3.327 \(\int \frac{x^3}{(d+e x) \sqrt{a+c x^2}} \, dx\)

Optimal. Leaf size=152 \[ \frac{\left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 c^{3/2} e^3}+\frac{d^3 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^3 \sqrt{a e^2+c d^2}}-\frac{3 d \sqrt{a+c x^2}}{2 c e^2}+\frac{\sqrt{a+c x^2} (d+e x)}{2 c e^2} \]

[Out]

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

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Rubi [A]  time = 0.273506, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{\left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 c^{3/2} e^3}+\frac{d^3 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^3 \sqrt{a e^2+c d^2}}-\frac{3 d \sqrt{a+c x^2}}{2 c e^2}+\frac{\sqrt{a+c x^2} (d+e x)}{2 c e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.236951, size = 131, normalized size = 0.86 \[ \frac{\left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )+\sqrt{c} \left (\frac{2 c d^3 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\sqrt{a e^2+c d^2}}+e \sqrt{a+c x^2} (e x-2 d)\right )}{2 c^{3/2} e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.261, size = 217, normalized size = 1.4 \begin{align*}{\frac{x}{2\,ce}\sqrt{c{x}^{2}+a}}-{\frac{a}{2\,e}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}-{\frac{d}{c{e}^{2}}\sqrt{c{x}^{2}+a}}+{\frac{{d}^{2}}{{e}^{3}}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}}+{\frac{{d}^{3}}{{e}^{4}}\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^3/(e*x+d)/(c*x^2+a)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 16.4052, size = 1917, normalized size = 12.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(c*d^2 + a*e^2)*c^2*d^3*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*sqr
t(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)) - (2*c^2*d^4 + a*c*d^2*e^2 - a^2*e^
4)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(2*c^2*d^3*e + 2*a*c*d*e^3 - (c^2*d^2*e^2 + a*c
*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^2*e^3 + a*c^2*e^5), 1/4*(4*sqrt(-c*d^2 - a*e^2)*c^2*d^3*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)) - (2*c^2*d^4 + a*c*d^2*e^2
- a^2*e^4)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(2*c^2*d^3*e + 2*a*c*d*e^3 - (c^2*d^2*
e^2 + a*c*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^2*e^3 + a*c^2*e^5), 1/2*(sqrt(c*d^2 + a*e^2)*c^2*d^3*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)) - (2*c^2*d^4 + a*c*d^2*e^2 - a^2*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (2
*c^2*d^3*e + 2*a*c*d*e^3 - (c^2*d^2*e^2 + a*c*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^2*e^3 + a*c^2*e^5), 1/2*(2*sqrt(
-c*d^2 - a*e^2)*c^2*d^3*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)) - (2*c^2*d^4 + a*c*d^2*e^2 - a^2*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (2*c^2*
d^3*e + 2*a*c*d*e^3 - (c^2*d^2*e^2 + a*c*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^2*e^3 + a*c^2*e^5)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\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**3/(e*x+d)/(c*x**2+a)**(1/2),x)

[Out]

Integral(x**3/(sqrt(a + c*x**2)*(d + e*x)), x)

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Giac [A]  time = 1.22742, size = 174, normalized size = 1.14 \begin{align*} -\frac{2 \, d^{3} \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 (-3\right )}}{\sqrt{-c d^{2} - a e^{2}}} + \frac{1}{2} \, \sqrt{c x^{2} + a}{\left (\frac{x e^{\left (-1\right )}}{c} - \frac{2 \, d e^{\left (-2\right )}}{c}\right )} - \frac{{\left (2 \, c d^{2} - a e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{2 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-2*d^3*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-3)/sqrt(-c*d^2 - a*e^2)
+ 1/2*sqrt(c*x^2 + a)*(x*e^(-1)/c - 2*d*e^(-2)/c) - 1/2*(2*c*d^2 - a*e^2)*e^(-3)*log(abs(-sqrt(c)*x + sqrt(c*
x^2 + a)))/c^(3/2)