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

Optimal. Leaf size=373 \[ -\frac {f x \sqrt {c+d x^2} (2 a d (c f+d e)+b c (d e-8 c f))}{3 c^2 d^3 \sqrt {e+f x^2}}+\frac {\sqrt {e} \sqrt {f} \sqrt {c+d x^2} (2 a d (c f+d e)+b c (d e-8 c f)) E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^3 \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {e^{3/2} \sqrt {f} \sqrt {c+d x^2} (4 b c-a d) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^2 \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {x \sqrt {e+f x^2} (a d (c f+2 d e)+b c (d e-4 c f))}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {x \left (e+f x^2\right )^{3/2} (b c-a d)}{3 c d \left (c+d x^2\right )^{3/2}} \]

[Out]

-1/3*(-a*d+b*c)*x*(f*x^2+e)^(3/2)/c/d/(d*x^2+c)^(3/2)-1/3*f*(b*c*(-8*c*f+d*e)+2*a*d*(c*f+d*e))*x*(d*x^2+c)^(1/
2)/c^2/d^3/(f*x^2+e)^(1/2)+1/3*(-a*d+4*b*c)*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/d^2/(e*(d*x^2+c)/c/(f*x^2+e))^(1/2)
/(f*x^2+e)^(1/2)+1/3*(b*c*(-8*c*f+d*e)+2*a*d*(c*f+d*e))*(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)*f^(1/2)*(d*x^2+c)^(1/2)/c^2/d^3/(e*(d*x^2+c)/c/(f*x
^2+e))^(1/2)/(f*x^2+e)^(1/2)+1/3*(b*c*(-4*c*f+d*e)+a*d*(c*f+2*d*e))*x*(f*x^2+e)^(1/2)/c^2/d^2/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 373, 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 = {526, 531, 418, 492, 411} \[ \frac {e^{3/2} \sqrt {f} \sqrt {c+d x^2} (4 b c-a d) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^2 \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {x \sqrt {e+f x^2} (a d (c f+2 d e)+b c (d e-4 c f))}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {f x \sqrt {c+d x^2} (2 a d (c f+d e)+b c (d e-8 c f))}{3 c^2 d^3 \sqrt {e+f x^2}}+\frac {\sqrt {e} \sqrt {f} \sqrt {c+d x^2} (2 a d (c f+d e)+b c (d e-8 c f)) E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^3 \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {x \left (e+f x^2\right )^{3/2} (b c-a d)}{3 c d \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(f*(b*c*(d*e - 8*c*f) + 2*a*d*(d*e + c*f))*x*Sqrt[c + d*x^2])/(3*c^2*d^3*Sqrt[e + f*x^2]) + ((b*c*(d*e - 4*c*
f) + a*d*(2*d*e + c*f))*x*Sqrt[e + f*x^2])/(3*c^2*d^2*Sqrt[c + d*x^2]) - ((b*c - a*d)*x*(e + f*x^2)^(3/2))/(3*
c*d*(c + d*x^2)^(3/2)) + (Sqrt[e]*Sqrt[f]*(b*c*(d*e - 8*c*f) + 2*a*d*(d*e + c*f))*Sqrt[c + d*x^2]*EllipticE[Ar
cTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*c^2*d^3*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])
+ ((4*b*c - a*d)*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*d^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 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}}{\left (c+d x^2\right )^{5/2}} \, dx &=-\frac {(b c-a d) x \left (e+f x^2\right )^{3/2}}{3 c d \left (c+d x^2\right )^{3/2}}-\frac {\int \frac {\sqrt {e+f x^2} \left (-(b c+2 a d) e-(4 b c-a d) f x^2\right )}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c d}\\ &=\frac {(b c (d e-4 c f)+a d (2 d e+c f)) x \sqrt {e+f x^2}}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {(b c-a d) x \left (e+f x^2\right )^{3/2}}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {\int \frac {c (4 b c-a d) e f-f (b c (d e-8 c f)+2 a d (d e+c f)) x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 c^2 d^2}\\ &=\frac {(b c (d e-4 c f)+a d (2 d e+c f)) x \sqrt {e+f x^2}}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {(b c-a d) x \left (e+f x^2\right )^{3/2}}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {((4 b c-a d) e f) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 c d^2}-\frac {(f (b c (d e-8 c f)+2 a d (d e+c f))) \int \frac {x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 c^2 d^2}\\ &=-\frac {f (b c (d e-8 c f)+2 a d (d e+c f)) x \sqrt {c+d x^2}}{3 c^2 d^3 \sqrt {e+f x^2}}+\frac {(b c (d e-4 c f)+a d (2 d e+c f)) x \sqrt {e+f x^2}}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {(b c-a d) x \left (e+f x^2\right )^{3/2}}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {(4 b c-a d) 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 d^2 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {(e f (b c (d e-8 c f)+2 a d (d e+c f))) \int \frac {\sqrt {c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{3 c^2 d^3}\\ &=-\frac {f (b c (d e-8 c f)+2 a d (d e+c f)) x \sqrt {c+d x^2}}{3 c^2 d^3 \sqrt {e+f x^2}}+\frac {(b c (d e-4 c f)+a d (2 d e+c f)) x \sqrt {e+f x^2}}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {(b c-a d) x \left (e+f x^2\right )^{3/2}}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {\sqrt {e} \sqrt {f} (b c (d e-8 c f)+2 a d (d e+c f)) \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d^3 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {(4 b c-a d) 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 d^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 = 296, normalized size = 0.79 \[ \frac {\left (\frac {d}{c}\right )^{3/2} \left (x \sqrt {\frac {d}{c}} \left (e+f x^2\right ) \left (a d \left (c^2 f+c d \left (3 e+2 f x^2\right )+2 d^2 e x^2\right )+b c \left (-4 c^2 f-5 c d f x^2+d^2 e x^2\right )\right )+i e \left (c+d x^2\right ) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (b c (4 c f-d e)-a d (c f+2 d e)) F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )-i e \left (c+d x^2\right ) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (b c (8 c f-d e)-2 a d (c f+d e)) E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )\right )}{3 d^4 \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d/c)^(3/2)*(Sqrt[d/c]*x*(e + f*x^2)*(b*c*(-4*c^2*f + d^2*e*x^2 - 5*c*d*f*x^2) + a*d*(c^2*f + 2*d^2*e*x^2 + c
*d*(3*e + 2*f*x^2))) - I*e*(-2*a*d*(d*e + c*f) + b*c*(-(d*e) + 8*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*e*(-(a*d*(2*d*e + c*f)) + b*c*(-(d*e) + 4*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*d^4*
(c + d*x^2)^(3/2)*Sqrt[e + f*x^2])

