3.921 \(\int \frac {(e x)^{3/2}}{(a-b x^2)^2 (c-d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=391 \[ \frac {3 \sqrt [4]{c} d^{3/4} e^{3/2} \sqrt {1-\frac {d x^2}{c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{2 \sqrt {c-d x^2} (b c-a d)^2}-\frac {\sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} (5 a d+b c) \Pi \left (-\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{4 a \sqrt [4]{d} \sqrt {c-d x^2} (b c-a d)^2}-\frac {\sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} (5 a d+b c) \Pi \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{4 a \sqrt [4]{d} \sqrt {c-d x^2} (b c-a d)^2}+\frac {3 d e \sqrt {e x}}{2 \sqrt {c-d x^2} (b c-a d)^2}+\frac {e \sqrt {e x}}{2 \left (a-b x^2\right ) \sqrt {c-d x^2} (b c-a d)} \]

[Out]

3/2*d*e*(e*x)^(1/2)/(-a*d+b*c)^2/(-d*x^2+c)^(1/2)+1/2*e*(e*x)^(1/2)/(-a*d+b*c)/(-b*x^2+a)/(-d*x^2+c)^(1/2)+3/2
*c^(1/4)*d^(3/4)*e^(3/2)*EllipticF(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2),I)*(1-d*x^2/c)^(1/2)/(-a*d+b*c)^2/(-d*x
^2+c)^(1/2)-1/4*c^(1/4)*(5*a*d+b*c)*e^(3/2)*EllipticPi(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2),-b^(1/2)*c^(1/2)/a^
(1/2)/d^(1/2),I)*(1-d*x^2/c)^(1/2)/a/d^(1/4)/(-a*d+b*c)^2/(-d*x^2+c)^(1/2)-1/4*c^(1/4)*(5*a*d+b*c)*e^(3/2)*Ell
ipticPi(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2),b^(1/2)*c^(1/2)/a^(1/2)/d^(1/2),I)*(1-d*x^2/c)^(1/2)/a/d^(1/4)/(-a
*d+b*c)^2/(-d*x^2+c)^(1/2)

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Rubi [A]  time = 0.65, antiderivative size = 391, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {466, 471, 527, 523, 224, 221, 409, 1219, 1218} \[ \frac {3 \sqrt [4]{c} d^{3/4} e^{3/2} \sqrt {1-\frac {d x^2}{c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{2 \sqrt {c-d x^2} (b c-a d)^2}-\frac {\sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} (5 a d+b c) \Pi \left (-\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{4 a \sqrt [4]{d} \sqrt {c-d x^2} (b c-a d)^2}-\frac {\sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} (5 a d+b c) \Pi \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{4 a \sqrt [4]{d} \sqrt {c-d x^2} (b c-a d)^2}+\frac {3 d e \sqrt {e x}}{2 \sqrt {c-d x^2} (b c-a d)^2}+\frac {e \sqrt {e x}}{2 \left (a-b x^2\right ) \sqrt {c-d x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d*e*Sqrt[e*x])/(2*(b*c - a*d)^2*Sqrt[c - d*x^2]) + (e*Sqrt[e*x])/(2*(b*c - a*d)*(a - b*x^2)*Sqrt[c - d*x^2]
) + (3*c^(1/4)*d^(3/4)*e^(3/2)*Sqrt[1 - (d*x^2)/c]*EllipticF[ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1
])/(2*(b*c - a*d)^2*Sqrt[c - d*x^2]) - (c^(1/4)*(b*c + 5*a*d)*e^(3/2)*Sqrt[1 - (d*x^2)/c]*EllipticPi[-((Sqrt[b
]*Sqrt[c])/(Sqrt[a]*Sqrt[d])), ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1])/(4*a*d^(1/4)*(b*c - a*d)^2*
Sqrt[c - d*x^2]) - (c^(1/4)*(b*c + 5*a*d)*e^(3/2)*Sqrt[1 - (d*x^2)/c]*EllipticPi[(Sqrt[b]*Sqrt[c])/(Sqrt[a]*Sq
rt[d]), ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1])/(4*a*d^(1/4)*(b*c - a*d)^2*Sqrt[c - d*x^2])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 409

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-(d/c), 2]*x^2)), x], x] + Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-(d/c), 2]*x^2)), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 471

