3.34 \(\int \frac{(a+b x^2) (e+f x^2)^{3/2}}{(c+d x^2)^{9/2}} \, dx\)

Optimal. Leaf size=531 \[ -\frac{e^{3/2} \sqrt{f} \sqrt{c+d x^2} \left (3 a d \left (c^2 f^2-11 c d e f+8 d^2 e^2\right )+2 b c \left (2 c^2 f^2-c d e f+2 d^2 e^2\right )\right ) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ),1-\frac{d e}{c f}\right )}{105 c^4 d^2 \sqrt{e+f x^2} (d e-c f)^2 \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x \sqrt{e+f x^2} \left (3 a d \left (-2 c^2 f^2-5 c d e f+8 d^2 e^2\right )+b c \left (-8 c^2 f^2+c d e f+4 d^2 e^2\right )\right )}{105 c^3 d^2 \left (c+d x^2\right )^{3/2} (d e-c f)}+\frac{\sqrt{e+f x^2} \left (6 a d \left (2 c^2 d e f^2+c^3 f^3-12 c d^2 e^2 f+8 d^3 e^3\right )+b c \left (-5 c^2 d e f^2+8 c^3 f^3-5 c d^2 e^2 f+8 d^3 e^3\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{c f}{d e}\right )}{105 c^{7/2} d^{5/2} \sqrt{c+d x^2} (d e-c f)^2 \sqrt{\frac{c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}+\frac{x \sqrt{e+f x^2} (d e (6 a d+b c)-c f (3 a d+4 b c))}{35 c^2 d^2 \left (c+d x^2\right )^{5/2}}-\frac{x \left (e+f x^2\right )^{3/2} (b c-a d)}{7 c d \left (c+d x^2\right )^{7/2}} \]

[Out]

((d*(b*c + 6*a*d)*e - c*(4*b*c + 3*a*d)*f)*x*Sqrt[e + f*x^2])/(35*c^2*d^2*(c + d*x^2)^(5/2)) + ((b*c*(4*d^2*e^
2 + c*d*e*f - 8*c^2*f^2) + 3*a*d*(8*d^2*e^2 - 5*c*d*e*f - 2*c^2*f^2))*x*Sqrt[e + f*x^2])/(105*c^3*d^2*(d*e - c
*f)*(c + d*x^2)^(3/2)) - ((b*c - a*d)*x*(e + f*x^2)^(3/2))/(7*c*d*(c + d*x^2)^(7/2)) + ((6*a*d*(8*d^3*e^3 - 12
*c*d^2*e^2*f + 2*c^2*d*e*f^2 + c^3*f^3) + b*c*(8*d^3*e^3 - 5*c*d^2*e^2*f - 5*c^2*d*e*f^2 + 8*c^3*f^3))*Sqrt[e
+ f*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (c*f)/(d*e)])/(105*c^(7/2)*d^(5/2)*(d*e - c*f)^2*Sqrt[c +
d*x^2]*Sqrt[(c*(e + f*x^2))/(e*(c + d*x^2))]) - (e^(3/2)*Sqrt[f]*(3*a*d*(8*d^2*e^2 - 11*c*d*e*f + c^2*f^2) + 2
*b*c*(2*d^2*e^2 - c*d*e*f + 2*c^2*f^2))*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)
])/(105*c^4*d^2*(d*e - c*f)^2*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

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Rubi [A]  time = 0.614756, antiderivative size = 531, normalized size of antiderivative = 1., 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 = {526, 527, 525, 418, 411} \[ \frac{x \sqrt{e+f x^2} \left (3 a d \left (-2 c^2 f^2-5 c d e f+8 d^2 e^2\right )+b c \left (-8 c^2 f^2+c d e f+4 d^2 e^2\right )\right )}{105 c^3 d^2 \left (c+d x^2\right )^{3/2} (d e-c f)}-\frac{e^{3/2} \sqrt{f} \sqrt{c+d x^2} \left (3 a d \left (c^2 f^2-11 c d e f+8 d^2 e^2\right )+2 b c \left (2 c^2 f^2-c d e f+2 d^2 e^2\right )\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{105 c^4 d^2 \sqrt{e+f x^2} (d e-c f)^2 \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{\sqrt{e+f x^2} \left (6 a d \left (2 c^2 d e f^2+c^3 f^3-12 c d^2 e^2 f+8 d^3 e^3\right )+b c \left (-5 c^2 d e f^2+8 c^3 f^3-5 c d^2 e^2 f+8 d^3 e^3\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{c f}{d e}\right )}{105 c^{7/2} d^{5/2} \sqrt{c+d x^2} (d e-c f)^2 \sqrt{\frac{c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}+\frac{x \sqrt{e+f x^2} (d e (6 a d+b c)-c f (3 a d+4 b c))}{35 c^2 d^2 \left (c+d x^2\right )^{5/2}}-\frac{x \left (e+f x^2\right )^{3/2} (b c-a d)}{7 c d \left (c+d x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*(e + f*x^2)^(3/2))/(c + d*x^2)^(9/2),x]

