3.9.71 \(\int \frac {\sqrt {c-d x^2}}{(e x)^{5/2} (a-b x^2)} \, dx\) [871]

Optimal. Leaf size=308 \[ -\frac {2 \sqrt {c-d x^2}}{3 a e (e x)^{3/2}}+\frac {2 \sqrt [4]{c} d^{3/4} \sqrt {1-\frac {d x^2}{c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{3 a e^{5/2} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} (b c-a d) \sqrt {1-\frac {d x^2}{c}} \Pi \left (-\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{a^2 \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} (b c-a d) \sqrt {1-\frac {d x^2}{c}} \Pi \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{a^2 \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.34, antiderivative size = 308, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {477, 486, 537, 230, 227, 418, 1233, 1232} \begin {gather*} \frac {\sqrt [4]{c} \sqrt {1-\frac {d x^2}{c}} (b c-a d) \Pi \left (-\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{a^2 \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} \sqrt {1-\frac {d x^2}{c}} (b c-a d) \Pi \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{a^2 \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}}+\frac {2 \sqrt [4]{c} d^{3/4} \sqrt {1-\frac {d x^2}{c}} F\left (\left .\text {ArcSin}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{3 a e^{5/2} \sqrt {c-d x^2}}-\frac {2 \sqrt {c-d x^2}}{3 a e (e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 227

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 230

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 418

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 477

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 486

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

Rule 537

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 1232

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

Rule 1233

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 {\sqrt {c-d x^2}}{(e x)^{5/2} \left (a-b x^2\right )} \, dx &=\frac {2 \text {Subst}\left (\int \frac {\sqrt {c-\frac {d x^4}{e^2}}}{x^4 \left (a-\frac {b x^4}{e^2}\right )} \, dx,x,\sqrt {e x}\right )}{e}\\ &=-\frac {2 \sqrt {c-d x^2}}{3 a e (e x)^{3/2}}+\frac {2 \text {Subst}\left (\int \frac {\frac {3 b c-2 a d}{e^2}-\frac {b 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 )}{3 a e}\\ &=-\frac {2 \sqrt {c-d x^2}}{3 a e (e x)^{3/2}}+\frac {(2 d) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{3 a e^3}+\frac {(2 (b c-a d)) \text {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 )}{a e^3}\\ &=-\frac {2 \sqrt {c-d x^2}}{3 a e (e x)^{3/2}}+\frac {(b c-a d) \text {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 )}{a^2 e^3}+\frac {(b c-a d) \text {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 )}{a^2 e^3}+\frac {\left (2 d \sqrt {1-\frac {d x^2}{c}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {d x^4}{c e^2}}} \, dx,x,\sqrt {e x}\right )}{3 a e^3 \sqrt {c-d x^2}}\\ &=-\frac {2 \sqrt {c-d x^2}}{3 a e (e x)^{3/2}}+\frac {2 \sqrt [4]{c} d^{3/4} \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 )}{3 a e^{5/2} \sqrt {c-d x^2}}+\frac {\left ((b c-a d) \sqrt {1-\frac {d x^2}{c}}\right ) \text {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 )}{a^2 e^3 \sqrt {c-d x^2}}+\frac {\left ((b c-a d) \sqrt {1-\frac {d x^2}{c}}\right ) \text {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 )}{a^2 e^3 \sqrt {c-d x^2}}\\ &=-\frac {2 \sqrt {c-d x^2}}{3 a e (e x)^{3/2}}+\frac {2 \sqrt [4]{c} d^{3/4} \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 )}{3 a e^{5/2} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} (b c-a d) \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 )}{a^2 \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} (b c-a d) \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 )}{a^2 \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 10.12, size = 146, normalized size = 0.47 \begin {gather*} \frac {x \left (10 (3 b c-2 a d) x^2 \sqrt {1-\frac {d x^2}{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 )-2 \left (5 a \left (c-d x^2\right )+b d x^4 \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 )\right )\right )}{15 a^2 (e x)^{5/2} \sqrt {c-d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(10*(3*b*c - 2*a*d)*x^2*Sqrt[1 - (d*x^2)/c]*AppellF1[1/4, 1/2, 1, 5/4, (d*x^2)/c, (b*x^2)/a] - 2*(5*a*(c -
d*x^2) + b*d*x^4*Sqrt[1 - (d*x^2)/c]*AppellF1[5/4, 1/2, 1, 9/4, (d*x^2)/c, (b*x^2)/a])))/(15*a^2*(e*x)^(5/2)*S
qrt[c - d*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1155\) vs. \(2(230)=460\).
time = 0.13, size = 1156, normalized size = 3.75 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-d*x^2+c)^(1/2)/(e*x)^(5/2)/(-b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

