3.1.36 \(\int \frac {(a+b x^2) (c+d x^2)^{3/2}}{\sqrt {e+f x^2}} \, dx\) [36]

Optimal. Leaf size=396 \[ -\frac {\left (10 a d f (d e-2 c f)-b \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )\right ) x \sqrt {c+d x^2}}{15 d f^2 \sqrt {e+f x^2}}-\frac {(4 b d e-3 b c f-5 a d f) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 f^2}+\frac {b x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 f}+\frac {\sqrt {e} \left (10 a d f (d e-2 c f)-b \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )\right ) \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 d f^{5/2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}-\frac {\sqrt {e} \left (5 a f (d e-3 c f)-b \left (4 d e^2-6 c e f\right )\right ) \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 f^{5/2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}} \]

[Out]

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

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Rubi [A]
time = 0.30, antiderivative size = 396, 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 = {542, 545, 429, 506, 422} \begin {gather*} \frac {\sqrt {e} \sqrt {c+d x^2} \left (10 a d f (d e-2 c f)-b \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )\right ) E\left (\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 d f^{5/2} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {\sqrt {e} \sqrt {c+d x^2} \left (5 a f (d e-3 c f)-b \left (4 d e^2-6 c e f\right )\right ) F\left (\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{15 f^{5/2} \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 (10 a d f (d e-2 c f)-b \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )\right )}{15 d f^2 \sqrt {e+f x^2}}-\frac {x \sqrt {c+d x^2} \sqrt {e+f x^2} (-5 a d f-3 b c f+4 b d e)}{15 f^2}+\frac {b x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*((10*a*d*f*(d*e - 2*c*f) - b*(8*d^2*e^2 - 13*c*d*e*f + 3*c^2*f^2))*x*Sqrt[c + d*x^2])/(d*f^2*Sqrt[e + f*
x^2]) - ((4*b*d*e - 3*b*c*f - 5*a*d*f)*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])/(15*f^2) + (b*x*(c + d*x^2)^(3/2)*Sq
rt[e + f*x^2])/(5*f) + (Sqrt[e]*(10*a*d*f*(d*e - 2*c*f) - b*(8*d^2*e^2 - 13*c*d*e*f + 3*c^2*f^2))*Sqrt[c + d*x
^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*d*f^(5/2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2)
)]*Sqrt[e + f*x^2]) - (Sqrt[e]*(5*a*f*(d*e - 3*c*f) - b*(4*d*e^2 - 6*c*e*f))*Sqrt[c + d*x^2]*EllipticF[ArcTan[
(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*f^(5/2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

Rule 422

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

Rule 429

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

Rule 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[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 542

