3.832 \(\int (e x)^{5/2} (a+b x^2)^2 (c+d x^2)^{3/2} \, dx\)

Optimal. Leaf size=530 \[ -\frac {4 c^{13/4} e^{5/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (51 a^2 d^2+b c (11 b c-42 a d)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{3315 d^{15/4} \sqrt {c+d x^2}}+\frac {8 c^{13/4} e^{5/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (51 a^2 d^2+b c (11 b c-42 a d)\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{3315 d^{15/4} \sqrt {c+d x^2}}-\frac {8 c^3 e^2 \sqrt {e x} \sqrt {c+d x^2} \left (51 a^2 d^2+b c (11 b c-42 a d)\right )}{3315 d^{7/2} \left (\sqrt {c}+\sqrt {d} x\right )}+\frac {8 c^2 e (e x)^{3/2} \sqrt {c+d x^2} \left (51 a^2 d^2+b c (11 b c-42 a d)\right )}{9945 d^3}+\frac {2 (e x)^{7/2} \left (c+d x^2\right )^{3/2} \left (51 a^2 d^2+b c (11 b c-42 a d)\right )}{663 d^2 e}+\frac {4 c (e x)^{7/2} \sqrt {c+d x^2} \left (51 a^2 d^2+b c (11 b c-42 a d)\right )}{1989 d^2 e}-\frac {2 b (e x)^{7/2} \left (c+d x^2\right )^{5/2} (11 b c-42 a d)}{357 d^2 e}+\frac {2 b^2 (e x)^{11/2} \left (c+d x^2\right )^{5/2}}{21 d e^3} \]

[Out]

2/663*(51*a^2*d^2+b*c*(-42*a*d+11*b*c))*(e*x)^(7/2)*(d*x^2+c)^(3/2)/d^2/e-2/357*b*(-42*a*d+11*b*c)*(e*x)^(7/2)
*(d*x^2+c)^(5/2)/d^2/e+2/21*b^2*(e*x)^(11/2)*(d*x^2+c)^(5/2)/d/e^3+8/9945*c^2*(51*a^2*d^2+b*c*(-42*a*d+11*b*c)
)*e*(e*x)^(3/2)*(d*x^2+c)^(1/2)/d^3+4/1989*c*(51*a^2*d^2+b*c*(-42*a*d+11*b*c))*(e*x)^(7/2)*(d*x^2+c)^(1/2)/d^2
/e-8/3315*c^3*(51*a^2*d^2+b*c*(-42*a*d+11*b*c))*e^2*(e*x)^(1/2)*(d*x^2+c)^(1/2)/d^(7/2)/(c^(1/2)+x*d^(1/2))+8/
3315*c^(13/4)*(51*a^2*d^2+b*c*(-42*a*d+11*b*c))*e^(5/2)*(cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))^2)
^(1/2)/cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))*EllipticE(sin(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e
^(1/2))),1/2*2^(1/2))*(c^(1/2)+x*d^(1/2))*((d*x^2+c)/(c^(1/2)+x*d^(1/2))^2)^(1/2)/d^(15/4)/(d*x^2+c)^(1/2)-4/3
315*c^(13/4)*(51*a^2*d^2+b*c*(-42*a*d+11*b*c))*e^(5/2)*(cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))^2)^
(1/2)/cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))*EllipticF(sin(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^
(1/2))),1/2*2^(1/2))*(c^(1/2)+x*d^(1/2))*((d*x^2+c)/(c^(1/2)+x*d^(1/2))^2)^(1/2)/d^(15/4)/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.56, antiderivative size = 530, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {464, 459, 279, 321, 329, 305, 220, 1196} \[ -\frac {8 c^3 e^2 \sqrt {e x} \sqrt {c+d x^2} \left (51 a^2 d^2+b c (11 b c-42 a d)\right )}{3315 d^{7/2} \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {4 c^{13/4} e^{5/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (51 a^2 d^2+b c (11 b c-42 a d)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{3315 d^{15/4} \sqrt {c+d x^2}}+\frac {8 c^{13/4} e^{5/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (51 a^2 d^2+b c (11 b c-42 a d)\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{3315 d^{15/4} \sqrt {c+d x^2}}+\frac {8 c^2 e (e x)^{3/2} \sqrt {c+d x^2} \left (51 a^2 d^2+b c (11 b c-42 a d)\right )}{9945 d^3}+\frac {2 (e x)^{7/2} \left (c+d x^2\right )^{3/2} \left (51 a^2 d^2+b c (11 b c-42 a d)\right )}{663 d^2 e}+\frac {4 c (e x)^{7/2} \sqrt {c+d x^2} \left (51 a^2 d^2+b c (11 b c-42 a d)\right )}{1989 d^2 e}-\frac {2 b (e x)^{7/2} \left (c+d x^2\right )^{5/2} (11 b c-42 a d)}{357 d^2 e}+\frac {2 b^2 (e x)^{11/2} \left (c+d x^2\right )^{5/2}}{21 d e^3} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^(5/2)*(a + b*x^2)^2*(c + d*x^2)^(3/2),x]

