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

Optimal. Leaf size=195 \[ \frac{\sqrt{a+c x^2} \left (11 c d^2-4 a e^2\right )}{6 c^2 e^3}-\frac{d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 c^{3/2} e^4}-\frac{d^4 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^4 \sqrt{a e^2+c d^2}}-\frac{7 d \sqrt{a+c x^2} (d+e x)}{6 c e^3}+\frac{\sqrt{a+c x^2} (d+e x)^2}{3 c e^3} \]

[Out]

((11*c*d^2 - 4*a*e^2)*Sqrt[a + c*x^2])/(6*c^2*e^3) - (7*d*(d + e*x)*Sqrt[a + c*x^2])/(6*c*e^3) + ((d + e*x)^2*
Sqrt[a + c*x^2])/(3*c*e^3) - (d*(2*c*d^2 - a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*c^(3/2)*e^4) - (d^4
*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(e^4*Sqrt[c*d^2 + a*e^2])

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Rubi [A]  time = 0.481779, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {1654, 844, 217, 206, 725} \[ \frac{\sqrt{a+c x^2} \left (11 c d^2-4 a e^2\right )}{6 c^2 e^3}-\frac{d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 c^{3/2} e^4}-\frac{d^4 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^4 \sqrt{a e^2+c d^2}}-\frac{7 d \sqrt{a+c x^2} (d+e x)}{6 c e^3}+\frac{\sqrt{a+c x^2} (d+e x)^2}{3 c e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*c*d^2 - 4*a*e^2)*Sqrt[a + c*x^2])/(6*c^2*e^3) - (7*d*(d + e*x)*Sqrt[a + c*x^2])/(6*c*e^3) + ((d + e*x)^2*
Sqrt[a + c*x^2])/(3*c*e^3) - (d*(2*c*d^2 - a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*c^(3/2)*e^4) - (d^4
*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(e^4*Sqrt[c*d^2 + a*e^2])

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) \sqrt{a+c x^2}} \, dx &=\frac{(d+e x)^2 \sqrt{a+c x^2}}{3 c e^3}+\frac{\int \frac{-2 a d^2 e^2-d e \left (c d^2+4 a e^2\right ) x-e^2 \left (5 c d^2+2 a e^2\right ) x^2-7 c d e^3 x^3}{(d+e x) \sqrt{a+c x^2}} \, dx}{3 c e^4}\\ &=-\frac{7 d (d+e x) \sqrt{a+c x^2}}{6 c e^3}+\frac{(d+e x)^2 \sqrt{a+c x^2}}{3 c e^3}+\frac{\int \frac{3 a c d^2 e^5+c d e^4 \left (5 c d^2-a e^2\right ) x+c e^5 \left (11 c d^2-4 a e^2\right ) x^2}{(d+e x) \sqrt{a+c x^2}} \, dx}{6 c^2 e^7}\\ &=\frac{\left (11 c d^2-4 a e^2\right ) \sqrt{a+c x^2}}{6 c^2 e^3}-\frac{7 d (d+e x) \sqrt{a+c x^2}}{6 c e^3}+\frac{(d+e x)^2 \sqrt{a+c x^2}}{3 c e^3}+\frac{\int \frac{3 a c^2 d^2 e^7-3 c^2 d e^6 \left (2 c d^2-a e^2\right ) x}{(d+e x) \sqrt{a+c x^2}} \, dx}{6 c^3 e^9}\\ &=\frac{\left (11 c d^2-4 a e^2\right ) \sqrt{a+c x^2}}{6 c^2 e^3}-\frac{7 d (d+e x) \sqrt{a+c x^2}}{6 c e^3}+\frac{(d+e x)^2 \sqrt{a+c x^2}}{3 c e^3}+\frac{d^4 \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{e^4}-\frac{\left (d \left (2 c d^2-a e^2\right )\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{2 c e^4}\\ &=\frac{\left (11 c d^2-4 a e^2\right ) \sqrt{a+c x^2}}{6 c^2 e^3}-\frac{7 d (d+e x) \sqrt{a+c x^2}}{6 c e^3}+\frac{(d+e x)^2 \sqrt{a+c x^2}}{3 c e^3}-\frac{d^4 \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}-\frac{\left (d \left (2 c d^2-a e^2\right )\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 (11 c d^2-4 a e^2\right ) \sqrt{a+c x^2}}{6 c^2 e^3}-\frac{7 d (d+e x) \sqrt{a+c x^2}}{6 c e^3}+\frac{(d+e x)^2 \sqrt{a+c x^2}}{3 c e^3}-\frac{d \left (2 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 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{c d^2+a e^2} \sqrt{a+c x^2}}\right )}{e^4 \sqrt{c d^2+a e^2}}\\ \end{align*}