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 527

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

Rule 1218

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

Rule 1219

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[Sqrt[1 + (c*x^4)/a]/Sqrt[a + c*x^4]
, Int[1/((d + e*x^2)*Sqrt[1 + (c*x^4)/a]), x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {(e x)^{3/2}}{\left (a-b x^2\right )^2 \left (c-d x^2\right )^{3/2}} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {x^4}{\left (a-\frac {b x^4}{e^2}\right )^2 \left (c-\frac {d x^4}{e^2}\right )^{3/2}} \, dx,x,\sqrt {e x}\right )}{e}\\ &=\frac {e \sqrt {e x}}{2 (b c-a d) \left (a-b x^2\right ) \sqrt {c-d x^2}}-\frac {e \operatorname {Subst}\left (\int \frac {c+\frac {5 d x^4}{e^2}}{\left (a-\frac {b x^4}{e^2}\right ) \left (c-\frac {d x^4}{e^2}\right )^{3/2}} \, dx,x,\sqrt {e x}\right )}{2 (b c-a d)}\\ &=\frac {3 d e \sqrt {e x}}{2 (b c-a d)^2 \sqrt {c-d x^2}}+\frac {e \sqrt {e x}}{2 (b c-a d) \left (a-b x^2\right ) \sqrt {c-d x^2}}+\frac {e^3 \operatorname {Subst}\left (\int \frac {-\frac {2 c (b c+2 a d)}{e^2}-\frac {6 b c d x^4}{e^4}}{\left (a-\frac {b x^4}{e^2}\right ) \sqrt {c-\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{4 c (b c-a d)^2}\\ &=\frac {3 d e \sqrt {e x}}{2 (b c-a d)^2 \sqrt {c-d x^2}}+\frac {e \sqrt {e x}}{2 (b c-a d) \left (a-b x^2\right ) \sqrt {c-d x^2}}+\frac {(3 d e) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{2 (b c-a d)^2}-\frac {((b c+5 a d) e) \operatorname {Subst}\left (\int \frac {1}{\left (a-\frac {b x^4}{e^2}\right ) \sqrt {c-\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{2 (b c-a d)^2}\\ &=\frac {3 d e \sqrt {e x}}{2 (b c-a d)^2 \sqrt {c-d x^2}}+\frac {e \sqrt {e x}}{2 (b c-a d) \left (a-b x^2\right ) \sqrt {c-d x^2}}-\frac {((b c+5 a d) e) \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {\sqrt {b} x^2}{\sqrt {a} e}\right ) \sqrt {c-\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{4 a (b c-a d)^2}-\frac {((b c+5 a d) e) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {\sqrt {b} x^2}{\sqrt {a} e}\right ) \sqrt {c-\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{4 a (b c-a d)^2}+\frac {\left (3 d e \sqrt {1-\frac {d x^2}{c}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {d x^4}{c e^2}}} \, dx,x,\sqrt {e x}\right )}{2 (b c-a d)^2 \sqrt {c-d x^2}}\\ &=\frac {3 d e \sqrt {e x}}{2 (b c-a d)^2 \sqrt {c-d x^2}}+\frac {e \sqrt {e x}}{2 (b c-a d) \left (a-b x^2\right ) \sqrt {c-d x^2}}+\frac {3 \sqrt [4]{c} d^{3/4} e^{3/2} \sqrt {1-\frac {d x^2}{c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{2 (b c-a d)^2 \sqrt {c-d x^2}}-\frac {\left ((b c+5 a d) e \sqrt {1-\frac {d x^2}{c}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {\sqrt {b} x^2}{\sqrt {a} e}\right ) \sqrt {1-\frac {d x^4}{c e^2}}} \, dx,x,\sqrt {e x}\right )}{4 a (b c-a d)^2 \sqrt {c-d x^2}}-\frac {\left ((b c+5 a d) e \sqrt {1-\frac {d x^2}{c}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {\sqrt {b} x^2}{\sqrt {a} e}\right ) \sqrt {1-\frac {d x^4}{c e^2}}} \, dx,x,\sqrt {e x}\right )}{4 a (b c-a d)^2 \sqrt {c-d x^2}}\\ &=\frac {3 d e \sqrt {e x}}{2 (b c-a d)^2 \sqrt {c-d x^2}}+\frac {e \sqrt {e x}}{2 (b c-a d) \left (a-b x^2\right ) \sqrt {c-d x^2}}+\frac {3 \sqrt [4]{c} d^{3/4} e^{3/2} \sqrt {1-\frac {d x^2}{c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{2 (b c-a d)^2 \sqrt {c-d x^2}}-\frac {\sqrt [4]{c} (b c+5 a d) e^{3/2} \sqrt {1-\frac {d x^2}{c}} \Pi \left (-\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{4 a \sqrt [4]{d} (b c-a d)^2 \sqrt {c-d x^2}}-\frac {\sqrt [4]{c} (b c+5 a d) e^{3/2} \sqrt {1-\frac {d x^2}{c}} \Pi \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{4 a \sqrt [4]{d} (b c-a d)^2 \sqrt {c-d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.21, size = 186, normalized size = 0.48 \[ -\frac {e \sqrt {e x} \left (3 b d x^2 \left (b x^2-a\right ) \sqrt {1-\frac {d x^2}{c}} F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\frac {d x^2}{c},\frac {b x^2}{a}\right )+5 \left (b x^2-a\right ) \sqrt {1-\frac {d x^2}{c}} (2 a d+b c) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\frac {d x^2}{c},\frac {b x^2}{a}\right )+5 a \left (2 a d+b \left (c-3 d x^2\right )\right )\right )}{10 a \left (b x^2-a\right ) \sqrt {c-d x^2} (b c-a d)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/10*(e*Sqrt[e*x]*(5*a*(2*a*d + b*(c - 3*d*x^2)) + 5*(b*c + 2*a*d)*(-a + b*x^2)*Sqrt[1 - (d*x^2)/c]*AppellF1[
1/4, 1/2, 1, 5/4, (d*x^2)/c, (b*x^2)/a] + 3*b*d*x^2*(-a + b*x^2)*Sqrt[1 - (d*x^2)/c]*AppellF1[5/4, 1/2, 1, 9/4
, (d*x^2)/c, (b*x^2)/a]))/(a*(b*c - a*d)^2*(-a + b*x^2)*Sqrt[c - d*x^2])