[Out]

(8*c^2*(51*a^2*d^2 + b*c*(11*b*c - 42*a*d))*e*(e*x)^(3/2)*Sqrt[c + d*x^2])/(9945*d^3) + (4*c*(51*a^2*d^2 + b*c
*(11*b*c - 42*a*d))*(e*x)^(7/2)*Sqrt[c + d*x^2])/(1989*d^2*e) - (8*c^3*(51*a^2*d^2 + b*c*(11*b*c - 42*a*d))*e^
2*Sqrt[e*x]*Sqrt[c + d*x^2])/(3315*d^(7/2)*(Sqrt[c] + Sqrt[d]*x)) + (2*(51*a^2*d^2 + b*c*(11*b*c - 42*a*d))*(e
*x)^(7/2)*(c + d*x^2)^(3/2))/(663*d^2*e) - (2*b*(11*b*c - 42*a*d)*(e*x)^(7/2)*(c + d*x^2)^(5/2))/(357*d^2*e) +
(2*b^2*(e*x)^(11/2)*(c + d*x^2)^(5/2))/(21*d*e^3) + (8*c^(13/4)*(51*a^2*d^2 + b*c*(11*b*c - 42*a*d))*e^(5/2)*
(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4
)*Sqrt[e])], 1/2])/(3315*d^(15/4)*Sqrt[c + d*x^2]) - (4*c^(13/4)*(51*a^2*d^2 + b*c*(11*b*c - 42*a*d))*e^(5/2)*
(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4
)*Sqrt[e])], 1/2])/(3315*d^(15/4)*Sqrt[c + d*x^2])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(d^2*(e*x)^
(m + n + 1)*(a + b*x^n)^(p + 1))/(b*e^(n + 1)*(m + n*(p + 2) + 1)), x] + Dist[1/(b*(m + n*(p + 2) + 1)), Int[(
e*x)^m*(a + b*x^n)^p*Simp[b*c^2*(m + n*(p + 2) + 1) + d*((2*b*c - a*d)*(m + n + 1) + 2*b*c*n*(p + 1))*x^n, x],
x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && NeQ[m + n*(p + 2) + 1, 0]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int (e x)^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx &=\frac {2 b^2 (e x)^{11/2} \left (c+d x^2\right )^{5/2}}{21 d e^3}+\frac {2 \int (e x)^{5/2} \left (c+d x^2\right )^{3/2} \left (\frac {21 a^2 d}{2}-\frac {1}{2} b (11 b c-42 a d) x^2\right ) \, dx}{21 d}\\ &=-\frac {2 b (11 b c-42 a d) (e x)^{7/2} \left (c+d x^2\right )^{5/2}}{357 d^2 e}+\frac {2 b^2 (e x)^{11/2} \left (c+d x^2\right )^{5/2}}{21 d e^3}+\frac {1}{51} \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) \int (e x)^{5/2} \left (c+d x^2\right )^{3/2} \, dx\\ &=\frac {2 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{663 e}-\frac {2 b (11 b c-42 a d) (e x)^{7/2} \left (c+d x^2\right )^{5/2}}{357 d^2 e}+\frac {2 b^2 (e x)^{11/2} \left (c+d x^2\right )^{5/2}}{21 d e^3}+\frac {1}{221} \left (2 c \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right )\right ) \int (e x)^{5/2} \sqrt {c+d x^2} \, dx\\ &=\frac {4 c \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) (e x)^{7/2} \sqrt {c+d x^2}}{1989 e}+\frac {2 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{663 e}-\frac {2 b (11 b c-42 a d) (e x)^{7/2} \left (c+d x^2\right )^{5/2}}{357 d^2 e}+\frac {2 b^2 (e x)^{11/2} \left (c+d x^2\right )^{5/2}}{21 d e^3}+\frac {\left (4 c^2 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right )\right ) \int \frac {(e x)^{5/2}}{\sqrt {c+d x^2}} \, dx}{1989}\\ &=\frac {8 c^2 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt {c+d x^2}}{9945 d}+\frac {4 c \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) (e x)^{7/2} \sqrt {c+d x^2}}{1989 e}+\frac {2 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{663 e}-\frac {2 b (11 b c-42 a d) (e x)^{7/2} \left (c+d x^2\right )^{5/2}}{357 d^2 e}+\frac {2 b^2 (e