3.220 \(\int \frac{(d+e x^2)^{5/2}}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx\)

Optimal. Leaf size=139 \[ -\frac{(2 c d-b e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} x \sqrt{2 c d-b e}}{\sqrt{d+e x^2} \sqrt{c d-b e}}\right )}{c^2 \sqrt{e} \sqrt{c d-b e}}+\frac{(5 c d-2 b e) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c^2 \sqrt{e}}+\frac{x \sqrt{d+e x^2}}{2 c} \]

[Out]

(x*Sqrt[d + e*x^2])/(2*c) + ((5*c*d - 2*b*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*c^2*Sqrt[e]) - ((2*c*d -
b*e)^(3/2)*ArcTanh[(Sqrt[e]*Sqrt[2*c*d - b*e]*x)/(Sqrt[c*d - b*e]*Sqrt[d + e*x^2])])/(c^2*Sqrt[e]*Sqrt[c*d -
b*e])

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Rubi [A]  time = 0.276271, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {1149, 416, 523, 217, 206, 377, 208} \[ -\frac{(2 c d-b e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} x \sqrt{2 c d-b e}}{\sqrt{d+e x^2} \sqrt{c d-b e}}\right )}{c^2 \sqrt{e} \sqrt{c d-b e}}+\frac{(5 c d-2 b e) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c^2 \sqrt{e}}+\frac{x \sqrt{d+e x^2}}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)^(5/2)/(-(c*d^2) + b*d*e + b*e^2*x^2 + c*e^2*x^4),x]

[Out]

(x*Sqrt[d + e*x^2])/(2*c) + ((5*c*d - 2*b*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*c^2*Sqrt[e]) - ((2*c*d -
b*e)^(3/2)*ArcTanh[(Sqrt[e]*Sqrt[2*c*d - b*e]*x)/(Sqrt[c*d - b*e]*Sqrt[d + e*x^2])])/(c^2*Sqrt[e]*Sqrt[c*d -
b*e])

Rule 1149

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[(d + e*x^2)^(p +
q)*(a/d + (c*x^2)/e)^p, x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2
, 0] && IntegerQ[p]

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^{5/2}}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx &=\int \frac{\left (d+e x^2\right )^{3/2}}{\frac{-c d^2+b d e}{d}+c e x^2} \, dx\\ &=\frac{x \sqrt{d+e x^2}}{2 c}+\frac{\int \frac{d e (3 c d-b e)+e^2 (5 c d-2 b e) x^2}{\sqrt{d+e x^2} \left (\frac{-c d^2+b d e}{d}+c e x^2\right )} \, dx}{2 c e}\\ &=\frac{x \sqrt{d+e x^2}}{2 c}+\frac{(5 c d-2 b e) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{2 c^2}+\frac{(2 c d-b e)^2 \int \frac{1}{\sqrt{d+e x^2} \left (\frac{-c d^2+b d e}{d}+c e x^2\right )} \, dx}{c^2}\\ &=\frac{x \sqrt{d+e x^2}}{2 c}+\frac{(5 c d-2 b e) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{2 c^2}+\frac{(2 c d-b e)^2 \operatorname{Subst}\left (\int \frac{1}{\frac{-c d^2+b d e}{d}-\left (-c d e+\frac{e \left (-c d^2+b d e\right )}{d}\right ) x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{c^2}\\ &=\frac{x \sqrt{d+e x^2}}{2 c}+\frac{(5 c d-2 b e) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c^2 \sqrt{e}}-\frac{(2 c d-b e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{2 c d-b e} x}{\sqrt{c d-b e} \sqrt{d+e x^2}}\right )}{c^2 \sqrt{e} \sqrt{c d-b e}}\\ \end{align*}

Mathematica [A]  time = 0.264286, size = 134, normalized size = 0.96 \[ -\frac{\frac{(2 b e-5 c d) \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )}{\sqrt{e}}-\frac{2 (b e-2 c d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} x \sqrt{b e-2 c d}}{\sqrt{d+e x^2} \sqrt{b e-c d}}\right )}{\sqrt{e} \sqrt{b e-c d}}-c x \sqrt{d+e x^2}}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^2)^(5/2)/(-(c*d^2) + b*d*e + b*e^2*x^2 + c*e^2*x^4),x]

[Out]

-(-(c*x*Sqrt[d + e*x^2]) - (2*(-2*c*d + b*e)^(3/2)*ArcTanh[(Sqrt[e]*Sqrt[-2*c*d + b*e]*x)/(Sqrt[-(c*d) + b*e]*
Sqrt[d + e*x^2])])/(Sqrt[e]*Sqrt[-(c*d) + b*e]) + ((-5*c*d + 2*b*e)*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/Sqrt[e
])/(2*c^2)

