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

Optimal. Leaf size=501 \[ -\frac {\sqrt {e} \sqrt {c+d x^2} \left (10 a d f (2 d e-3 c f)-b \left (15 c^2 f^2-41 c d e f+24 d^2 e^2\right )\right ) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 f^{7/2} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {\sqrt {c+d x^2} \left (5 a f \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )-2 b e \left (19 c^2 f^2-44 c d e f+24 d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 \sqrt {e} f^{7/2} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {x \sqrt {c+d x^2} \left (5 a f \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )-2 b e \left (19 c^2 f^2-44 c d e f+24 d^2 e^2\right )\right )}{15 e f^3 \sqrt {e+f x^2}}-\frac {d x \sqrt {c+d x^2} \sqrt {e+f x^2} (b e (24 d e-23 c f)-5 a f (4 d e-3 c f))}{15 e f^3}+\frac {d x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2} (6 b e-5 a f)}{5 e f^2}-\frac {x \left (c+d x^2\right )^{5/2} (b e-a f)}{e f \sqrt {e+f x^2}} \]

[Out]

-(-a*f+b*e)*x*(d*x^2+c)^(5/2)/e/f/(f*x^2+e)^(1/2)-1/15*(5*a*f*(3*c^2*f^2-13*c*d*e*f+8*d^2*e^2)-2*b*e*(19*c^2*f
^2-44*c*d*e*f+24*d^2*e^2))*x*(d*x^2+c)^(1/2)/e/f^3/(f*x^2+e)^(1/2)+1/15*(5*a*f*(3*c^2*f^2-13*c*d*e*f+8*d^2*e^2
)-2*b*e*(19*c^2*f^2-44*c*d*e*f+24*d^2*e^2))*(1/(1+f*x^2/e))^(1/2)*(1+f*x^2/e)^(1/2)*EllipticE(x*f^(1/2)/e^(1/2
)/(1+f*x^2/e)^(1/2),(1-d*e/c/f)^(1/2))*(d*x^2+c)^(1/2)/f^(7/2)/e^(1/2)/(e*(d*x^2+c)/c/(f*x^2+e))^(1/2)/(f*x^2+
e)^(1/2)-1/15*(10*a*d*f*(-3*c*f+2*d*e)-b*(15*c^2*f^2-41*c*d*e*f+24*d^2*e^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))*e^(1/2)*(d*x^2+c)^(1/2)/f^(7/2)/(e*(d*
x^2+c)/c/(f*x^2+e))^(1/2)/(f*x^2+e)^(1/2)+1/5*d*(-5*a*f+6*b*e)*x*(d*x^2+c)^(3/2)*(f*x^2+e)^(1/2)/e/f^2-1/15*d*
(b*e*(-23*c*f+24*d*e)-5*a*f*(-3*c*f+4*d*e))*x*(d*x^2+c)^(1/2)*(f*x^2+e)^(1/2)/e/f^3

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Rubi [A]  time = 0.60, antiderivative size = 501, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {526, 528, 531, 418, 492, 411} \[ -\frac {x \sqrt {c+d x^2} \left (5 a f \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )-2 b e \left (19 c^2 f^2-44 c d e f+24 d^2 e^2\right )\right )}{15 e f^3 \sqrt {e+f x^2}}-\frac {\sqrt {e} \sqrt {c+d x^2} \left (10 a d f (2 d e-3 c f)-b \left (15 c^2 f^2-41 c d e f+24 d^2 e^2\right )\right ) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 f^{7/2} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {\sqrt {c+d x^2} \left (5 a f \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )-2 b e \left (19 c^2 f^2-44 c d e f+24 d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 \sqrt {e} f^{7/2} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {d x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2} (6 b e-5 a f)}{5 e f^2}-\frac {d x \sqrt {c+d x^2} \sqrt {e+f x^2} (b e (24 d e-23 c f)-5 a f (4 d e-3 c f))}{15 e f^3}-\frac {x \left (c+d x^2\right )^{5/2} (b e-a f)}{e f \sqrt {e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*a*f*(8*d^2*e^2 - 13*c*d*e*f + 3*c^2*f^2) - 2*b*e*(24*d^2*e^2 - 44*c*d*e*f + 19*c^2*f^2))*x*Sqrt[c + d*x^2
])/(15*e*f^3*Sqrt[e + f*x^2]) - ((b*e - a*f)*x*(c + d*x^2)^(5/2))/(e*f*Sqrt[e + f*x^2]) - (d*(b*e*(24*d*e - 23
*c*f) - 5*a*f*(4*d*e - 3*c*f))*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])/(15*e*f^3) + (d*(6*b*e - 5*a*f)*x*(c + d*x^2
)^(3/2)*Sqrt[e + f*x^2])/(5*e*f^2) + ((5*a*f*(8*d^2*e^2 - 13*c*d*e*f + 3*c^2*f^2) - 2*b*e*(24*d^2*e^2 - 44*c*d
*e*f + 19*c^2*f^2))*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*Sqrt[e]*f^(7/
2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) - (Sqrt[e]*(10*a*d*f*(2*d*e - 3*c*f) - b*(24*d^2*e^2
- 41*c*d*e*f + 15*c^2*f^2))*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*f^(7
/2)*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 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, 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 528

