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

Optimal. Leaf size=400 \[ \frac {e^{3/2} \sqrt {c+d x^2} \left (5 a d (3 d e-c f)-b \left (6 c d e-4 c^2 f\right )\right ) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 c d^2 \sqrt {f} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {x \sqrt {c+d x^2} \left (10 a d f (2 d e-c f)+b \left (8 c^2 f^2-13 c d e f+3 d^2 e^2\right )\right )}{15 d^3 \sqrt {e+f x^2}}-\frac {\sqrt {e} \sqrt {c+d x^2} \left (10 a d f (2 d e-c f)+b \left (8 c^2 f^2-13 c d e f+3 d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 d^3 \sqrt {f} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {x \sqrt {c+d x^2} \sqrt {e+f x^2} (5 a d f-4 b c f+3 b d e)}{15 d^2}+\frac {b x \sqrt {c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d} \]

[Out]

1/5*b*x*(f*x^2+e)^(3/2)*(d*x^2+c)^(1/2)/d+1/15*(10*a*d*f*(-c*f+2*d*e)+b*(8*c^2*f^2-13*c*d*e*f+3*d^2*e^2))*x*(d
*x^2+c)^(1/2)/d^3/(f*x^2+e)^(1/2)+1/15*e^(3/2)*(5*a*d*(-c*f+3*d*e)-b*(-4*c^2*f+6*c*d*e))*(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))*(d*x^2+c)^(1/2)/c/d^2/f^(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*(-c*f+2*d*e)+b*(8*c^2*f^2-13*c*d*e*f+3*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))
*e^(1/2)*(d*x^2+c)^(1/2)/d^3/f^(1/2)/(e*(d*x^2+c)/c/(f*x^2+e))^(1/2)/(f*x^2+e)^(1/2)+1/15*(5*a*d*f-4*b*c*f+3*b
*d*e)*x*(d*x^2+c)^(1/2)*(f*x^2+e)^(1/2)/d^2

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Rubi [A]  time = 0.44, antiderivative size = 400, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {528, 531, 418, 492, 411} \[ \frac {x \sqrt {c+d x^2} \left (10 a d f (2 d e-c f)+b \left (8 c^2 f^2-13 c d e f+3 d^2 e^2\right )\right )}{15 d^3 \sqrt {e+f x^2}}-\frac {\sqrt {e} \sqrt {c+d x^2} \left (10 a d f (2 d e-c f)+b \left (8 c^2 f^2-13 c d e f+3 d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 d^3 \sqrt {f} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {e^{3/2} \sqrt {c+d x^2} \left (5 a d (3 d e-c f)-b \left (6 c d e-4 c^2 f\right )\right ) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 c d^2 \sqrt {f} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {x \sqrt {c+d x^2} \sqrt {e+f x^2} (5 a d f-4 b c f+3 b d e)}{15 d^2}+\frac {b x \sqrt {c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((10*a*d*f*(2*d*e - c*f) + b*(3*d^2*e^2 - 13*c*d*e*f + 8*c^2*f^2))*x*Sqrt[c + d*x^2])/(15*d^3*Sqrt[e + f*x^2])
+ ((3*b*d*e - 4*b*c*f + 5*a*d*f)*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])/(15*d^2) + (b*x*Sqrt[c + d*x^2]*(e + f*x^
2)^(3/2))/(5*d) - (Sqrt[e]*(10*a*d*f*(2*d*e - c*f) + b*(3*d^2*e^2 - 13*c*d*e*f + 8*c^2*f^2))*Sqrt[c + d*x^2]*E
llipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*d^3*Sqrt[f]*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*
Sqrt[e + f*x^2]) + (e^(3/2)*(5*a*d*(3*d*e - c*f) - b*(6*c*d*e - 4*c^2*f))*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sq
rt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*c*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 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 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 (e+f x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx &=\frac {b x \sqrt {c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d}+\frac {\int \frac {\sqrt {e+f x^2} \left (-(b c-5 a d) e+(3 b d e-4 b c f+5 a d f) x^2\right )}{\sqrt {c+d x^2}} \, dx}{5 d}\\ &=\frac {(3 b d e-4 b c f+5 a d f) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 d^2}+\frac {b x \sqrt {c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d}+\frac {\int \frac {-e (2 b c (3 d e-2 c f)-5 a d (3 d e-c f))+\left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 c^2 f^2\right )\right ) x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{15 d^2}\\ &=\frac {(3 b d e-4 b c f+5 a d f) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 d^2}+\frac {b x \sqrt {c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d}-\frac {(e (2 b c (3 d e-2 c f)-5 a d (3 d e-c f))) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{15 d^2}+\frac {\left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 c^2 f^2\right )\right ) \int \frac {x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{15 d^2}\\ &=\frac {\left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 c^2 f^2\right )\right ) x \sqrt {c+d x^2}}{15 d^3 \sqrt {e+f x^2}}+\frac {(3 b d e-4 b c f+5 a d f) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 d^2}+\frac {b x \sqrt {c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d}+\frac {e^{3/2} \left (5 a d (3 d e-c f)-b \left (6 c d e-4 c^2 f\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 c 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 {\left (e \left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 c^2 f^2\right )\right )\right ) \int \frac {\sqrt {c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{15 d^3}\\ &=\frac {\left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 c^2 f^2\right )\right ) x \sqrt {c+d x^2}}{15 d^3 \sqrt {e+f x^2}}+\frac {(3 b d e-4 b c f+5 a d f) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 d^2}+\frac {b x \sqrt {c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d}-\frac {\sqrt {e} \left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 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 d^3 \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^{3/2} \left (5 a d (3 d e-c f)-b \left (6 c d e-4 c^2 f\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 c 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 = 1.18, size = 275, normalized size = 0.69 \[ \frac {-i e \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \left (10 a d f (2 d e-c f)+b \left (8 c^2 f^2-13 c d e f+3 d^2 e^2\right )\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+f x \left (-\sqrt {\frac {d}{c}}\right ) \left (c+d x^2\right ) \left (e+f x^2\right ) \left (-5 a d f+4 b c f-3 b d \left (2 e+f x^2\right )\right )+i e \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (c f-d e) (-5 a d f+4 b c f-3 b d e) F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )}{15 c^2 f \left (\frac {d}{c}\right )^{5/2} \sqrt {c+d x^2} \sqrt {e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[d/c]*f*x*(c + d*x^2)*(e + f*x^2)*(4*b*c*f - 5*a*d*f - 3*b*d*(2*e + f*x^2))) - I*e*(10*a*d*f*(2*d*e - c
*f) + b*(3*d^2*e^2 - 13*c*d*e*f + 8*c^2*f^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*e*(-(d*e) + c*f)*(-3*b*d*e + 4*b*c*f - 5*a*d*f)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2
)/e]*EllipticF[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)])/(15*c^2*(d/c)^(5/2)*f*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])