[Out]

((d*(b*c + 6*a*d)*e - c*(4*b*c + 3*a*d)*f)*x*Sqrt[e + f*x^2])/(35*c^2*d^2*(c + d*x^2)^(5/2)) + ((b*c*(4*d^2*e^
2 + c*d*e*f - 8*c^2*f^2) + 3*a*d*(8*d^2*e^2 - 5*c*d*e*f - 2*c^2*f^2))*x*Sqrt[e + f*x^2])/(105*c^3*d^2*(d*e - c
*f)*(c + d*x^2)^(3/2)) - ((b*c - a*d)*x*(e + f*x^2)^(3/2))/(7*c*d*(c + d*x^2)^(7/2)) + ((6*a*d*(8*d^3*e^3 - 12
*c*d^2*e^2*f + 2*c^2*d*e*f^2 + c^3*f^3) + b*c*(8*d^3*e^3 - 5*c*d^2*e^2*f - 5*c^2*d*e*f^2 + 8*c^3*f^3))*Sqrt[e
+ f*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (c*f)/(d*e)])/(105*c^(7/2)*d^(5/2)*(d*e - c*f)^2*Sqrt[c +
d*x^2]*Sqrt[(c*(e + f*x^2))/(e*(c + d*x^2))]) - (e^(3/2)*Sqrt[f]*(3*a*d*(8*d^2*e^2 - 11*c*d*e*f + c^2*f^2) + 2
*b*c*(2*d^2*e^2 - c*d*e*f + 2*c^2*f^2))*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)
])/(105*c^4*d^2*(d*e - c*f)^2*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

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]