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

Rule 531

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{\left (e+f x^2\right )^{3/2}} \, dx &=-\frac {(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt {e+f x^2}}-\frac {\int \frac {\left (c+d x^2\right )^{3/2} \left (-b c e-d (6 b e-5 a f) x^2\right )}{\sqrt {e+f x^2}} \, dx}{e f}\\ &=-\frac {(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt {e+f x^2}}+\frac {d (6 b e-5 a f) x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 e f^2}-\frac {\int \frac {\sqrt {c+d x^2} \left (c e (6 b d e-5 b c f-5 a d f)+d (b e (24 d e-23 c f)-5 a f (4 d e-3 c f)) x^2\right )}{\sqrt {e+f x^2}} \, dx}{5 e f^2}\\ &=-\frac {(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt {e+f x^2}}-\frac {d (b e (24 d e-23 c f)-5 a f (4 d e-3 c f)) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 e f^3}+\frac {d (6 b e-5 a f) x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 e f^2}-\frac {\int \frac {c e \left (10 a d f (2 d e-3 c f)-b \left (24 d^2 e^2-41 c d e f+15 c^2 f^2\right )\right )+d \left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right ) x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{15 e f^3}\\ &=-\frac {(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt {e+f x^2}}-\frac {d (b e (24 d e-23 c f)-5 a f (4 d e-3 c f)) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 e f^3}+\frac {d (6 b e-5 a f) x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 e f^2}-\frac {\left (c \left (10 a d f (2 d e-3 c f)-b \left (24 d^2 e^2-41 c d e f+15 c^2 f^2\right )\right )\right ) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{15 f^3}-\frac {\left (d \left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right )\right ) \int \frac {x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{15 e f^3}\\ &=-\frac {\left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right ) x \sqrt {c+d x^2}}{15 e f^3 \sqrt {e+f x^2}}-\frac {(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt {e+f x^2}}-\frac {d (b e (24 d e-23 c f)-5 a f (4 d e-3 c f)) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 e f^3}+\frac {d (6 b e-5 a f) x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 e f^2}-\frac {\sqrt {e} \left (10 a d f (2 d e-3 c f)-b \left (24 d^2 e^2-41 c d e f+15 c^2 f^2\right )\right ) \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 f^{7/2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {\left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right ) \int \frac {\sqrt {c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{15 f^3}\\ &=-\frac {\left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right ) x \sqrt {c+d x^2}}{15 e f^3 \sqrt {e+f x^2}}-\frac {(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt {e+f x^2}}-\frac {d (b e (24 d e-23 c f)-5 a f (4 d e-3 c f)) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 e f^3}+\frac {d (6 b e-5 a f) x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 e f^2}+\frac {\left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right ) \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 \sqrt {e} f^{7/2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}-\frac {\sqrt {e} \left (10 a d f (2 d e-3 c f)-b \left (24 d^2 e^2-41 c d e f+15 c^2 f^2\right )\right ) \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 f^{7/2} \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 = 1.23, size = 369, normalized size = 0.74 \[ \frac {-i e \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (c f-d e) \left (5 a d f (9 c f-8 d e)+b \left (15 c^2 f^2-64 c d e f+48 d^2 