrt(d*x + c)*cos((d*x
+ c)^(3/2)*b + a) + sqrt(d*x + c)*(((sqrt(3) - I)*gamma(1/3, I*(d*x + c)^(3/2)*b) + (sqrt(3) + I)*gamma(1/3,
-I*(d*x + c)^(3/2)*b))*cos(a) + ((-I*sqrt(3) - 1)*gamma(1/3, I*(d*x + c)^(3/2)*b) + (I*sqrt(3) - 1)*gamma(1/3,
-I*(d*x + c)^(3/2)*b))*sin(a)))*e*f/(((d*x + c)^(3/2)*b)^(1/3)*b*d) - 2*(12*((d*x + c)^(3/2)*b)^(1/3)*sqrt(d*
x + c)*cos((d*x + c)^(3/2)*b + a) + sqrt(d*x + c)*(((sqrt(3) - I)*gamma(1/3, I*(d*x + c)^(3/2)*b) + (sqrt(3) +
I)*gamma(1/3, -I*(d*x + c)^(3/2)*b))*cos(a) + ((-I*sqrt(3) - 1)*gamma(1/3, I*(d*x + c)^(3/2)*b) + (I*sqrt(3)
- 1)*gamma(1/3, -I*(d*x + c)^(3/2)*b))*sin(a)))*c*f^2/(((d*x + c)^(3/2)*b)^(1/3)*b*d^2) + 12*((d*x + c)^(3/2)*
b*cos((d*x + c)^(3/2)*b + a) - sin((d*x + c)^(3/2)*b + a))*f^2/(b^2*d^2))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e + f x\right )^{2} \sin {\left (a + b c \sqrt {c + d x} + b d x \sqrt {c + d x} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e + f*x)**2*sin(a + b*c*sqrt(c + d*x) + b*d*x*sqrt(c + d*x)), x)

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