3.343 \(\int \frac{x^4}{(d+e x)^2 \sqrt{a+c x^2}} \, dx\)

Optimal. Leaf size=204 \[ \frac{\left (6 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 c^{3/2} e^4}-\frac{d^4 \sqrt{a+c x^2}}{e^3 (d+e x) \left (a e^2+c d^2\right )}+\frac{d^3 \left (4 a e^2+3 c d^2\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^4 \left (a e^2+c d^2\right )^{3/2}}-\frac{5 d \sqrt{a+c x^2}}{2 c e^3}+\frac{\sqrt{a+c x^2} (d+e x)}{2 c e^3} \]

[Out]

(-5*d*Sqrt[a + c*x^2])/(2*c*e^3) - (d^4*Sqrt[a + c*x^2])/(e^3*(c*d^2 + a*e^2)*(d + e*x)) + ((d + e*x)*Sqrt[a +
c*x^2])/(2*c*e^3) + ((6*c*d^2 - a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*c^(3/2)*e^4) + (d^3*(3*c*d^2
+ 4*a*e^2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(e^4*(c*d^2 + a*e^2)^(3/2))

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Rubi [A]  time = 0.523424, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {1651, 1654, 844, 217, 206, 725} \[ \frac{\left (6 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 c^{3/2} e^4}-\frac{d^4 \sqrt{a+c x^2}}{e^3 (d+e x) \left (a e^2+c d^2\right )}+\frac{d^3 \left (4 a e^2+3 c d^2\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^4 \left (a e^2+c d^2\right )^{3/2}}-\frac{5 d \sqrt{a+c x^2}}{2 c e^3}+\frac{\sqrt{a+c x^2} (d+e x)}{2 c e^3} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((d + e*x)^2*Sqrt[a + c*x^2]),x]

[Out]

(-5*d*Sqrt[a + c*x^2])/(2*c*e^3) - (d^4*Sqrt[a + c*x^2])/(e^3*(c*d^2 + a*e^2)*(d + e*x)) + ((d + e*x)*Sqrt[a +
c*x^2])/(2*c*e^3) + ((6*c*d^2 - a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*c^(3/2)*e^4) + (d^3*(3*c*d^2
+ 4*a*e^2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(e^4*(c*d^2 + a*e^2)^(3/2))

Rule 1651

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, d
+ e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1
)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p*ExpandToSum[(m
+ 1)*(c*d^2 + a*e^2)*Q + c*d*R*(m + 1) - c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e, p}, x] && Po
lyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 1654

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

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 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rubi steps

\begin{align*} \int \frac{x^4}{(d+e x)^2 \sqrt{a+c x^2}} \, dx &=-\frac{d^4 \sqrt{a+c x^2}}{e^3 \left (c d^2+a e^2\right ) (d+e x)}-\frac{\int \frac{\frac{a d^3}{e^2}-\frac{d^2 \left (c d^2+a e^2\right ) x}{e^3}+d \left (a+\frac{c d^2}{e^2}\right ) x^2-\frac{\left (c d^2+a e^2\right ) x^3}{e}}{(d+e x) \sqrt{a+c x^2}} \, dx}{c d^2+a e^2}\\ &=-\frac{d^4 \sqrt{a+c x^2}}{e^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \sqrt{a+c x^2}}{2 c e^3}-\frac{\int \frac{a d e \left (3 c d^2+a e^2\right )-\left (c^2 d^4-a^2 e^4\right ) x+5 c d e \left (c d^2+a e^2\right ) x^2}{(d+e x) \sqrt{a+c x^2}} \, dx}{2 c e^3 \left (c d^2+a e^2\right )}\\ &=-\frac{5 d \sqrt{a+c x^2}}{2 c e^3}-\frac{d^4 \sqrt{a+c x^2}}{e^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \sqrt{a+c x^2}}{2 c e^3}-\frac{\int \frac{a c d e^3 \left (3 c d^2+a e^2\right )-c e^2 \left (6 c d^2-a e^2\right ) \left (c d^2+a e^2\right ) x}{(d+e x) \sqrt{a+c x^2}} \, dx}{2 c^2 e^5 \left (c d^2+a e^2\right )}\\ &=-\frac{5 d \sqrt{a+c x^2}}{2 c e^3}-\frac{d^4 \sqrt{a+c x^2}}{e^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \sqrt{a+c x^2}}{2 c e^3}+\frac{\left (6 c d^2-a e^2\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{2 c e^4}-\frac{\left (d^3 \left (3 c d^2+4 a e^2\right )\right ) \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{e^4 \left (c d^2+a e^2\right )}\\ &=-\frac{5 d \sqrt{a+c x^2}}{2 c e^3}-\frac{d^4 \sqrt{a+c x^2}}{e^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \sqrt{a+c x^2}}{2 c e^3}+\frac{\left (6 c d^2-a e^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{2 c e^4}+\frac{\left (d^3 \left (3 c d^2+4 a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c d^2+a e^2-x^2} \, dx,x,\frac{a e-c d x}{\sqrt{a+c x^2}}\right )}{e^4 \left (c d^2+a e^2\right )}\\ &=-\frac{5 d \sqrt{a+c x^2}}{2 c e^3}-\frac{d^4 \sqrt{a+c x^2}}{e^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \sqrt{a+c x^2}}{2 c e^3}+\frac{\left (6 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 c^{3/2} e^4}+\frac{d^3 \left (3 c d^2+4 a e^2\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{c d^2+a e^2} \sqrt{a+c x^2}}\right )}{e^4 \left (c d^2+a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.429782, size = 208, normalized size = 1.02 \[ \frac{\frac{\left (6 c d^2-a e^2\right ) \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )}{c^{3/2}}+e \sqrt{a+c x^2} \left (\frac{e x-4 d}{c}-\frac{2 d^4}{(d+e x) \left (a e^2+c d^2\right )}\right )+\frac{2 d^3 \left (4 a e^2+3 c d^2\right ) \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )}{\left (a e^2+c d^2\right )^{3/2}}-\frac{2 d^3 \left (4 a e^2+3 c d^2\right ) \log (d+e x)}{\left (a e^2+c d^2\right )^{3/2}}}{2 e^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/((d + e*x)^2*Sqrt[a + c*x^2]),x]