e^2\right )\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )-i d e \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \left (2 b e \left (19 c^2 f^2-44 c d e f+24 d^2 e^2\right )-5 a f \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+f x \sqrt {\frac {d}{c}} \left (c+d x^2\right ) \left (5 a f \left (3 c^2 f^2-6 c d e f+d^2 e \left (4 e+f x^2\right )\right )+b e \left (-15 c^2 f^2+c d f \left (41 e+11 f x^2\right )-3 d^2 \left (8 e^2+2 e f x^2-f^2 x^4\right )\right )\right )}{15 e f^4 \sqrt {\frac {d}{c}} \sqrt {c+d x^2} \sqrt {e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d/c]*f*x*(c + d*x^2)*(5*a*f*(-6*c*d*e*f + 3*c^2*f^2 + d^2*e*(4*e + f*x^2)) + b*e*(-15*c^2*f^2 + c*d*f*(4
1*e + 11*f*x^2) - 3*d^2*(8*e^2 + 2*e*f*x^2 - f^2*x^4))) - I*d*e*(-5*a*f*(8*d^2*e^2 - 13*c*d*e*f + 3*c^2*f^2) +
2*b*e*(24*d^2*e^2 - 44*c*d*e*f + 19*c^2*f^2))*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticE[I*ArcSinh[Sqr
t[d/c]*x], (c*f)/(d*e)] - I*e*(-(d*e) + c*f)*(5*a*d*f*(-8*d*e + 9*c*f) + b*(48*d^2*e^2 - 64*c*d*e*f + 15*c^2*f
^2))*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticF[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)])/(15*Sqrt[d/c]*e*f
^4*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b d^{2} x^{6} + {\left (2 \, b c d + a d^{2}\right )} x^{4} + a c^{2} + {\left (b c^{2} + 2 \, a c d\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {f x^{2} + e}}{f^{2} x^{4} + 2 \, e f x^{2} + e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*d^2*x^6 + (2*b*c*d + a*d^2)*x^4 + a*c^2 + (b*c^2 + 2*a*c*d)*x^2)*sqrt(d*x^2 + c)*sqrt(f*x^2 + e)/(
f^2*x^4 + 2*e*f*x^2 + e^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 1169, normalized size = 2.33 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(d*x^2+c)^(1/2)*(f*x^2+e)^(1/2)*(-24*(-1/c*d)^(1/2)*b*d^3*e^3*f*x^3+15*x*a*c^3*f^4*(-1/c*d)^(1/2)-15*(-1/
c*d)^(1/2)*b*c^3*e*f^3*x+3*(-1/c*d)^(1/2)*b*d^3*e*f^3*x^7+5*(-1/c*d)^(1/2)*a*d^3*e*f^3*x^5-6*(-1/c*d)^(1/2)*b*
d^3*e^2*f^2*x^5+15*(-1/c*d)^(1/2)*a*c^2*d*f^4*x^3+20*(-1/c*d)^(1/2)*a*d^3*e^2*f^2*x^3-48*((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))*b*d^3*e^4+48*((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))*b*d^3*e^4-15*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*Ellipti
cE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a*c^2*d*e*f^3-30*(-1/c*d)^(1/2)*a*c^2*d*e*f^3*x+20*(-1/c*d)^(1/2)*a*c*d^2
*e^2*f^2*x+41*(-1/c*d)^(1/2)*b*c^2*d*e^2*f^2*x-24*(-1/c*d)^(1/2)*b*c*d^2*e^3*f*x+40*((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*d^3*e^3*f+15*((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))*b*c^3*e*f^3-40*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*Elliptic
E((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a*d^3*e^3*f+65*((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*c*d^2*e^2*f^2+38*((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))*b*c^2*d*e^2*f^2-88*((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))*b*c*d^2*e^3*f+45*((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*c^2*d*e*f^3-85*((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*c*d^2
*e^2*f^2-79*((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))*b*c^2*d*e^2*f^
2+112*((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))*b*c*d^2*e^3*f+14*(-1
/c*d)^(1/2)*b*c*d^2*e*f^3*x^5-25*(-1/c*d)^(1/2)*a*c*d^2*e*f^3*x^3-4*(-1/c*d)^(1/2)*b*c^2*d*e*f^3*x^3+35*(-1/c*
d)^(1/2)*b*c*d^2*e^2*f^2*x^3)/f^4/(d*f*x^4+c*f*x^2+d*e*x^2+c*e)/(-1/c*d)^(1/2)/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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