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

Optimal. Leaf size=88 \[ \frac{d e \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}}-\frac{d-e x}{\sqrt{a+c x^2} \left (a e^2+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0519365, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {823, 12, 725, 206} \[ \frac{d e \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}}-\frac{d-e x}{\sqrt{a+c x^2} \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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])

Rubi steps

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

Mathematica [A]  time = 0.0583673, size = 88, normalized size = 1. \[ \frac{e x-d}{\sqrt{a+c x^2} \left (a e^2+c d^2\right )}+\frac{d e \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.23, size = 283, normalized size = 3.2 \begin{align*}{\frac{x}{ae}{\frac{1}{\sqrt{c{x}^{2}+a}}}}-{\frac{d}{a{e}^{2}+c{d}^{2}}{\frac{1}{\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}}}}}}}-{\frac{{d}^{2}xc}{e \left ( a{e}^{2}+c{d}^{2} \right ) a}{\frac{1}{\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}}}}}}}+{\frac{d}{a{e}^{2}+c{d}^{2}}\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/(e*x+d)/(c*x^2+a)^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.5926, size = 863, normalized size = 9.81 \begin{align*} \left [\frac{{\left (c d e x^{2} + a d e\right )} \sqrt{c d^{2} + a e^{2}} \log \left (\frac{2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} -{\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} + 2 \, \sqrt{c d^{2} + a e^{2}}{\left (c d x - a e\right )} \sqrt{c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \,{\left (c d^{3} + a d e^{2} -{\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt{c x^{2} + a}}{2 \,{\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4} +{\left (c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )} x^{2}\right )}}, \frac{{\left (c d e x^{2} + a d e\right )} \sqrt{-c d^{2} - a e^{2}} \arctan \left (\frac{\sqrt{-c d^{2} - a e^{2}}{\left (c d x - a e\right )} \sqrt{c x^{2} + a}}{a c d^{2} + a^{2} e^{2} +{\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) -{\left (c d^{3} + a d e^{2} -{\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt{c x^{2} + a}}{a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4} +{\left (c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.14151, size = 219, normalized size = 2.49 \begin{align*} \frac{2 \, d \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 (c d^{2} + a e^{2}\right )} \sqrt{-c d^{2} - a e^{2}}} + \frac{\frac{{\left (c d^{2} e + a e^{3}\right )} x}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} - \frac{c d^{3} + a d e^{2}}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}}}{\sqrt{c x^{2} + a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d*arctan(((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e/((c*d^2 + a*e^2)*sqrt(-c*d^2
- a*e^2)) + ((c*d^2*e + a*e^3)*x/(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4) - (c*d^3 + a*d*e^2)/(c^2*d^4 + 2*a*c*d^2*
e^2 + a^2*e^4))/sqrt(c*x^2 + a)