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

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

[Out]

b*e^(3/2)*(-a*f+b*e)*(1/(1+f*x^2/e))^(1/2)*(1+f*x^2/e)^(1/2)*EllipticPi(x*f^(1/2)/e^(1/2)/(1+f*x^2/e)^(1/2),1-
b*e/a/f,(1-d*e/c/f)^(1/2))*(d*x^2+c)^(1/2)/a/c/(-a*d+b*c)^2/f^(1/2)/(e*(d*x^2+c)/c/(f*x^2+e))^(1/2)/(f*x^2+e)^
(1/2)+1/3*e^(3/2)*(1/(1+f*x^2/e))^(1/2)*(1+f*x^2/e)^(1/2)*EllipticF(x*f^(1/2)/e^(1/2)/(1+f*x^2/e)^(1/2),(1-d*e
/c/f)^(1/2))*f^(1/2)*(d*x^2+c)^(1/2)/c^2/(-a*d+b*c)/(e*(d*x^2+c)/c/(f*x^2+e))^(1/2)/(f*x^2+e)^(1/2)-1/3*(-c*f+
d*e)*x*(f*x^2+e)^(1/2)/c/(-a*d+b*c)/(d*x^2+c)^(3/2)-1/3*(b*c*(-c*f+5*d*e)-2*a*d*(c*f+d*e))*(1/(1+d*x^2/c))^(1/
2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(1-c*f/d/e)^(1/2))*(f*x^2+e)^(1/2)/c^(3/2)/
(-a*d+b*c)^2/d^(1/2)/(d*x^2+c)^(1/2)/(c*(f*x^2+e)/e/(d*x^2+c))^(1/2)

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Rubi [A]  time = 0.40, antiderivative size = 391, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {544, 539, 526, 525, 418, 411} \[ \frac {e^{3/2} \sqrt {f} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 \sqrt {e+f x^2} (b c-a d) \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {\sqrt {e+f x^2} (b c (5 d e-c f)-2 a d (c f+d e)) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{3 c^{3/2} \sqrt {d} \sqrt {c+d x^2} (b c-a d)^2 \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}+\frac {b e^{3/2} \sqrt {c+d x^2} (b e-a f) \Pi \left (1-\frac {b e}{a f};\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{a c \sqrt {f} \sqrt {e+f x^2} (b c-a d)^2 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {x \sqrt {e+f x^2} (d e-c f)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] && PosQ[d/c]

Rule 526

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)/(a*b*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n
)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + b*e - a*f) + d*(b*e*n*(p + 1) + (b*e - a*f)*(n*q + 1))*x
^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rule 539

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(c*Sqrt[e +
f*x^2]*EllipticPi[1 - (b*c)/(a*d), ArcTan[Rt[d/c, 2]*x], 1 - (c*f)/(d*e)])/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sq
rt[(c*(e + f*x^2))/(e*(c + d*x^2))]), x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c]

Rule 544

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

Rubi steps

\begin {align*} \int \frac {\left (e+f x^2\right )^{3/2}}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx &=-\frac {\int \frac {\sqrt {e+f x^2} \left (2 b c d e-a d^2 e-b c^2 f+d^2 (b e-a f) x^2\right )}{\left (c+d x^2\right )^{5/2}} \, dx}{(b c-a d)^2}+\frac {(b (b e-a f)) \int \frac {\sqrt {e+f x^2}}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{(b c-a d)^2}\\ &=-\frac {(d e-c f) x \sqrt {e+f x^2}}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b e^{3/2} (b e-a f) \sqrt {c+d x^2} \Pi \left (1-\frac {b e}{a f};\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{a c (b c-a d)^2 \sqrt {f} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {\int \frac {-d e (b c (5 d e-2 c f)-a d (2 d e+c f))-d f (b c (4 d e-c f)-a d (d e+2 c f)) x^2}{\left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}} \, dx}{3 c d (b c-a d)^2}\\ &=-\frac {(d e-c f) x \sqrt {e+f x^2}}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b e^{3/2} (b e-a f) \sqrt {c+d x^2} \Pi \left (1-\frac {b e}{a f};\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{a c (b c-a d)^2 \sqrt {f} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {(e f) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 c (b c-a d)}-\frac {(b c (5 d e-c f)-2 a d (d e+c f)) \int \frac {\sqrt {e+f x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c (b c-a d)^2}\\ &=-\frac {(d e-c f) x \sqrt {e+f x^2}}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac {(b c (5 d e-c f)-2 a d (d e+c f)) \sqrt {e+f x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{3 c^{3/2} \sqrt {d} (b c-a d)^2 \sqrt {c+d x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}+\frac {e^{3/2} \sqrt {f} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 (b c-a d) \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {b e^{3/2} (b e-a f) \sqrt {c+d x^2} \Pi \left (1-\frac {b e}{a f};\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{a c (b c-a d)^2 \sqrt {f} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}\\ \end {align*}