-1/6*b*d*(2*2^(1/2)*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*d*x*(a*b)^(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)*(-d*x/(c*d)^(1/2))^(1/2)-2*2^(1/
2)*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2))*b*c*x*(a*b)^(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)*(-d*x/(c*d)^(1/2))^(1/2)+3*2^(1/2)*EllipticPi(((d
*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)*b/((c*d)^(1/2)*b+(a*b)^(1/2)*d),1/2*2^(1/2))*a*b*c*d*x*((d*x+(c
*d)^(1/2))/(c*d)^(1/2))^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-d*x/(c*d)^(1/2))^(1/2)-3*2^(1/2)*Ellipt
icPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)*b/((c*d)^(1/2)*b+(a*b)^(1/2)*d),1/2*2^(1/2))*a*d*x*(a*b
)^(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)*(-d*x/(c*d)^(
1/2))^(1/2)-3*2^(1/2)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)*b/((c*d)^(1/2)*b+(a*b)^(1/2
)*d),1/2*2^(1/2))*b^2*c^2*x*((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-d*x
/(c*d)^(1/2))^(1/2)+3*2^(1/2)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)*b/((c*d)^(1/2)*b+(a
*b)^(1/2)*d),1/2*2^(1/2))*b*c*x*(a*b)^(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)*(-d*x/(c*d)^(1/2))^(1/2)-3*2^(1/2)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*
d)^(1/2)*b/((c*d)^(1/2)*b-(a*b)^(1/2)*d),1/2*2^(1/2))*a*b*c*d*x*((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*((-d*x+(
c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-d*x/(c*d)^(1/2))^(1/2)-3*2^(1/2)*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(
1/2),(c*d)^(1/2)*b/((c*d)^(1/2)*b-(a*b)^(1/2)*d),1/2*2^(1/2))*a*d*x*(a*b)^(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)*(-d*x/(c*d)^(1/2))^(1/2)+3*2^(1/2)*EllipticPi(((d*x
+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)*b/((c*d)^(1/2)*b-(a*b)^(1/2)*d),1/2*2^(1/2))*b^2*c^2*x*((d*x+(c*d
)^(1/2))/(c*d)^(1/2))^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-d*x/(c*d)^(1/2))^(1/2)+3*2^(1/2)*Elliptic
Pi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)*b/((c*d)^(1/2)*b-(a*b)^(1/2)*d),1/2*2^(1/2))*b*c*x*(a*b)^
(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)*(-d*x/(c*d)^(1/
2))^(1/2)+4*a*d^2*x^2*(a*b)^(1/2)-4*b*c*d*x^2*(a*b)^(1/2)-4*a*c*d*(a*b)^(1/2)+4*b*c^2*(a*b)^(1/2))/(-d*x^2+c)^
(1/2)/x/a/e^2/(e*x)^(1/2)/(a*b)^(1/2)/((c*d)^(1/2)*b+(a*b)^(1/2)*d)/((c*d)^(1/2)*b-(a*b)^(1/2)*d)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-d*x^2+c)^(1/2)/(e*x)^(5/2)/(-b*x^2+a),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-d*x^2+c)^(1/2)/(e*x)^(5/2)/(-b*x^2+a),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\sqrt {c - d x^{2}}}{- a \left (e x\right )^{\frac {5}{2}} + b x^{2} \left (e x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-d*x**2+c)**(1/2)/(e*x)**(5/2)/(-b*x**2+a),x)

[Out]

-Integral(sqrt(c - d*x**2)/(-a*(e*x)**(5/2) + b*x**2*(e*x)**(5/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-d*x^2+c)^(1/2)/(e*x)^(5/2)/(-b*x^2+a),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {c-d\,x^2}}{{\left (e\,x\right )}^{5/2}\,\left (a-b\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - d*x^2)^(1/2)/((e*x)^(5/2)*(a - b*x^2)),x)

[Out]

int((c - d*x^2)^(1/2)/((e*x)^(5/2)*(a - b*x^2)), x)

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