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b f x^{4} + {\left (b e + a f\right )} x^{2} + a e\right )} \sqrt {d x^{2} + c} \sqrt {f x^{2} + e}}{d^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}}, x\right ) \]

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)^(5/2),x, algorithm="fricas")

[Out]

integral((b*f*x^4 + (b*e + a*f)*x^2 + a*e)*sqrt(d*x^2 + c)*sqrt(f*x^2 + e)/(d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d*x^
2 + c^3), x)

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

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)^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.03, size = 1231, normalized size = 3.30 \[ \text {result too large to display} \]

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)^(5/2),x)

[Out]

-1/3*(-3*(-1/c*d)^(1/2)*a*c*d^2*e^2*x-EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2*b*c*d^2*e^2*((d*x^2+c)/c
)^(1/2)*((f*x^2+e)/e)^(1/2)-2*(-1/c*d)^(1/2)*a*d^3*e^2*x^3+4*(-1/c*d)^(1/2)*b*c^3*e*f*x-2*(-1/c*d)^(1/2)*a*c*d
^2*f^2*x^5-2*(-1/c*d)^(1/2)*a*d^3*e*f*x^5+5*(-1/c*d)^(1/2)*b*c^2*d*f^2*x^5-(-1/c*d)^(1/2)*a*c^2*d*f^2*x^3-(-1/
c*d)^(1/2)*b*c*d^2*e^2*x^3+4*(-1/c*d)^(1/2)*b*c^3*f^2*x^3+EllipticE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2*b*c*
d^2*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-(-1/c*d)^(1/2)*b*c*d^2*e*f*x^5-5*(-1/c*d)^(1/2)*a*c*d^2*e*f*x^
3+5*(-1/c*d)^(1/2)*b*c^2*d*e*f*x^3-(-1/c*d)^(1/2)*a*c^2*d*e*f*x-2*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*
x^2*a*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*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*c*d^2*e^2*((d*x^2+
c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+4*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*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))*b*c^2*d*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*c*d^2*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-8*Ellipt
icE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*b*c^3*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))*b*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*c^2*d*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*c^
2*d*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*c*d^2*e*f*((
d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+4*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2*b*c^2*d*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))*x^2*a*c*d^2*e*f*((d*x^2+c)/c)^(1/2
)*((f*x^2+e)/e)^(1/2)-8*EllipticE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2*b*c^2*d*e*f*((d*x^2+c)/c)^(1/2)*((f*x^
2+e)/e)^(1/2))/(f*x^2+e)^(1/2)/c^2/(-1/c*d)^(1/2)/(d*x^2+c)^(3/2)/d^2

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

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)^(5/2),x, algorithm="maxima")

[Out]

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

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

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)^(5/2),x)

[Out]

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

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

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)**(5/2),x)

[Out]

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

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