Mathematica [A]  time = 0.266167, size = 149, normalized size = 0.76 \[ \frac{\frac{e \sqrt{a+c x^2} \left (-4 a e^2+6 c d^2-3 c d e x+2 c e^2 x^2\right )}{c^2}-\frac{3 d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{c^{3/2}}-\frac{6 d^4 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\sqrt{a e^2+c d^2}}}{6 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.237, size = 260, normalized size = 1.3 \begin{align*}{\frac{{x}^{2}}{3\,ce}\sqrt{c{x}^{2}+a}}-{\frac{2\,a}{3\,{c}^{2}e}\sqrt{c{x}^{2}+a}}-{\frac{dx}{2\,c{e}^{2}}\sqrt{c{x}^{2}+a}}+{\frac{ad}{2\,{e}^{2}}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{{d}^{2}}{{e}^{3}c}\sqrt{c{x}^{2}+a}}-{\frac{{d}^{3}}{{e}^{4}}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{d}^{4}}{{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}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/e*x^2/c*(c*x^2+a)^(1/2)-2/3/e*a/c^2*(c*x^2+a)^(1/2)-1/2*d/e^2*x/c*(c*x^2+a)^(1/2)+1/2*d/e^2*a/c^(3/2)*ln(x
*c^(1/2)+(c*x^2+a)^(1/2))+d^2/e^3/c*(c*x^2+a)^(1/2)-d^3/e^4*ln(x*c^(1/2)+(c*x^2+a)^(1/2))/c^(1/2)-d^4/e^5/((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)/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 16.2924, size = 2218, normalized size = 11.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(6*sqrt(c*d^2 + a*e^2)*c^2*d^4*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 - 2*sq
rt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) - 3*(2*c^2*d^5 + a*c*d^3*e^2 - a^2
*d*e^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(6*c^2*d^4*e + 2*a*c*d^2*e^3 - 4*a^2*e^5 +
2*(c^2*d^2*e^3 + a*c*e^5)*x^2 - 3*(c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^2*e^4 + a*c^2*e^6), -1
/12*(12*sqrt(-c*d^2 - a*e^2)*c^2*d^4*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*
e^2 + (c^2*d^2 + a*c*e^2)*x^2)) + 3*(2*c^2*d^5 + a*c*d^3*e^2 - a^2*d*e^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2
+ a)*sqrt(c)*x - a) - 2*(6*c^2*d^4*e + 2*a*c*d^2*e^3 - 4*a^2*e^5 + 2*(c^2*d^2*e^3 + a*c*e^5)*x^2 - 3*(c^2*d^3*
e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^2*e^4 + a*c^2*e^6), 1/6*(3*sqrt(c*d^2 + a*e^2)*c^2*d^4*log((2*a*c*
d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))
/(e^2*x^2 + 2*d*e*x + d^2)) + 3*(2*c^2*d^5 + a*c*d^3*e^2 - a^2*d*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 +
a)) + (6*c^2*d^4*e + 2*a*c*d^2*e^3 - 4*a^2*e^5 + 2*(c^2*d^2*e^3 + a*c*e^5)*x^2 - 3*(c^2*d^3*e^2 + a*c*d*e^4)*x
)*sqrt(c*x^2 + a))/(c^3*d^2*e^4 + a*c^2*e^6), -1/6*(6*sqrt(-c*d^2 - a*e^2)*c^2*d^4*arctan(sqrt(-c*d^2 - a*e^2)
*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - 3*(2*c^2*d^5 + a*c*d^3*e^2 - a
^2*d*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (6*c^2*d^4*e + 2*a*c*d^2*e^3 - 4*a^2*e^5 + 2*(c^2*d^2*
e^3 + a*c*e^5)*x^2 - 3*(c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^2*e^4 + a*c^2*e^6)]

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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 )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/(sqrt(a + c*x**2)*(d + e*x)), x)

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Giac [A]  time = 1.23821, size = 220, normalized size = 1.13 \begin{align*} \frac{2 \, d^{4} \arctan \left (-\frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} e + \sqrt{c} d}{\sqrt{-c d^{2} - a e^{2}}}\right ) e^{\left (-4\right )}}{\sqrt{-c d^{2} - a e^{2}}} + \frac{1}{6} \, \sqrt{c x^{2} + a}{\left (x{\left (\frac{2 \, x e^{\left (-1\right )}}{c} - \frac{3 \, d e^{\left (-2\right )}}{c}\right )} + \frac{2 \,{\left (3 \, c^{2} d^{2} e^{7} - 2 \, a c e^{9}\right )} e^{\left (-10\right )}}{c^{3}}\right )} + \frac{{\left (2 \, c^{\frac{3}{2}} d^{3} - a \sqrt{c} d e^{2}\right )} e^{\left (-4\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d^4*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-4)/sqrt(-c*d^2 - a*e^2)
+ 1/6*sqrt(c*x^2 + a)*(x*(2*x*e^(-1)/c - 3*d*e^(-2)/c) + 2*(3*c^2*d^2*e^7 - 2*a*c*e^9)*e^(-10)/c^3) + 1/2*(2*c
^(3/2)*d^3 - a*sqrt(c)*d*e^2)*e^(-4)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^2