x)^{11/2} \left (c+d x^2\right )^{5/2}}{21 d e^3}-\frac {\left (4 c^3 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) e^2\right ) \int \frac {\sqrt {e x}}{\sqrt {c+d x^2}} \, dx}{3315 d}\\ &=\frac {8 c^2 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt {c+d x^2}}{9945 d}+\frac {4 c \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) (e x)^{7/2} \sqrt {c+d x^2}}{1989 e}+\frac {2 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{663 e}-\frac {2 b (11 b c-42 a d) (e x)^{7/2} \left (c+d x^2\right )^{5/2}}{357 d^2 e}+\frac {2 b^2 (e x)^{11/2} \left (c+d x^2\right )^{5/2}}{21 d e^3}-\frac {\left (8 c^3 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) e\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{3315 d}\\ &=\frac {8 c^2 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt {c+d x^2}}{9945 d}+\frac {4 c \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) (e x)^{7/2} \sqrt {c+d x^2}}{1989 e}+\frac {2 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{663 e}-\frac {2 b (11 b c-42 a d) (e x)^{7/2} \left (c+d x^2\right )^{5/2}}{357 d^2 e}+\frac {2 b^2 (e x)^{11/2} \left (c+d x^2\right )^{5/2}}{21 d e^3}-\frac {\left (8 c^{7/2} \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{3315 d^{3/2}}+\frac {\left (8 c^{7/2} \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) e^2\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {d} x^2}{\sqrt {c} e}}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{3315 d^{3/2}}\\ &=\frac {8 c^2 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt {c+d x^2}}{9945 d}+\frac {4 c \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) (e x)^{7/2} \sqrt {c+d x^2}}{1989 e}-\frac {8 c^3 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) e^2 \sqrt {e x} \sqrt {c+d x^2}}{3315 d^{3/2} \left (\sqrt {c}+\sqrt {d} x\right )}+\frac {2 \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{663 e}-\frac {2 b (11 b c-42 a d) (e x)^{7/2} \left (c+d x^2\right )^{5/2}}{357 d^2 e}+\frac {2 b^2 (e x)^{11/2} \left (c+d x^2\right )^{5/2}}{21 d e^3}+\frac {8 c^{13/4} \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) e^{5/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{3315 d^{7/4} \sqrt {c+d x^2}}-\frac {4 c^{13/4} \left (51 a^2+\frac {b c (11 b c-42 a d)}{d^2}\right ) e^{5/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{3315 d^{7/4} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.18, size = 210, normalized size = 0.40 \[ \frac {2 e (e x)^{3/2} \left (\left (c+d x^2\right ) \left (357 a^2 d^2 \left (4 c^2+25 c d x^2+15 d^2 x^4\right )+42 a b d \left (-28 c^3+20 c^2 d x^2+285 c d^2 x^4+195 d^3 x^6\right )+b^2 \left (308 c^4-220 c^3 d x^2+180 c^2 d^2 x^4+4485 c d^3 x^6+3315 d^4 x^8\right )\right )-84 c^3 \sqrt {\frac {c}{d x^2}+1} \left (51 a^2 d^2-42 a b c d+11 b^2 c^2\right ) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {c}{d x^2}\right )\right )}{69615 d^3 \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^(5/2)*(a + b*x^2)^2*(c + d*x^2)^(3/2),x]

