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

Optimal. Leaf size=274 \[ \frac {\sqrt {e+f x^2} (d e (2 a d+b c)-c f (a d+2 b c)) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{3 c^{3/2} d^{3/2} \sqrt {c+d x^2} (d e-c f) \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}+\frac {e^{3/2} \sqrt {f} \sqrt {c+d x^2} (b c-a d) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d \sqrt {e+f x^2} (d e-c f) \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {x \sqrt {e+f x^2} (b c-a d)}{3 c d \left (c+d x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {526, 525, 418, 411} \[ \frac {\sqrt {e+f x^2} (d e (2 a d+b c)-c f (a d+2 b c)) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{3 c^{3/2} d^{3/2} \sqrt {c+d x^2} (d e-c f) \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}+\frac {e^{3/2} \sqrt {f} \sqrt {c+d x^2} (b c-a d) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 d \sqrt {e+f x^2} (d e-c f) \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {x \sqrt {e+f x^2} (b c-a d)}{3 c d \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c - a*d)*x*Sqrt[e + f*x^2])/(3*c*d*(c + d*x^2)^(3/2)) + ((d*(b*c + 2*a*d)*e - c*(2*b*c + a*d)*f)*Sqrt[e +
f*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (c*f)/(d*e)])/(3*c^(3/2)*d^(3/2)*(d*e - c*f)*Sqrt[c + d*x^2
]*Sqrt[(c*(e + f*x^2))/(e*(c + d*x^2))]) + ((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*(d*e - c*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 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] && PosQ[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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right ) \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx &=-\frac {(b c-a d) x \sqrt {e+f x^2}}{3 c d \left (c+d x^2\right )^{3/2}}-\frac {\int \frac {-(b c+2 a d) e-(2 b c+a d) f x^2}{\left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}} \, dx}{3 c d}\\ &=-\frac {(b c-a d) x \sqrt {e+f x^2}}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {((b c-a d) e f) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 c d (d e-c f)}+\frac {(d (b c+2 a d) e-c (2 b c+a d) f) \int \frac {\sqrt {e+f x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c d (d e-c f)}\\ &=-\frac {(b c-a d) x \sqrt {e+f x^2}}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {(d (b c+2 a d) e-c (2 b c+a d) f) \sqrt {e+f x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{3 c^{3/2} d^{3/2} (d e-c f) \sqrt {c+d x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}+\frac {(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 (d e-c 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 = 1.27, size = 297, normalized size = 1.08 \[ \frac {x \sqrt {\frac {d}{c}} \left (e+f x^2\right ) \left (a d \left (2 c^2 f-3 c d e+c d f x^2-2 d^2 e x^2\right )+b c \left (c^2 f+2 c d f x^2-d^2 e x^2\right )\right )-i e \left (c+d x^2\right ) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (2 a d+b c) (c f-d e) F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+i e \left (c+d x^2\right ) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (a d (c f-2 d e)+b c (2 c f-d e)) E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )}{3 c^3 \left (\frac {d}{c}\right )^{3/2} \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2} (c f-d e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c} \sqrt {f x^{2} + e}}{d^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(1/2)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)*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 \[ \int \frac {{\left (b x^{2} + a\right )} \sqrt {f x^{2} + e}}{{\left (d x^{2} + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(1/2)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 1236, normalized size = 4.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(f*x^2+e)^(1/2)/(d*x^2+c)^(5/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(1/2)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)*(e + f*x^2)^(1/2))/(c + d*x^2)^(5/2),x)

[Out]

int(((a + b*x^2)*(e + f*x^2)^(1/2))/(c + d*x^2)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(f*x**2+e)**(1/2)/(d*x**2+c)**(5/2),x)

[Out]

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

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