[Out]

(e*Sqrt[a + c*x^2]*((-4*d + e*x)/c - (2*d^4)/((c*d^2 + a*e^2)*(d + e*x))) - (2*d^3*(3*c*d^2 + 4*a*e^2)*Log[d +
e*x])/(c*d^2 + a*e^2)^(3/2) + ((6*c*d^2 - a*e^2)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/c^(3/2) + (2*d^3*(3*c*d^
2 + 4*a*e^2)*Log[a*e - c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]])/(c*d^2 + a*e^2)^(3/2))/(2*e^4)

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Maple [B]  time = 0.255, size = 435, normalized size = 2.1 \begin{align*}{\frac{x}{2\,c{e}^{2}}\sqrt{c{x}^{2}+a}}-{\frac{a}{2\,{e}^{2}}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}-2\,{\frac{d\sqrt{c{x}^{2}+a}}{c{e}^{3}}}+3\,{\frac{{d}^{2}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ) }{{e}^{4}\sqrt{c}}}+4\,{\frac{{d}^{3}}{{e}^{5}}\ln \left ({ \left ( 2\,{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}-{\frac{{d}^{4}}{{e}^{4} \left ( a{e}^{2}+c{d}^{2} \right ) }\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \left ({\frac{d}{e}}+x \right ) ^{-1}}-{\frac{{d}^{5}c}{{e}^{5} \left ( a{e}^{2}+c{d}^{2} \right ) }\ln \left ({ \left ( 2\,{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(e*x+d)^2/(c*x^2+a)^(1/2),x)

[Out]

1/2/e^2*x/c*(c*x^2+a)^(1/2)-1/2/e^2*a/c^(3/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))-2*d*(c*x^2+a)^(1/2)/c/e^3+3*d^2/e^
4*ln(x*c^(1/2)+(c*x^2+a)^(1/2))/c^(1/2)+4/e^5*d^3/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(d
/e+x)+2*((a*e^2+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))/(d/e+x))-d^4/e^4/(a*e
^2+c*d^2)/(d/e+x)*((d/e+x)^2*c-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)-d^5/e^5*c/(a*e^2+c*d^2)/((a*e^2+c*d^2)
/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(d/e+x)+2*((a*e^2+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c-2*c*d/e*(d/e+x)+(
a*e^2+c*d^2)/e^2)^(1/2))/(d/e+x))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*x+d)^2/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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(x^4/(e*x+d)^2/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{a + c x^{2}} \left (d + e x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(e*x+d)**2/(c*x**2+a)**(1/2),x)

[Out]

Integral(x**4/(sqrt(a + c*x**2)*(d + 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(x^4/(e*x+d)^2/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

Timed out