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Maple [B]  time = 0.05, size = 7043, normalized size = 50.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^(5/2)/(c*e^2*x^4+b*e^2*x^2+b*d*e-c*d^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 (e x^{2} + d\right )}^{\frac{5}{2}}}{c e^{2} x^{4} + b e^{2} x^{2} - c d^{2} + b d e}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(5/2)/(c*e^2*x^4+b*e^2*x^2+b*d*e-c*d^2),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)^(5/2)/(c*e^2*x^4 + b*e^2*x^2 - c*d^2 + b*d*e), x)

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Fricas [A]  time = 4.7133, size = 2303, normalized size = 16.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(5/2)/(c*e^2*x^4+b*e^2*x^2+b*d*e-c*d^2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(e*x^2 + d)*c*e*x - (5*c*d - 2*b*e)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - (2*c
*d*e - b*e^2)*sqrt((2*c*d - b*e)/(c*d*e - b*e^2))*log((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + (17*c^2*d^2*e^2 -
24*b*c*d*e^3 + 8*b^2*e^4)*x^4 + 2*(7*c^2*d^3*e - 11*b*c*d^2*e^2 + 4*b^2*d*e^3)*x^2 + 4*((3*c^2*d^2*e^2 - 5*b*
c*d*e^3 + 2*b^2*e^4)*x^3 + (c^2*d^3*e - 2*b*c*d^2*e^2 + b^2*d*e^3)*x)*sqrt(e*x^2 + d)*sqrt((2*c*d - b*e)/(c*d*
e - b*e^2)))/(c^2*e^2*x^4 + c^2*d^2 - 2*b*c*d*e + b^2*e^2 - 2*(c^2*d*e - b*c*e^2)*x^2)))/(c^2*e), 1/4*(2*sqrt(
e*x^2 + d)*c*e*x - 2*(5*c*d - 2*b*e)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (2*c*d*e - b*e^2)*sqrt((2*c
*d - b*e)/(c*d*e - b*e^2))*log((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + (17*c^2*d^2*e^2 - 24*b*c*d*e^3 + 8*b^2*e
^4)*x^4 + 2*(7*c^2*d^3*e - 11*b*c*d^2*e^2 + 4*b^2*d*e^3)*x^2 + 4*((3*c^2*d^2*e^2 - 5*b*c*d*e^3 + 2*b^2*e^4)*x^
3 + (c^2*d^3*e - 2*b*c*d^2*e^2 + b^2*d*e^3)*x)*sqrt(e*x^2 + d)*sqrt((2*c*d - b*e)/(c*d*e - b*e^2)))/(c^2*e^2*x
^4 + c^2*d^2 - 2*b*c*d*e + b^2*e^2 - 2*(c^2*d*e - b*c*e^2)*x^2)))/(c^2*e), 1/4*(2*sqrt(e*x^2 + d)*c*e*x + 2*(2
*c*d*e - b*e^2)*sqrt(-(2*c*d - b*e)/(c*d*e - b*e^2))*arctan(1/2*(c*d^2 - b*d*e + (3*c*d*e - 2*b*e^2)*x^2)*sqrt
(e*x^2 + d)*sqrt(-(2*c*d - b*e)/(c*d*e - b*e^2))/((2*c*d*e - b*e^2)*x^3 + (2*c*d^2 - b*d*e)*x)) - (5*c*d - 2*b
*e)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d))/(c^2*e), 1/2*(sqrt(e*x^2 + d)*c*e*x - (5*c*d - 2*
b*e)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (2*c*d*e - b*e^2)*sqrt(-(2*c*d - b*e)/(c*d*e - b*e^2))*arct
an(1/2*(c*d^2 - b*d*e + (3*c*d*e - 2*b*e^2)*x^2)*sqrt(e*x^2 + d)*sqrt(-(2*c*d - b*e)/(c*d*e - b*e^2))/((2*c*d*
e - b*e^2)*x^3 + (2*c*d^2 - b*d*e)*x)))/(c^2*e)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**(5/2)/(c*e**2*x**4+b*e**2*x**2+b*d*e-c*d**2),x)

[Out]

Integral((d + e*x**2)**(3/2)/(b*e - c*d + c*e*x**2), x)

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Giac [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((e*x^2+d)^(5/2)/(c*e^2*x^4+b*e^2*x^2+b*d*e-c*d^2),x, algorithm="giac")

[Out]

Timed out