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Mathematica [C]  time = 1.86, size = 999, normalized size = 2.55 \[ \frac {a b c^3 \left (\frac {d}{c}\right )^{3/2} f^2 x^5+2 a^2 c d^2 \sqrt {\frac {d}{c}} f^2 x^5+2 a^2 d^3 \sqrt {\frac {d}{c}} e f x^5-5 a b c d^2 \sqrt {\frac {d}{c}} e f x^5+2 a^2 d^3 \sqrt {\frac {d}{c}} e^2 x^3-5 a b c d^2 \sqrt {\frac {d}{c}} e^2 x^3+a^2 c^3 \left (\frac {d}{c}\right )^{3/2} f^2 x^3+2 a b c^3 \sqrt {\frac {d}{c}} f^2 x^3-5 a b c^3 \left (\frac {d}{c}\right )^{3/2} e f x^3+5 a^2 c d^2 \sqrt {\frac {d}{c}} e f x^3-3 i b^2 c^2 d e^2 \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \Pi \left (\frac {b c}{a d};i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right ) x^2-3 i a^2 c^2 d f^2 \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \Pi \left (\frac {b c}{a d};i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right ) x^2+6 i a b c^2 d e f \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \Pi \left (\frac {b c}{a d};i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right ) x^2-6 a b c^3 \left (\frac {d}{c}\right )^{3/2} e^2 x+3 a^2 c d^2 \sqrt {\frac {d}{c}} e^2 x+a^2 c^3 \left (\frac {d}{c}\right )^{3/2} e f x+2 a b c^3 \sqrt {\frac {d}{c}} e f x+i a e (b c (c f-5 d e)+2 a d (d e+c f)) \left (d x^2+c\right ) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )-i a (c f-d e) (5 b c e-2 a d e-3 a c f) \left (d x^2+c\right ) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )-3 i b^2 c^3 e^2 \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \Pi \left (\frac {b c}{a d};i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )-3 i a^2 c^3 f^2 \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \Pi \left (\frac {b c}{a d};i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+6 i a b c^3 e f \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \Pi \left (\frac {b c}{a d};i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )}{3 a c^2 \sqrt {\frac {d}{c}} (b c-a d)^2 \left (d x^2+c\right )^{3/2} \sqrt {f x^2+e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*c*d^2*Sqrt[d/c]*e^2*x - 6*a*b*c^3*(d/c)^(3/2)*e^2*x + 2*a*b*c^3*Sqrt[d/c]*e*f*x + a^2*c^3*(d/c)^(3/2)*e
*f*x - 5*a*b*c*d^2*Sqrt[d/c]*e^2*x^3 + 2*a^2*d^3*Sqrt[d/c]*e^2*x^3 + 5*a^2*c*d^2*Sqrt[d/c]*e*f*x^3 - 5*a*b*c^3
*(d/c)^(3/2)*e*f*x^3 + 2*a*b*c^3*Sqrt[d/c]*f^2*x^3 + a^2*c^3*(d/c)^(3/2)*f^2*x^3 - 5*a*b*c*d^2*Sqrt[d/c]*e*f*x
^5 + 2*a^2*d^3*Sqrt[d/c]*e*f*x^5 + 2*a^2*c*d^2*Sqrt[d/c]*f^2*x^5 + a*b*c^3*(d/c)^(3/2)*f^2*x^5 + I*a*e*(b*c*(-
5*d*e + c*f) + 2*a*d*(d*e + c*f))*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticE[I*ArcSinh[Sqrt
[d/c]*x], (c*f)/(d*e)] - I*a*(-(d*e) + c*f)*(5*b*c*e - 2*a*d*e - 3*a*c*f)*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt
[1 + (f*x^2)/e]*EllipticF[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] - (3*I)*b^2*c^3*e^2*Sqrt[1 + (d*x^2)/c]*Sqrt[1
+ (f*x^2)/e]*EllipticPi[(b*c)/(a*d), I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] + (6*I)*a*b*c^3*e*f*Sqrt[1 + (d*x^2)
/c]*Sqrt[1 + (f*x^2)/e]*EllipticPi[(b*c)/(a*d), I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] - (3*I)*a^2*c^3*f^2*Sqrt[
1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticPi[(b*c)/(a*d), I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] - (3*I)*b^2*c^
2*d*e^2*x^2*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticPi[(b*c)/(a*d), I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e
)] + (6*I)*a*b*c^2*d*e*f*x^2*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticPi[(b*c)/(a*d), I*ArcSinh[Sqrt[d/
c]*x], (c*f)/(d*e)] - (3*I)*a^2*c^2*d*f^2*x^2*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticPi[(b*c)/(a*d),