[Out]

(2*e*(e*x)^(3/2)*((c + d*x^2)*(357*a^2*d^2*(4*c^2 + 25*c*d*x^2 + 15*d^2*x^4) + 42*a*b*d*(-28*c^3 + 20*c^2*d*x^
2 + 285*c*d^2*x^4 + 195*d^3*x^6) + b^2*(308*c^4 - 220*c^3*d*x^2 + 180*c^2*d^2*x^4 + 4485*c*d^3*x^6 + 3315*d^4*
x^8)) - 84*c^3*(11*b^2*c^2 - 42*a*b*c*d + 51*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*Hypergeometric2F1[-1/4, 1/2, 3/4, -(
c/(d*x^2))]))/(69615*d^3*Sqrt[c + d*x^2])

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} d e^{2} x^{8} + {\left (b^{2} c + 2 \, a b d\right )} e^{2} x^{6} + a^{2} c e^{2} x^{2} + {\left (2 \, a b c + a^{2} d\right )} e^{2} x^{4}\right )} \sqrt {d x^{2} + c} \sqrt {e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(b*x^2+a)^2*(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

integral((b^2*d*e^2*x^8 + (b^2*c + 2*a*b*d)*e^2*x^6 + a^2*c*e^2*x^2 + (2*a*b*c + a^2*d)*e^2*x^4)*sqrt(d*x^2 +
c)*sqrt(e*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(b*x^2+a)^2*(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2*(d*x^2 + c)^(3/2)*(e*x)^(5/2), x)

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maple [A]  time = 0.05, size = 743, normalized size = 1.40 \[ -\frac {2 \sqrt {e x}\, \left (-3315 b^{2} d^{6} x^{12}-8190 a b \,d^{6} x^{10}-7800 b^{2} c \,d^{5} x^{10}-5355 a^{2} d^{6} x^{8}-20160 a b c \,d^{5} x^{8}-4665 b^{2} c^{2} d^{4} x^{8}-14280 a^{2} c \,d^{5} x^{6}-12810 a b \,c^{2} d^{4} x^{6}+40 b^{2} c^{3} d^{3} x^{6}-10353 a^{2} c^{2} d^{4} x^{4}+336 a b \,c^{3} d^{3} x^{4}-88 b^{2} c^{4} d^{2} x^{4}-1428 a^{2} c^{3} d^{3} x^{2}+1176 a b \,c^{4} d^{2} x^{2}-308 b^{2} c^{5} d \,x^{2}+4284 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c^{4} d^{2} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-2142 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c^{4} d^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-3528 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{5} d \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+1764 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{5} d \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+924 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{6} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-462 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{6} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )\right ) e^{2}}{69615 \sqrt {d \,x^{2}+c}\, d^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(5/2)*(b*x^2+a)^2*(d*x^2+c)^(3/2),x)

[Out]