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 545

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 )^{3/2}}{\sqrt {e+f x^2}} \, dx &=\frac {b x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 f}+\frac {\int \frac {\sqrt {c+d x^2} \left (-c (b e-5 a f)+(-4 b d e+3 b c f+5 a d f) x^2\right )}{\sqrt {e+f x^2}} \, dx}{5 f}\\ &=-\frac {(4 b d e-3 b c f-5 a d f) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 f^2}+\frac {b x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 f}+\frac {\int \frac {-c (5 a f (d e-3 c f)-2 b e (2 d e-3 c f))+\left (-10 a d f (d e-2 c f)+b \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )\right ) x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{15 f^2}\\ &=-\frac {(4 b d e-3 b c f-5 a d f) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 f^2}+\frac {b x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 f}-\frac {\left (c \left (5 a f (d e-3 c f)-b \left (4 d e^2-6 c e f\right )\right )\right ) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{15 f^2}-\frac {\left (10 a d f (d e-2 c f)-b \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )\right ) \int \frac {x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{15 f^2}\\ &=-\frac {\left (10 a d f (d e-2 c f)-b \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )\right ) x \sqrt {c+d x^2}}{15 d f^2 \sqrt {e+f x^2}}-\frac {(4 b d e-3 b c f-5 a d f) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 f^2}+\frac {b x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 f}-\frac {\sqrt {e} \left (5 a f (d e-3 c f)-b \left (4 d e^2-6 c e 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 f^{5/2} \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 (d e-2 c f)-b \left (8 d^2 e^2-13 c d e f+3 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 f^2}\\ &=-\frac {\left (10 a d f (d e-2 c f)-b \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )\right ) x \sqrt {c+d x^2}}{15 d f^2 \sqrt {e+f x^2}}-\frac {(4 b d e-3 b c f-5 a d f) x \sqrt {c+d x^2} \sqrt {e+f x^2}}{15 f^2}+\frac {b x \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}}{5 f}+\frac {\sqrt {e} \left (10 a d f (d e-2 c f)-b \left (8 d^2 e^2-13 c d e f+3 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 f^{5/2} \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 (5 a f (d e-3 c f)-b \left (4 d e^2-6 c e 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 f^{5/2} \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] Result contains complex when optimal does not.
time = 3.30, size = 279, normalized size = 0.70 \begin {gather*} \frac {\sqrt {\frac {d}{c}} f x \left (c+d x^2\right ) \left (e+f x^2\right ) \left (5 a d f+b \left (-4 d e+6 c f+3 d f x^2\right )\right )-i e \left (-10 a d f (d e-2 c f)+b \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )\right ) \sqrt {1+\frac {d x^2}{c}} \sqrt {1+\frac {f x^2}{e}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+i (-d e+c f) (5 a f (2 d e-3 c f)+b e (-8 d e+9 c f)) \sqrt {1+\frac {d x^2}{c}} \sqrt {1+\frac {f x^2}{e}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )}{15 \sqrt {\frac {d}{c}} f^3 \sqrt {c+d x^2} \sqrt {e+f x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d/c]*f*x*(c + d*x^2)*(e + f*x^2)*(5*a*d*f + b*(-4*d*e + 6*c*f + 3*d*f*x^2)) - I*e*(-10*a*d*f*(d*e - 2*c*
f) + b*(8*d^2*e^2 - 13*c*d*e*f + 3*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*(-(d*e) + c*f)*(5*a*f*(2*d*e - 3*c*f) + b*e*(-8*d*e + 9*c*f))*Sqrt[1 + (d*x^2)/c]*Sq
rt[1 + (f*x^2)/e]*EllipticF[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)])/(15*Sqrt[d/c]*f^3*Sqrt[c + d*x^2]*Sqrt[e + f
*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(923\) vs. \(2(424)=848\).
time = 0.14, size = 924, normalized size = 2.33 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(d*x^2+c)^(1/2)*(f*x^2+e)^(1/2)*(3*(-d/c)^(1/2)*b*d^2*f^3*x^7+5*(-d/c)^(1/2)*a*d^2*f^3*x^5+9*(-d/c)^(1/2)
*b*c*d*f^3*x^5-(-d/c)^(1/2)*b*d^2*e*f^2*x^5+5*(-d/c)^(1/2)*a*c*d*f^3*x^3+5*(-d/c)^(1/2)*a*d^2*e*f^2*x^3+6*(-d/
c)^(1/2)*b*c^2*f^3*x^3+5*(-d/c)^(1/2)*b*c*d*e*f^2*x^3-4*(-d/c)^(1/2)*b*d^2*e^2*f*x^3+15*((d*x^2+c)/c)^(1/2)*((
f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c^2*f^3-25*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/
2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c*d*e*f^2+10*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(
x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*d^2*e^2*f-9*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2)
,(c*f/d/e)^(1/2))*b*c^2*e*f^2+17*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1
/2))*b*c*d*e^2*f-8*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*d^2*e^3
+20*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c*d*e*f^2-10*((d*x^2+c
)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*d^2*e^2*f+3*((d*x^2+c)/c)^(1/2)*((f
*x^2+e)/e)^(1/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c^2*e*f^2-13*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1
/2)*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c*d*e^2*f+8*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticE(
x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*d^2*e^3+5*(-d/c)^(1/2)*a*c*d*e*f^2*x+6*(-d/c)^(1/2)*b*c^2*e*f^2*x-4*(-d/c)^(
1/2)*b*c*d*e^2*f*x)/f^3/(d*f*x^4+c*f*x^2+d*e*x^2+c*e)/(-d/c)^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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