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right ) \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{9/2}} \, dx &=-\frac{(b c-a d) x \left (e+f x^2\right )^{3/2}}{7 c d \left (c+d x^2\right )^{7/2}}-\frac{\int \frac{\sqrt{e+f x^2} \left (-(b c+6 a d) e-(4 b c+3 a d) f x^2\right )}{\left (c+d x^2\right )^{7/2}} \, dx}{7 c d}\\ &=\frac{(d (b c+6 a d) e-c (4 b c+3 a d) f) x \sqrt{e+f x^2}}{35 c^2 d^2 \left (c+d x^2\right )^{5/2}}-\frac{(b c-a d) x \left (e+f x^2\right )^{3/2}}{7 c d \left (c+d x^2\right )^{7/2}}+\frac{\int \frac{e (4 b c (d e+c f)+3 a d (8 d e+c f))+f (6 a d (3 d e+c f)+b c (3 d e+8 c f)) x^2}{\left (c+d x^2\right )^{5/2} \sqrt{e+f x^2}} \, dx}{35 c^2 d^2}\\ &=\frac{(d (b c+6 a d) e-c (4 b c+3 a d) f) x \sqrt{e+f x^2}}{35 c^2 d^2 \left (c+d x^2\right )^{5/2}}+\frac{\left (b c \left (4 d^2 e^2+c d e f-8 c^2 f^2\right )+3 a d \left (8 d^2 e^2-5 c d e f-2 c^2 f^2\right )\right ) x \sqrt{e+f x^2}}{105 c^3 d^2 (d e-c f) \left (c+d x^2\right )^{3/2}}-\frac{(b c-a d) x \left (e+f x^2\right )^{3/2}}{7 c d \left (c+d x^2\right )^{7/2}}-\frac{\int \frac{-e \left (b c \left (8 d^2 e^2-c d e f-4 c^2 f^2\right )+3 a d \left (16 d^2 e^2-16 c d e f-c^2 f^2\right )\right )-f \left (b c \left (4 d^2 e^2+c d e f-8 c^2 f^2\right )+3 a d \left (8 d^2 e^2-5 c d e f-2 c^2 f^2\right )\right ) x^2}{\left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}} \, dx}{105 c^3 d^2 (d e-c f)}\\ &=\frac{(d (b c+6 a d) e-c (4 b c+3 a d) f) x \sqrt{e+f x^2}}{35 c^2 d^2 \left (c+d x^2\right )^{5/2}}+\frac{\left (b c \left (4 d^2 e^2+c d e f-8 c^2 f^2\right )+3 a d \left (8 d^2 e^2-5 c d e f-2 c^2 f^2\right )\right ) x \sqrt{e+f x^2}}{105 c^3 d^2 (d e-c f) \left (c+d x^2\right )^{3/2}}-\frac{(b c-a d) x \left (e+f x^2\right )^{3/2}}{7 c d \left (c+d x^2\right )^{7/2}}-\frac{\left (e f \left (3 a d \left (8 d^2 e^2-11 c d e f+c^2 f^2\right )+2 b c \left (2 d^2 e^2-c d e f+2 c^2 f^2\right )\right )\right ) \int \frac{1}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{105 c^3 d^2 (d e-c f)^2}+\frac{\left (6 a d \left (8 d^3 e^3-12 c d^2 e^2 f+2 c^2 d e f^2+c^3 f^3\right )+b c \left (8 d^3 e^3-5 c d^2 e^2 f-5 c^2 d e f^2+8 c^3 f^3\right )\right ) \int \frac{\sqrt{e+f x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{105 c^3 d^2 (d e-c f)^2}\\ &=\frac{(d (b c+6 a d) e-c (4 b c+3 a d) f) x \sqrt{e+f x^2}}{35 c^2 d^2 \left (c+d x^2\right )^{5/2}}+\frac{\left (b c \left (4 d^2 e^2+c d e f-8 c^2 f^2\right )+3 a d \left (8 d^2 e^2-5 c d e f-2 c^2 f^2\right )\right ) x \sqrt{e+f x^2}}{105 c^3 d^2 (d e-c f) \left (c+d x^2\right )^{3/2}}-\frac{(b c-a d) x \left (e+f x^2\right )^{3/2}}{7 c d \left (c+d x^2\right )^{7/2}}+\frac{\left (6 a d \left (8 d^3 e^3-12 c d^2 e^2 f+2 c^2 d e f^2+c^3 f^3\right )+b c \left (8 d^3 e^3-5 c d^2 e^2 f-5 c^2 d e f^2+8 c^3 f^3\right )\right ) \sqrt{e+f x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{c f}{d e}\right )}{105 c^{7/2} d^{5/2} (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{e^{3/2} \sqrt{f} \left (3 a d \left (8 d^2 e^2-11 c d e f+c^2 f^2\right )+2 b c \left (2 d^2 e^2-c d e f+2 c^2 f^2\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 )}{105 c^4 d^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*}

Mathematica [C]  time = 1.83476, size = 545, normalized size = 1.03 \[ \frac{\sqrt{\frac{d}{c}} \left (-x \sqrt{\frac{d}{c}} \left (e+f x^2\right ) \left (-c \left (c+d x^2\right )^2 (d e-c f) \left (3 a d \left (-2 c^2 f^2-5 c d e f+8 d^2 e^2\right )+b c \left (-8 c^2 f^2+c d e f+4 d^2 e^2\right )\right )-\left (c+d x^2\right )^3 \left (6 a d \left (2 c^2 d e f^2+c^3 f^3-12 c d^2 e^2 f+8 d^3 e^3\right )+b c \left (-5 c^2 d e f^2+8 c^3 f^3-5 c d^2 e^2 f+8 d^3 e^3\right )\right )-3 c^2 \left (c+d x^2\right ) (d e-c f)^2 (2 a d (c f+3 d e)+b c (d e-9 c f))+15 c^3 (b c-a d) (d e-c f)^3\right )+i e \left (c+d x^2\right )^3 \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} \left (\left (6 a d \left (2 c^2 d e f^2+c^3 f^3-12 c d^2 e^2 f+8 d^3 e^3\right )+b c \left (-5 c^2 d e f^2+8 c^3 f^3-5 c d^2 e^2 f+8 d^3 e^3\right )\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )-(c f-d e) \left (3 a d \left (c^2 f^2+16 c d e f-16 d^2 e^2\right )+b c \left (4 c^2 f^2+c d e f-8 d^2 e^2\right )\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{d}{c}}\right ),\frac{c f}{d e}\right )\right )\right )}{105 c^3 d^3 \left (c+d x^2\right )^{7/2} \sqrt{e+f x^2} (d e-c f)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*(e + f*x^2)^(3/2))/(c + d*x^2)^(9/2),x]