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 2277, normalized size = 5.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b*d*(8*x*a^2*d^2*(a*b)^(1/2)-4*(a*b)^(1/2)*b^2*c^2*x-4*(a*b)^(1/2)*a*b*c*d*x-12*(a*b)^(1/2)*a*b*d^2*x^3+(
(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)
*(a*b)^(1/2)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b+(a*b)^(1/2)*d)*b,1/2*
2^(1/2))*(c*d)^(1/2)*x^2*b^2*c+((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)*(a*b)^(1/2)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*
d)^(1/2)*b-(a*b)^(1/2)*d)*b,1/2*2^(1/2))*(c*d)^(1/2)*x^2*b^2*c-5*((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)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^
(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b+(a*b)^(1/2)*d)*b,1/2*2^(1/2))*x^2*a*b^2*c*d-5*((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)*(a*b)^(1/2)*EllipticPi(((d*x+
(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b+(a*b)^(1/2)*d)*b,1/2*2^(1/2))*(c*d)^(1/2)*a^2*d-5*(
(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)
*(a*b)^(1/2)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b-(a*b)^(1/2)*d)*b,1/2*
2^(1/2))*(c*d)^(1/2)*a^2*d+5*((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)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b+(a*
b)^(1/2)*d)*b,1/2*2^(1/2))*a^2*b*c*d-5*((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)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d)^(
1/2)*b-(a*b)^(1/2)*d)*b,1/2*2^(1/2))*a^2*b*c*d-6*2^(1/2)*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2
^(1/2))*a*b*c*(a*b)^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-1/(c*d)^(1/2)*d*x)^(1/2)*(c*d)^(1/2)*((d*x+
(c*d)^(1/2))/(c*d)^(1/2))^(1/2)+((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)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b+
(a*b)^(1/2)*d)*b,1/2*2^(1/2))*a*b^2*c^2-((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)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d)^
(1/2)*b-(a*b)^(1/2)*d)*b,1/2*2^(1/2))*a*b^2*c^2+6*2^(1/2)*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*
2^(1/2))*x^2*b^2*c*(a*b)^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-1/(c*d)^(1/2)*d*x)^(1/2)*(c*d)^(1/2)*(
(d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)+6*2^(1/2)*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2))*a^
2*d*(a*b)^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-1/(c*d)^(1/2)*d*x)^(1/2)*(c*d)^(1/2)*((d*x+(c*d)^(1/2
))/(c*d)^(1/2))^(1/2)-((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)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b+(a*b)^(1/2
)*d)*b,1/2*2^(1/2))*x^2*b^3*c^2+((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)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b-
(a*b)^(1/2)*d)*b,1/2*2^(1/2))*x^2*b^3*c^2-6*2^(1/2)*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2
))*x^2*a*b*d*(a*b)^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-1/(c*d)^(1/2)*d*x)^(1/2)*(c*d)^(1/2)*((d*x+(
c*d)^(1/2))/(c*d)^(1/2))^(1/2)+12*(a*b)^(1/2)*b^2*c*d*x^3+5*((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)*(a*b)^(1/2)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^
(1/2))^(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b+(a*b)^(1/2)*d)*b,1/2*2^(1/2))*(c*d)^(1/2)*x^2*a*b*d+5*((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)*(a*b)^(1/2)*El
lipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b-(a*b)^(1/2)*d)*b,1/2*2^(1/2))*(c*d)^
(1/2)*x^2*a*b*d+5*((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)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b-(a*b)^(1/2)*d)
*b,1/2*2^(1/2))*x^2*a*b^2*c*d-((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)*(a*b)^(1/2)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)/((c*d
)^(1/2)*b+(a*b)^(1/2)*d)*b,1/2*2^(1/2))*(c*d)^(1/2)*a*b*c-((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)*(a*b)^(1/2)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1
/2))^(1/2),(c*d)^(1/2)/((c*d)^(1/2)*b-(a*b)^(1/2)*d)*b,1/2*2^(1/2))*(c*d)^(1/2)*a*b*c)*(-d*x^2+c)^(1/2)*e*(e*x
)^(1/2)/x/((c*d)^(1/2)*b-(a*b)^(1/2)*d)/((c*d)^(1/2)*b+(a*b)^(1/2)*d)/(a*b)^(1/2)/(b*x^2-a)/(a*d-b*c)^2/(d*x^2
-c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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