3.878 \(\int \frac {(c-d x^2)^{3/2}}{(e x)^{5/2} (a-b x^2)} \, dx\)

Optimal. Leaf size=330 \[ \frac {\sqrt [4]{c} \sqrt {1-\frac {d x^2}{c}} (b c-a d)^2 \Pi \left (-\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{a^2 b \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} \sqrt {1-\frac {d x^2}{c}} (b c-a d)^2 \Pi \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{a^2 b \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}}+\frac {2 \sqrt [4]{c} d^{3/4} \sqrt {1-\frac {d x^2}{c}} (b c-3 a d) F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{3 a b e^{5/2} \sqrt {c-d x^2}}-\frac {2 c \sqrt {c-d x^2}}{3 a e (e x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.59, antiderivative size = 330, 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 = {466, 474, 523, 224, 221, 409, 1219, 1218} \[ \frac {\sqrt [4]{c} \sqrt {1-\frac {d x^2}{c}} (b c-a d)^2 \Pi \left (-\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{a^2 b \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} \sqrt {1-\frac {d x^2}{c}} (b c-a d)^2 \Pi \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{a^2 b \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}}+\frac {2 \sqrt [4]{c} d^{3/4} \sqrt {1-\frac {d x^2}{c}} (b c-3 a d) F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )\right |-1\right )}{3 a b e^{5/2} \sqrt {c-d x^2}}-\frac {2 c \sqrt {c-d x^2}}{3 a e (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*Sqrt[c - d*x^2])/(3*a*e*(e*x)^(3/2)) + (2*c^(1/4)*d^(3/4)*(b*c - 3*a*d)*Sqrt[1 - (d*x^2)/c]*EllipticF[Ar
cSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1])/(3*a*b*e^(5/2)*Sqrt[c - d*x^2]) + (c^(1/4)*(b*c - a*d)^2*Sqr
t[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])/(a^2*b*d^(1/4)*e^(5/2)*Sqrt[c - d*x^2]) + (c^(1/4)*(b*c - a*d)^2*Sqrt[1 - (d*x^2)/c]*EllipticPi[(Sqr
t[b]*Sqrt[c])/(Sqrt[a]*Sqrt[d]), ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1])/(a^2*b*d^(1/4)*e^(5/2)*Sq
rt[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 474

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(c*(e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(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 - 2)*Simp[c*(c*b - a*d)*(m + 1) + c*n*(b*c*(p + 1) + a*d*(q - 1)) + d*((c*b - a*
d)*(m + 1) + c*b*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0]
 && GtQ[q, 1] && LtQ[m, -1] && 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 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 {\left (c-d x^2\right )^{3/2}}{(e x)^{5/2} \left (a-b x^2\right )} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {\left (c-\frac {d x^4}{e^2}\right )^{3/2}}{x^4 \left (a-\frac {b x^4}{e^2}\right )} \, dx,x,\sqrt {e x}\right )}{e}\\ &=-\frac {2 c \sqrt {c-d x^2}}{3 a e (e x)^{3/2}}+\frac {2 \operatorname {Subst}\left (\int \frac {\frac {c (3 b c-5 a d)}{e^2}-\frac {d (b c-3 a 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 c \sqrt {c-d x^2}}{3 a e (e x)^{3/2}}+\frac {(2 d (b c-3 a d)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{3 a b e^3}+\frac {\left (2 (b c-a d)^2\right ) \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 )}{a b e^3}\\ &=-\frac {2 c \sqrt {c-d x^2}}{3 a e (e x)^{3/2}}+\frac {(b c-a d)^2 \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 )}{a^2 b e^3}+\frac {(b c-a d)^2 \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 )}{a^2 b e^3}+\frac {\left (2 d (b c-3 a d) \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 )}{3 a b e^3 \sqrt {c-d x^2}}\\ &=-\frac {2 c \sqrt {c-d x^2}}{3 a e (e x)^{3/2}}+\frac {2 \sqrt [4]{c} d^{3/4} (b c-3 a d) \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 b e^{5/2} \sqrt {c-d x^2}}+\frac {\left ((b c-a d)^2 \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 )}{a^2 b e^3 \sqrt {c-d x^2}}+\frac {\left ((b c-a d)^2 \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 )}{a^2 b e^3 \sqrt {c-d x^2}}\\ &=-\frac {2 c \sqrt {c-d x^2}}{3 a e (e x)^{3/2}}+\frac {2 \sqrt [4]{c} d^{3/4} (b c-3 a d) \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 b e^{5/2} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} (b c-a d)^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 )}{a^2 b \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}}+\frac {\sqrt [4]{c} (b c-a d)^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 )}{a^2 b \sqrt [4]{d} e^{5/2} \sqrt {c-d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 153, normalized size = 0.46 \[ \frac {x \left (10 c x^2 \sqrt {1-\frac {d x^2}{c}} (3 b c-5 a d) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\frac {d x^2}{c},\frac {b x^2}{a}\right )-2 d x^4 \sqrt {1-\frac {d x^2}{c}} (b c-3 a d) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\frac {d x^2}{c},\frac {b x^2}{a}\right )-10 a c \left (c-d x^2\right )\right )}{15 a^2 (e x)^{5/2} \sqrt {c-d x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(-10*a*c*(c - d*x^2) + 10*c*(3*b*c - 5*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*d*(b*c - 3*a*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)*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((-d*x^2+c)^(3/2)/(e*x)^(5/2)/(-b*x^2+a),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 1740, normalized size = 5.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(-d*x^2+c)^(1/2)*d*(6*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2))*2^(1/2)*x*a^2*d^2*(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)*((-d*x+(c*d)^(1/2))/(c*d)
^(1/2))^(1/2)-8*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2))*2^(1/2)*x*a*b*c*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)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/
2)+2*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2))*2^(1/2)*x*b^2*c^2*(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)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)+3*Ellipt
icPi(((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))*2^(1/2)*x*
a^2*b*c*d^2*(-1/(c*d)^(1/2)*d*x)^(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)-3*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))*2^(1/2)*x*a^2*d^2*(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)
*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)-6*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))*2^(1/2)*x*a*b^2*c^2*d*(-1/(c*d)^(1/2)*d*x)^(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)+6*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))*2^(1/2)*x*a*b*c*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)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)+3*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))*2^(1/2)*x*b^3*c^3*(-1/(c
*d)^(1/2)*d*x)^(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)-3*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))*2^(1/2)*x*b^2*
c^2*(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)*((-d*x+(c*d)^(1/2
))/(c*d)^(1/2))^(1/2)-3*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))*2^(1/2)*x*a^2*b*c*d^2*(-1/(c*d)^(1/2)*d*x)^(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)-3*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))*2^(1/2)*x*a^2*d^2*(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)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)+6*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))*2^(1/2)*x*a*b^2*c^2*d*(-1/(c*d)^(1/2)*d*x)
^(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)+6*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))*2^(1/2)*x*a*b*c*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)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2)
)^(1/2)-3*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))*2^(1/2)*x*b^3*c^3*(-1/(c*d)^(1/2)*d*x)^(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)-3*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))*2^(1/2)*x*b^2*c^2*(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)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)-4*x^2*a*b*c*d^2*(a*b)^(1/2)+4*x^2*b^2*c^2*d*(a*b)^(1/2)+4*
a*b*c^2*d*(a*b)^(1/2)-4*b^2*c^3*(a*b)^(1/2))/x/a/e^2/(e*x)^(1/2)/(d*x^2-c)/(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 \[ -\int \frac {{\left (-d x^{2} + c\right )}^{\frac {3}{2}}}{{\left (b x^{2} - a\right )} \left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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