3.1.25 \(\int \frac {(a+b x^2) \sqrt {e+f x^2}}{\sqrt {c+d x^2}} \, dx\) [25]

Optimal. Leaf size=283 \[ \frac {(b d e-2 b c f+3 a d f) x \sqrt {c+d x^2}}{3 d^2 \sqrt {e+f x^2}}+\frac {b x \sqrt {c+d x^2} \sqrt {e+f x^2}}{3 d}-\frac {\sqrt {e} (b d e-2 b c f+3 a d f) \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 d^2 \sqrt {f} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}-\frac {(b c-3 a d) e^{3/2} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c d \sqrt {f} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 5, 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} (3 a d f-2 b c f+b d e) E\left (\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 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 {e^{3/2} \sqrt {c+d x^2} (b c-3 a d) F\left (\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c d \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} (3 a d f-2 b c f+b d e)}{3 d^2 \sqrt {e+f x^2}}+\frac {b x \sqrt {c+d x^2} \sqrt {e+f x^2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d*e - 2*b*c*f + 3*a*d*f)*x*Sqrt[c + d*x^2])/(3*d^2*Sqrt[e + f*x^2]) + (b*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2]
)/(3*d) - (Sqrt[e]*(b*d*e - 2*b*c*f + 3*a*d*f)*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e
)/(c*f)])/(3*d^2*Sqrt[f]*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) - ((b*c - 3*a*d)*e^(3/2)*Sqrt[
c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*c*d*Sqrt[f]*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 ) \sqrt {e+f x^2}}{\sqrt {c+d x^2}} \, dx &=\frac {b x \sqrt {c+d x^2} \sqrt {e+f x^2}}{3 d}+\frac {\int \frac {-(b c-3 a d) e+(b d e-2 b c f+3 a d f) x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 d}\\ &=\frac {b x \sqrt {c+d x^2} \sqrt {e+f x^2}}{3 d}-\frac {((b c-3 a d) e) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 d}+\frac {(b d e-2 b c f+3 a d f) \int \frac {x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 d}\\ &=\frac {(b d e-2 b c f+3 a d f) x \sqrt {c+d x^2}}{3 d^2 \sqrt {e+f x^2}}+\frac {b x \sqrt {c+d x^2} \sqrt {e+f x^2}}{3 d}-\frac {(b c-3 a d) e^{3/2} \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 d \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 (b d e-2 b c f+3 a d f)) \int \frac {\sqrt {c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=\frac {(b d e-2 b c f+3 a d f) x \sqrt {c+d x^2}}{3 d^2 \sqrt {e+f x^2}}+\frac {b x \sqrt {c+d x^2} \sqrt {e+f x^2}}{3 d}-\frac {\sqrt {e} (b d e-2 b c f+3 a d 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 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 {(b c-3 a d) e^{3/2} \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 d \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] Result contains complex when optimal does not.
time = 1.53, size = 212, normalized size = 0.75 \begin {gather*} \frac {b \sqrt {\frac {d}{c}} f x \left (c+d x^2\right ) \left (e+f x^2\right )+i e (-b d e+2 b c f-3 a d f) \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 b e (-d e+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 )}{3 d \sqrt {\frac {d}{c}} f \sqrt {c+d x^2} \sqrt {e+f x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d/c]*f*x*(c + d*x^2)*(e + f*x^2) + I*e*(-(b*d*e) + 2*b*c*f - 3*a*d*f)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*
x^2)/e]*EllipticE[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] - I*b*e*(-(d*e) + c*f)*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*Sqrt[d/c]*f*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])

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Maple [A]
time = 0.13, size = 394, normalized size = 1.39

method	result	size
elliptic	\(\frac {\sqrt {\left (d \,x^{2}+c \right ) \left (f \,x^{2}+e \right )}\, \left (\frac {b x \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}}{3 d}+\frac {\left (a e -\frac {c e b}{3 d}\right ) \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}}-\frac {\left (a f +b e -\frac {b \left (2 c f +2 d e \right )}{3 d}\right ) e \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )-\EllipticE \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}\, f}\right )}{\sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}}\)	\(312\)
risch	\(\frac {b x \sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}}{3 d}+\frac {\left (-\frac {\left (3 a d f -2 b c f +b d e \right ) e \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )-\EllipticE \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}\, f}+\frac {3 a d e \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}}-\frac {b c e \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}}\right ) \sqrt {\left (d \,x^{2}+c \right ) \left (f \,x^{2}+e \right )}}{3 d \sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}}\)	\(386\)
default	\(\frac {\sqrt {f \,x^{2}+e}\, \sqrt {d \,x^{2}+c}\, \left (\sqrt {-\frac {d}{c}}\, b d \,f^{2} x^{5}+\sqrt {-\frac {d}{c}}\, b c \,f^{2} x^{3}+\sqrt {-\frac {d}{c}}\, b d e f \,x^{3}+\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) b c e f -\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) b d \,e^{2}+3 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, \EllipticE \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) a d e f -2 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, \EllipticE \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) b c e f +\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, \EllipticE \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) b d \,e^{2}+\sqrt {-\frac {d}{c}}\, b c e f x \right )}{3 \left (d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e \right ) d \sqrt {-\frac {d}{c}}\, f}\)	\(394\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(f*x^2+e)^(1/2)*(d*x^2+c)^(1/2)*((-d/c)^(1/2)*b*d*f^2*x^5+(-d/c)^(1/2)*b*c*f^2*x^3+(-d/c)^(1/2)*b*d*e*f*x^
3+((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*e*f-((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*e^2+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))*a*d*e*f-2*((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*e*f+((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*e^2+(-d/c)^(1/2)*b*c*e*f*x)/(d*f*x^4+c*f*x^2+d*e*x^2+c*e)/d/(-d/c)^(1/2)/f

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

[Out]

integrate((b*x^2 + a)*sqrt(f*x^2 + e)/sqrt(d*x^2 + c), 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)*(f*x^2+e)^(1/2)/(d*x^2+c)^(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 ) \sqrt {e + f x^{2}}}{\sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*x^2 + a)*sqrt(f*x^2 + e)/sqrt(d*x^2 + c), 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 )\,\sqrt {f\,x^2+e}}{\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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