I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)])/(3*a*c^2*Sqrt[d/c]*(b*c - a*d)^2*(c + d*x^2)^(3/2)*Sqrt[e + f*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((f*x^2+e)^(3/2)/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 1879, normalized size = 4.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-2*x^5*a^2*c*d^2*f^2*(-1/c*d)^(1/2)-2*x^3*a^2*d^3*e^2*(-1/c*d)^(1/2)-3*EllipticPi((-1/c*d)^(1/2)*x,1/a*b
*c/d,(-1/e*f)^(1/2)/(-1/c*d)^(1/2))*b^2*c^3*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-2*x^5*a^2*d^3*e*f*(-1/
c*d)^(1/2)-x^3*a^2*c^2*d*f^2*(-1/c*d)^(1/2)-2*x^3*a*b*c^3*f^2*(-1/c*d)^(1/2)-3*x*a^2*c*d^2*e^2*(-1/c*d)^(1/2)+
3*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a^2*c^3*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-3*EllipticPi
((-1/c*d)^(1/2)*x,1/a*b*c/d,(-1/e*f)^(1/2)/(-1/c*d)^(1/2))*a^2*c^3*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)
-5*EllipticE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a*b*c^2*d*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+5*x^5*a*b
*c*d^2*e*f*(-1/c*d)^(1/2)+5*x^3*a*b*c^2*d*e*f*(-1/c*d)^(1/2)+3*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2
*a^2*c^2*d*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-3*EllipticPi((-1/c*d)^(1/2)*x,1/a*b*c/d,(-1/e*f)^(1/2)/
(-1/c*d)^(1/2))*x^2*a^2*c^2*d*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-3*EllipticPi((-1/c*d)^(1/2)*x,1/a*b*
c/d,(-1/e*f)^(1/2)/(-1/c*d)^(1/2))*x^2*b^2*c^2*d*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-EllipticF((-1/c*d
)^(1/2)*x,(c/d/e*f)^(1/2))*a^2*c^2*d*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-5*EllipticF((-1/c*d)^(1/2)*x,
(c/d/e*f)^(1/2))*a*b*c^3*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+5*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1
/2))*a*b*c^2*d*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+6*EllipticPi((-1/c*d)^(1/2)*x,1/a*b*c/d,(-1/e*f)^(1
/2)/(-1/c*d)^(1/2))*a*b*c^3*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+2*EllipticE((-1/c*d)^(1/2)*x,(c/d/e*f)
^(1/2))*a^2*c^2*d*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+6*EllipticPi((-1/c*d)^(1/2)*x,1/a*b*c/d,(-1/e*f)
^(1/2)/(-1/c*d)^(1/2))*x^2*a*b*c^2*d*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+EllipticE((-1/c*d)^(1/2)*x,(c
/d/e*f)^(1/2))*x^2*a*b*c^2*d*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-5*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f
)^(1/2))*x^2*a*b*c^2*d*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+EllipticE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))
*a*b*c^3*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2*a^2*c*d^2
*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+5*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2*a*b*c*d^2*e^2*(
(d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+2*EllipticE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2*a^2*c*d^2*e*f*((d*x^2
+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-5*EllipticE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2*a*b*c*d^2*e^2*((d*x^2+c)/c)
^(1/2)*((f*x^2+e)/e)^(1/2)-2*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2*a^2*d^3*e^2*((d*x^2+c)/c)^(1/2)*(
(f*x^2+e)/e)^(1/2)+2*EllipticE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2*a^2*d^3*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e
)/e)^(1/2)-2*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a^2*c*d^2*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)
+2*EllipticE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a^2*c*d^2*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-x^5*a*b*c
^2*d*f^2*(-1/c*d)^(1/2)-5*x^3*a^2*c*d^2*e*f*(-1/c*d)^(1/2)+5*x^3*a*b*c*d^2*e^2*(-1/c*d)^(1/2)-x*a^2*c^2*d*e*f*
(-1/c*d)^(1/2)-2*x*a*b*c^3*e*f*(-1/c*d)^(1/2)+6*x*a*b*c^2*d*e^2*(-1/c*d)^(1/2))/(f*x^2+e)^(1/2)/(a*d-b*c)^2/(-
1/c*d)^(1/2)/a/c^2/(d*x^2+c)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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