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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b f x^{4} + {\left (b e + a f\right )} x^{2} + a e\right )} \sqrt {f x^{2} + e}}{\sqrt {d x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 870, normalized size = 2.18 \[ \frac {\sqrt {f \,x^{2}+e}\, \sqrt {d \,x^{2}+c}\, \left (3 \sqrt {-\frac {d}{c}}\, b \,d^{2} f^{3} x^{7}+5 \sqrt {-\frac {d}{c}}\, a \,d^{2} f^{3} x^{5}-\sqrt {-\frac {d}{c}}\, b c d \,f^{3} x^{5}+9 \sqrt {-\frac {d}{c}}\, b \,d^{2} e \,f^{2} x^{5}+5 \sqrt {-\frac {d}{c}}\, a c d \,f^{3} x^{3}+5 \sqrt {-\frac {d}{c}}\, a \,d^{2} e \,f^{2} x^{3}-4 \sqrt {-\frac {d}{c}}\, b \,c^{2} f^{3} x^{3}+5 \sqrt {-\frac {d}{c}}\, b c d e \,f^{2} x^{3}+6 \sqrt {-\frac {d}{c}}\, b \,d^{2} e^{2} f \,x^{3}+5 \sqrt {-\frac {d}{c}}\, a c d e \,f^{2} x -10 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, a c d e \,f^{2} \EllipticE \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )+5 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, a c d e \,f^{2} \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )+20 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, a \,d^{2} e^{2} f \EllipticE \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-5 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, a \,d^{2} e^{2} f \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-4 \sqrt {-\frac {d}{c}}\, b \,c^{2} e \,f^{2} x +8 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b \,c^{2} e \,f^{2} \EllipticE \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-4 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b \,c^{2} e \,f^{2} \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )+6 \sqrt {-\frac {d}{c}}\, b c d \,e^{2} f x -13 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b c d \,e^{2} f \EllipticE \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )+7 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b c d \,e^{2} f \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )+3 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b \,d^{2} e^{3} \EllipticE \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-3 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b \,d^{2} e^{3} \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )\right )}{15 \left (d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e \right ) \sqrt {-\frac {d}{c}}\, d^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(f*x^2+e)^(1/2)*(d*x^2+c)^(1/2)*(3*(-1/c*d)^(1/2)*b*d^2*f^3*x^7+5*(-1/c*d)^(1/2)*a*d^2*f^3*x^5-(-1/c*d)^(
1/2)*b*c*d*f^3*x^5+9*(-1/c*d)^(1/2)*b*d^2*e*f^2*x^5+5*(-1/c*d)^(1/2)*a*c*d*f^3*x^3+5*(-1/c*d)^(1/2)*a*d^2*e*f^
2*x^3-4*(-1/c*d)^(1/2)*b*c^2*f^3*x^3+5*(-1/c*d)^(1/2)*b*c*d*e*f^2*x^3+6*(-1/c*d)^(1/2)*b*d^2*e^2*f*x^3+5*((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*e*f^2-5*((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^2*e^2*f-4*((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*e*f^2+7*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*E
llipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*b*c*d*e^2*f-3*((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^2*e^3-10*((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*e*f^2+20*((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*d^2*e^2*f+8*((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
*e*f^2-13*((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*e^2*f+3*((
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^2*e^3+5*(-1/c*d)^(1/2)*a
*c*d*e*f^2*x-4*(-1/c*d)^(1/2)*b*c^2*e*f^2*x+6*(-1/c*d)^(1/2)*b*c*d*e^2*f*x)/d^2/f/(d*f*x^4+c*f*x^2+d*e*x^2+c*e
)/(-1/c*d)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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