3.1.44 \(\int \frac {(a+b x^2) \sqrt {c+d x^2}}{(e+f x^2)^{3/2}} \, dx\) [44]

Optimal. Leaf size=258 \[ -\frac {(b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}+\frac {(2 b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}-\frac {(2 b e-a f) \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{\sqrt {e} f^{3/2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {b \sqrt {e} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{f^{3/2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 258, 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 = {540, 545, 429, 506, 422} \begin {gather*} -\frac {\sqrt {c+d x^2} (2 b e-a f) E\left (\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{\sqrt {e} f^{3/2} \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} (b e-a f)}{e f \sqrt {e+f x^2}}+\frac {x \sqrt {c+d x^2} (2 b e-a f)}{e f \sqrt {e+f x^2}}+\frac {b \sqrt {e} \sqrt {c+d x^2} F\left (\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{f^{3/2} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*e - a*f)*x*Sqrt[c + d*x^2])/(e*f*Sqrt[e + f*x^2])) + ((2*b*e - a*f)*x*Sqrt[c + d*x^2])/(e*f*Sqrt[e + f*x
^2]) - ((2*b*e - a*f)*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(Sqrt[e]*f^(3/2
)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) + (b*Sqrt[e]*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f
]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(f^(3/2)*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 540

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 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 {c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx &=-\frac {(b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}-\frac {\int \frac {-b c e-d (2 b e-a f) x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{e f}\\ &=-\frac {(b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}+\frac {(b c) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{f}+\frac {(d (2 b e-a f)) \int \frac {x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{e f}\\ &=-\frac {(b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}+\frac {(2 b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}+\frac {b \sqrt {e} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{f^{3/2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}-\frac {(2 b e-a f) \int \frac {\sqrt {c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{f}\\ &=-\frac {(b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}+\frac {(2 b e-a f) x \sqrt {c+d x^2}}{e f \sqrt {e+f x^2}}-\frac {(2 b e-a 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 )}{\sqrt {e} f^{3/2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {b \sqrt {e} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{f^{3/2} \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 = 2.75, size = 208, normalized size = 0.81 \begin {gather*} \frac {\sqrt {\frac {d}{c}} f (-b e+a f) x \left (c+d x^2\right )-i d e (2 b e-a 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 e (-2 b d e+b c f+a d 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 )}{\sqrt {\frac {d}{c}} e f^2 \sqrt {c+d x^2} \sqrt {e+f x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 393, normalized size = 1.52

method	result	size
elliptic	\(\frac {\sqrt {\left (d \,x^{2}+c \right ) \left (f \,x^{2}+e \right )}\, \left (\frac {\left (d f \,x^{2}+c f \right ) \left (a f -b e \right ) x}{f^{2} e \sqrt {\left (x^{2}+\frac {e}{f}\right ) \left (d f \,x^{2}+c f \right )}}+\frac {\left (\frac {a d f +b c f -b d e}{f^{2}}+\frac {\left (a f -b e \right ) \left (c f -d e \right )}{f^{2} e}-\frac {c \left (a f -b e \right )}{f e}\right ) \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}}-\frac {\left (\frac {b d}{f}-\frac {\left (a f -b e \right ) d}{f e}\right ) e \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )-\EllipticE \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}\, f}\right )}{\sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}}\)	\(378\)
default	\(\frac {\sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}\, \left (\sqrt {-\frac {d}{c}}\, a d \,f^{2} x^{3}-\sqrt {-\frac {d}{c}}\, b d e f \,x^{3}+\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) a d e f +\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) b c e f -2 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) b d \,e^{2}-\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, \EllipticE \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) a d e f +2 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, \EllipticE \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) b d \,e^{2}+\sqrt {-\frac {d}{c}}\, a c \,f^{2} x -\sqrt {-\frac {d}{c}}\, b c e f x \right )}{\left (d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e \right ) f^{2} e \sqrt {-\frac {d}{c}}}\)	\(393\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x^2+c)^(1/2)*(f*x^2+e)^(1/2)*((-d/c)^(1/2)*a*d*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))*a*d*e*f+((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*Ellipti
cF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c*e*f-2*((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-((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*d*e^2+(-d/c)^(1/2)*
a*c*f^2*x-(-d/c)^(1/2)*b*c*e*f*x)/(d*f*x^4+c*f*x^2+d*e*x^2+c*e)/f^2/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)^(1/2)/(f*x^2+e)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)*sqrt(d*x^2 + c)/(f*x^2 + e)^(3/2), 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)^(1/2)/(f*x^2+e)^(3/2),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 {c + d x^{2}}}{\left (e + f x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*x^2 + a)*sqrt(d*x^2 + c)/(f*x^2 + e)^(3/2), 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 {d\,x^2+c}}{{\left (f\,x^2+e\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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