[Out]

(Sqrt[d/c]*(-(Sqrt[d/c]*x*(e + f*x^2)*(15*c^3*(b*c - a*d)*(d*e - c*f)^3 - 3*c^2*(d*e - c*f)^2*(b*c*(d*e - 9*c*
f) + 2*a*d*(3*d*e + c*f))*(c + d*x^2) - c*(d*e - c*f)*(b*c*(4*d^2*e^2 + c*d*e*f - 8*c^2*f^2) + 3*a*d*(8*d^2*e^
2 - 5*c*d*e*f - 2*c^2*f^2))*(c + d*x^2)^2 - (6*a*d*(8*d^3*e^3 - 12*c*d^2*e^2*f + 2*c^2*d*e*f^2 + c^3*f^3) + b*
c*(8*d^3*e^3 - 5*c*d^2*e^2*f - 5*c^2*d*e*f^2 + 8*c^3*f^3))*(c + d*x^2)^3)) + I*e*(c + d*x^2)^3*Sqrt[1 + (d*x^2
)/c]*Sqrt[1 + (f*x^2)/e]*((6*a*d*(8*d^3*e^3 - 12*c*d^2*e^2*f + 2*c^2*d*e*f^2 + c^3*f^3) + b*c*(8*d^3*e^3 - 5*c
*d^2*e^2*f - 5*c^2*d*e*f^2 + 8*c^3*f^3))*EllipticE[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] - (-(d*e) + c*f)*(3*a*
d*(-16*d^2*e^2 + 16*c*d*e*f + c^2*f^2) + b*c*(-8*d^2*e^2 + c*d*e*f + 4*c^2*f^2))*EllipticF[I*ArcSinh[Sqrt[d/c]
*x], (c*f)/(d*e)])))/(105*c^3*d^3*(d*e - c*f)^2*(c + d*x^2)^(7/2)*Sqrt[e + f*x^2])

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Maple [B]  time = 0.069, size = 5113, normalized size = 9.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(f*x^2+e)^(3/2)/(d*x^2+c)^(9/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}{\left (f x^{2} + e\right )}^{\frac{3}{2}}}{{\left (d x^{2} + c\right )}^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(3/2)/(d*x^2+c)^(9/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)*(f*x^2 + e)^(3/2)/(d*x^2 + c)^(9/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b f x^{4} +{\left (b e + a f\right )} x^{2} + a e\right )} \sqrt{d x^{2} + c} \sqrt{f x^{2} + e}}{d^{5} x^{10} + 5 \, c d^{4} x^{8} + 10 \, c^{2} d^{3} x^{6} + 10 \, c^{3} d^{2} x^{4} + 5 \, c^{4} d x^{2} + c^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(3/2)/(d*x^2+c)^(9/2),x, algorithm="fricas")

[Out]

integral((b*f*x^4 + (b*e + a*f)*x^2 + a*e)*sqrt(d*x^2 + c)*sqrt(f*x^2 + e)/(d^5*x^10 + 5*c*d^4*x^8 + 10*c^2*d^
3*x^6 + 10*c^3*d^2*x^4 + 5*c^4*d*x^2 + c^5), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(f*x**2+e)**(3/2)/(d*x**2+c)**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}{\left (f x^{2} + e\right )}^{\frac{3}{2}}}{{\left (d x^{2} + c\right )}^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(3/2)/(d*x^2+c)^(9/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)*(f*x^2 + e)^(3/2)/(d*x^2 + c)^(9/2), x)