-2/69615*e^2/x*(e*x)^(1/2)/(d*x^2+c)^(1/2)/d^4*(-3315*x^12*b^2*d^6-8190*x^10*a*b*d^6-7800*x^10*b^2*c*d^5-5355*
x^8*a^2*d^6-20160*x^8*a*b*c*d^5-4665*x^8*b^2*c^2*d^4-14280*x^6*a^2*c*d^5-12810*x^6*a*b*c^2*d^4+40*x^6*b^2*c^3*
d^3+4284*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(
1/2)*d*x)^(1/2)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a^2*c^4*d^2-3528*((d*x+(-c*d)^(
1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*Ellipti
cE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*b*c^5*d+924*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*
2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticE(((d*x+(-c*d)^(1/2))/(-c
*d)^(1/2))^(1/2),1/2*2^(1/2))*b^2*c^6-2142*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2)
)/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2
))*a^2*c^4*d^2+1764*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(
-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*b*c^5*d-462*((d*x+
(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)
*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*b^2*c^6-10353*x^4*a^2*c^2*d^4+336*x^4*a*b*c^3*
d^3-88*x^4*b^2*c^4*d^2-1428*x^2*a^2*c^3*d^3+1176*x^2*a*b*c^4*d^2-308*x^2*b^2*c^5*d)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(b*x^2+a)^2*(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*(d*x^2 + c)^(3/2)*(e*x)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e\,x\right )}^{5/2}\,{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(5/2)*(a + b*x^2)^2*(c + d*x^2)^(3/2),x)

[Out]

int((e*x)^(5/2)*(a + b*x^2)^2*(c + d*x^2)^(3/2), x)

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sympy [C]  time = 148.82, size = 306, normalized size = 0.58 \[ \frac {a^{2} c^{\frac {3}{2}} e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \Gamma \left (\frac {11}{4}\right )} + \frac {a^{2} \sqrt {c} d e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \Gamma \left (\frac {15}{4}\right )} + \frac {a b c^{\frac {3}{2}} e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{\Gamma \left (\frac {15}{4}\right )} + \frac {a b \sqrt {c} d e^{\frac {5}{2}} x^{\frac {15}{2}} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{\Gamma \left (\frac {19}{4}\right )} + \frac {b^{2} c^{\frac {3}{2}} e^{\frac {5}{2}} x^{\frac {15}{2}} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \Gamma \left (\frac {19}{4}\right )} + \frac {b^{2} \sqrt {c} d e^{\frac {5}{2}} x^{\frac {19}{2}} \Gamma \left (\frac {19}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {19}{4} \\ \frac {23}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \Gamma \left (\frac {23}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(5/2)*(b*x**2+a)**2*(d*x**2+c)**(3/2),x)

[Out]

a**2*c**(3/2)*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), d*x**2*exp_polar(I*pi)/c)/(2*gamma(11/4
)) + a**2*sqrt(c)*d*e**(5/2)*x**(11/2)*gamma(11/4)*hyper((-1/2, 11/4), (15/4,), d*x**2*exp_polar(I*pi)/c)/(2*g
amma(15/4)) + a*b*c**(3/2)*e**(5/2)*x**(11/2)*gamma(11/4)*hyper((-1/2, 11/4), (15/4,), d*x**2*exp_polar(I*pi)/
c)/gamma(15/4) + a*b*sqrt(c)*d*e**(5/2)*x**(15/2)*gamma(15/4)*hyper((-1/2, 15/4), (19/4,), d*x**2*exp_polar(I*
pi)/c)/gamma(19/4) + b**2*c**(3/2)*e**(5/2)*x**(15/2)*gamma(15/4)*hyper((-1/2, 15/4), (19/4,), d*x**2*exp_pola
r(I*pi)/c)/(2*gamma(19/4)) + b**2*sqrt(c)*d*e**(5/2)*x**(19/2)*gamma(19/4)*hyper((-1/2, 19/4), (23/4,), d*x**2
*exp_polar(I*pi)/c)/(2*gamma(23/4))

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