3.40 \(\int \frac {a+b x^2}{(c+d x^2)^{5/2} \sqrt {e+f x^2}} \, dx\)
Optimal. Leaf size=284 \[ \frac {\sqrt {e+f x^2} (2 a d (d e-2 c f)+b c (c f+d e)) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{3 c^{3/2} \sqrt {d} \sqrt {c+d x^2} (d e-c f)^2 \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}-\frac {\sqrt {e} \sqrt {f} \sqrt {c+d x^2} (-3 a c f+a d e+2 b c e) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 \sqrt {e+f x^2} (d e-c f)^2 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {x \sqrt {e+f x^2} (b c-a d)}{3 c \left (c+d x^2\right )^{3/2} (d e-c f)} \]
[Out]
-1/3*(-3*a*c*f+a*d*e+2*b*c*e)*(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)*f^(1/2)*(d*x^2+c)^(1/2)/c^2/(-c*f+d*e)^2/(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/(-c*f+d*e)/(d*x^2+c)^(3/2)+1/3*(2*a*d*(-2*c*f+d*e)+b*c*(c*f+d*e
))*(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)/(-c*f+d*e)^2/d^(1/2)/(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 = 284, 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 =
{527, 525, 418, 411} \[ -\frac {\sqrt {e} \sqrt {f} \sqrt {c+d x^2} (-3 a c f+a d e+2 b c e) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{3 c^2 \sqrt {e+f x^2} (d e-c f)^2 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {\sqrt {e+f x^2} (2 a d (d e-2 c f)+b c (c f+d e)) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{3 c^{3/2} \sqrt {d} \sqrt {c+d x^2} (d e-c f)^2 \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}-\frac {x \sqrt {e+f x^2} (b c-a d)}{3 c \left (c+d x^2\right )^{3/2} (d e-c f)} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*x^2)/((c + d*x^2)^(5/2)*Sqrt[e + f*x^2]),x]
[Out]
-((b*c - a*d)*x*Sqrt[e + f*x^2])/(3*c*(d*e - c*f)*(c + d*x^2)^(3/2)) + ((2*a*d*(d*e - 2*c*f) + b*c*(d*e + c*f)
)*Sqrt[e + f*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (c*f)/(d*e)])/(3*c^(3/2)*Sqrt[d]*(d*e - c*f)^2*Sq
rt[c + d*x^2]*Sqrt[(c*(e + f*x^2))/(e*(c + d*x^2))]) - (Sqrt[e]*Sqrt[f]*(2*b*c*e + a*d*e - 3*a*c*f)*Sqrt[c + d
*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*c^2*(d*e - c*f)^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 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 527
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 + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
Rubi steps
\begin {align*} \int \frac {a+b x^2}{\left (c+d x^2\right )^{5/2} \sqrt {e+f x^2}} \, dx &=-\frac {(b c-a d) x \sqrt {e+f x^2}}{3 c (d e-c f) \left (c+d x^2\right )^{3/2}}-\frac {\int \frac {-b c e-2 a d e+3 a c f+(b c-a d) f x^2}{\left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}} \, dx}{3 c (d e-c f)}\\ &=-\frac {(b c-a d) x \sqrt {e+f x^2}}{3 c (d e-c f) \left (c+d x^2\right )^{3/2}}-\frac {(f (2 b c e+a d e-3 a c f)) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{3 c (d e-c f)^2}+\frac {(2 a d (d e-2 c f)+b c (d e+c f)) \int \frac {\sqrt {e+f x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c (d e-c f)^2}\\ &=-\frac {(b c-a d) x \sqrt {e+f x^2}}{3 c (d e-c f) \left (c+d x^2\right )^{3/2}}+\frac {(2 a d (d e-2 c f)+b c (d e+c 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} \sqrt {d} (d e-c f)^2 \sqrt {c+d x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}-\frac {\sqrt {e} \sqrt {f} (2 b c e+a d e-3 a c 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 e-c f)^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 = 302, normalized size = 1.06 \[ \frac {x \sqrt {\frac {d}{c}} \left (e+f x^2\right ) \left (a d \left (-5 c^2 f+c d \left (3 e-4 f x^2\right )+2 d^2 e x^2\right )+b c \left (2 c^2 f+c d f x^2+d^2 e x^2\right )\right )+i \left (c+d x^2\right ) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (c f-d e) (-3 a c f+2 a d e+b c e) F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+i e \left (c+d x^2\right ) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (2 a d (d e-2 c f)+b c (c f+d e)) E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )}{3 c^2 \sqrt {\frac {d}{c}} \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2} (d e-c f)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*x^2)/((c + d*x^2)^(5/2)*Sqrt[e + f*x^2]),x]
