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

Optimal. Leaf size=261 \[ \frac {\sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} \Pi \left (-\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{b \sqrt [4]{d} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} \Pi \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{b \sqrt [4]{d} \sqrt {c-d x^2}}-\frac {2 \sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{b \sqrt [4]{d} \sqrt {c-d x^2}} \]

[Out]

-2*c^(1/4)*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)/b/d^(1/4)/(-d*x^2+c)^(1/
2)+c^(1/4)*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)/b/d^(1/4)/(-d*x^2+c)^(1/2)+c^(1/4)*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)/b/d^(1/4)/(-d*x^2+c)^(1/2)

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Rubi [A]  time = 0.36, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {466, 483, 224, 221, 409, 1219, 1218} \[ \frac {\sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} \Pi \left (-\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{b \sqrt [4]{d} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} \Pi \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{b \sqrt [4]{d} \sqrt {c-d x^2}}-\frac {2 \sqrt [4]{c} e^{3/2} \sqrt {1-\frac {d x^2}{c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{b \sqrt [4]{d} \sqrt {c-d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c^(1/4)*e^(3/2)*Sqrt[1 - (d*x^2)/c]*EllipticF[ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1])/(b*d^(1/
4)*Sqrt[c - d*x^2]) + (c^(1/4)*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])/(b*d^(1/4)*Sqrt[c - d*x^2]) + (c^(1/4)*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])/(
b*d^(1/4)*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 483

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

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 ) \sqrt {c-d x^2}} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {x^4}{\left (a-\frac {b x^4}{e^2}\right ) \sqrt {c-\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{e}\\ &=-\frac {(2 e) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{b}+\frac {(2 a 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 )}{b}\\ &=\frac {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 )}{b}+\frac {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 )}{b}-\frac {\left (2 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 )}{b \sqrt {c-d x^2}}\\ &=-\frac {2 \sqrt [4]{c} 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 )}{b \sqrt [4]{d} \sqrt {c-d x^2}}+\frac {\left (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 )}{b \sqrt {c-d x^2}}+\frac {\left (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 )}{b \sqrt {c-d x^2}}\\ &=-\frac {2 \sqrt [4]{c} 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 )}{b \sqrt [4]{d} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} 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 )}{b \sqrt [4]{d} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} 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 )}{b \sqrt [4]{d} \sqrt {c-d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 70, normalized size = 0.27 \[ \frac {2 x (e x)^{3/2} \sqrt {\frac {c-d x^2}{c}} F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\frac {d x^2}{c},\frac {b x^2}{a}\right )}{5 a \sqrt {c-d x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)/(-b*x^2+a)/(-d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.03, size = 417, normalized size = 1.60 \[ -\frac {\left (-\sqrt {c d}\, a b \EllipticPi \left (\sqrt {\frac {d x +\sqrt {c d}}{\sqrt {c d}}}, \frac {\sqrt {c d}\, b}{\sqrt {c d}\, b -\sqrt {a b}\, d}, \frac {\sqrt {2}}{2}\right )+\sqrt {c d}\, a b \EllipticPi \left (\sqrt {\frac {d x +\sqrt {c d}}{\sqrt {c d}}}, \frac {\sqrt {c d}\, b}{\sqrt {c d}\, b +\sqrt {a b}\, d}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {a b}\, a d \EllipticF \left (\sqrt {\frac {d x +\sqrt {c d}}{\sqrt {c d}}}, \frac {\sqrt {2}}{2}\right )-\sqrt {a b}\, a d \EllipticPi \left (\sqrt {\frac {d x +\sqrt {c d}}{\sqrt {c d}}}, \frac {\sqrt {c d}\, b}{\sqrt {c d}\, b -\sqrt {a b}\, d}, \frac {\sqrt {2}}{2}\right )-\sqrt {a b}\, a d \EllipticPi \left (\sqrt {\frac {d x +\sqrt {c d}}{\sqrt {c d}}}, \frac {\sqrt {c d}\, b}{\sqrt {c d}\, b +\sqrt {a b}\, d}, \frac {\sqrt {2}}{2}\right )-2 \sqrt {a b}\, b c \EllipticF \left (\sqrt {\frac {d x +\sqrt {c d}}{\sqrt {c d}}}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {c d}\, \sqrt {-\frac {d x}{\sqrt {c d}}}\, \sqrt {\frac {-d x +\sqrt {c d}}{\sqrt {c d}}}\, \sqrt {2}\, \sqrt {\frac {d x +\sqrt {c d}}{\sqrt {c d}}}\, \sqrt {-d \,x^{2}+c}\, \sqrt {e x}\, e}{2 \left (\sqrt {c d}\, b -\sqrt {a b}\, d \right ) \left (\sqrt {c d}\, b +\sqrt {a b}\, d \right ) \sqrt {a b}\, \left (d \,x^{2}-c \right ) x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)/(-b*x^2+a)/(-d*x^2+c)^(1/2),x)

[Out]

-1/2*(2*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*d*(a*b)^(1/2)-2*EllipticF(((d*x+(c*d)^(
1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2))*b*c*(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))*a*b*(c*d)^(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*d*(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))*a*b*(c*d)^(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*d*(a*b)^(1/2))*(c*d)^(
1/2)*(-1/(c*d)^(1/2)*d*x)^(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)*(-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
)/(d*x^2-c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)/(-b*x^2+a)/(-d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)/((a - b*x^2)*(c - d*x^2)^(1/2)),x)

[Out]

int((e*x)^(3/2)/((a - b*x^2)*(c - d*x^2)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(3/2)/(-b*x**2+a)/(-d*x**2+c)**(1/2),x)

[Out]

-Integral((e*x)**(3/2)/(-a*sqrt(c - d*x**2) + b*x**2*sqrt(c - d*x**2)), x)

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