[Out]
(Sqrt[d/c]*x*(e + f*x^2)*(b*c*(2*c^2*f + d^2*e*x^2 + c*d*f*x^2) + a*d*(-5*c^2*f + 2*d^2*e*x^2 + c*d*(3*e - 4*f
*x^2))) + I*e*(2*a*d*(d*e - 2*c*f) + b*c*(d*e + c*f))*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*Elli
pticE[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] + I*(-(d*e) + c*f)*(b*c*e + 2*a*d*e - 3*a*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^2*Sqrt[d/c]*(d*e - c*f)^2
*(c + d*x^2)^(3/2)*Sqrt[e + f*x^2])
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fricas [F] time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c} \sqrt {f x^{2} + e}}{d^{3} f x^{8} + {\left (d^{3} e + 3 \, c d^{2} f\right )} x^{6} + 3 \, {\left (c d^{2} e + c^{2} d f\right )} x^{4} + c^{3} e + {\left (3 \, c^{2} d e + c^{3} f\right )} x^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x^2+a)/(d*x^2+c)^(5/2)/(f*x^2+e)^(1/2),x, algorithm="fricas")
[Out]
integral((b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(f*x^2 + e)/(d^3*f*x^8 + (d^3*e + 3*c*d^2*f)*x^6 + 3*(c*d^2*e + c^2*d
*f)*x^4 + c^3*e + (3*c^2*d*e + c^3*f)*x^2), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b x^{2} + a}{{\left (d x^{2} + c\right )}^{\frac {5}{2}} \sqrt {f x^{2} + e}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x^2+a)/(d*x^2+c)^(5/2)/(f*x^2+e)^(1/2),x, algorithm="giac")
[Out]
integrate((b*x^2 + a)/((d*x^2 + c)^(5/2)*sqrt(f*x^2 + e)), x)
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maple [B] time = 0.04, size = 1352, normalized size = 4.76 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x^2+a)/(d*x^2+c)^(5/2)/(f*x^2+e)^(1/2),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+2*(-1/c*d)^(1/2)*b*c^3*e*f*x-4*(-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+(-1/c*d)^(1/2)*b*c^2*d*f^2*x^5-5*(-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+3*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a*c^3*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e
)/e)^(1/2)+2*(-1/c*d)^(1/2)*b*c^3*f^2*x^3+3*EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*x^2*a*c^2*d*f^2*((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*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-(-1/c*d)^(1/2)*a*c*d^2*e*f*x^3+(-1/c*d)^(1/2)*b*c^2*d*e
*f*x^3-5*(-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)-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)-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)-5*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)+4*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)-5*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
)+4*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)-Ellipt
icE((-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*f-d*e)^2/c^2/(-1/c*d)^(1/2)/(d*x^2+c)^(3/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b x^{2} + a}{{\left (d x^{2} + c\right )}^{\frac {5}{2}} \sqrt {f x^{2} + e}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x^2+a)/(d*x^2+c)^(5/2)/(f*x^2+e)^(1/2),x, algorithm="maxima")
[Out]
integrate((b*x^2 + a)/((d*x^2 + c)^(5/2)*sqrt(f*x^2 + e)), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {b\,x^2+a}{{\left (d\,x^2+c\right )}^{5/2}\,\sqrt {f\,x^2+e}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*x^2)/((c + d*x^2)^(5/2)*(e + f*x^2)^(1/2)),x)
[Out]
int((a + b*x^2)/((c + d*x^2)^(5/2)*(e + f*x^2)^(1/2)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b x^{2}}{\left (c + d x^{2}\right )^{\frac {5}{2}} \sqrt {e + f x^{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x**2+a)/(d*x**2+c)**(5/2)/(f*x**2+e)**(1/2),x)
[Out]
Integral((a + b*x**2)/((c + d*x**2)**(5/2)*sqrt(e + f*x**2)), x)
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