3.574 \(\int \frac{1}{(d+e x)^2 (a+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=151 \[ \frac{e \sqrt{a+c x^2} \left (c d^2-2 a e^2\right )}{a (d+e x) \left (a e^2+c d^2\right )^2}+\frac{a e+c d x}{a \sqrt{a+c x^2} (d+e x) \left (a e^2+c d^2\right )}-\frac{3 c d e^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0844861, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {741, 807, 725, 206} \[ \frac{e \sqrt{a+c x^2} \left (c d^2-2 a e^2\right )}{a (d+e x) \left (a e^2+c d^2\right )^2}+\frac{a e+c d x}{a \sqrt{a+c x^2} (d+e x) \left (a e^2+c d^2\right )}-\frac{3 c d e^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 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]

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

Mathematica [A]  time = 0.125062, size = 139, normalized size = 0.92 \[ \frac{-a^2 e^3+a c e \left (2 d^2+d e x-2 e^2 x^2\right )+c^2 d^2 x (d+e x)}{a \sqrt{a+c x^2} (d+e x) \left (a e^2+c d^2\right )^2}-\frac{3 c d e^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.194, size = 400, normalized size = 2.7 \begin{align*} -{\frac{1}{a{e}^{2}+c{d}^{2}} \left ({\frac{d}{e}}+x \right ) ^{-1}{\frac{1}{\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}+3\,{\frac{ced}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}{\frac{1}{\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}+3\,{\frac{{c}^{2}{d}^{2}x}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}a}{\frac{1}{\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}-3\,{\frac{ced}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{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{c \left ({\frac{d}{e}}+x \right ) ^{2}-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}}}}}}}-2\,{\frac{cx}{ \left ( a{e}^{2}+c{d}^{2} \right ) a}{\frac{1}{\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^2/(c*x^2+a)^(3/2),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.41863, size = 1777, normalized size = 11.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*(a*c^2*d*e^3*x^3 + a*c^2*d^2*e^2*x^2 + a^2*c*d*e^3*x + a^2*c*d^2*e^2)*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*(2*a*c^2*d^4*e + a^2*c*d^2*e^3 - a^3*e^5 + (c^3*d^4*e - a*c^2*d^2*e^3 - 2*a^2*c
*e^5)*x^2 + (c^3*d^5 + 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*x)*sqrt(c*x^2 + a))/(a^2*c^3*d^7 + 3*a^3*c^2*d^5*e^2 + 3
*a^4*c*d^3*e^4 + a^5*d*e^6 + (a*c^4*d^6*e + 3*a^2*c^3*d^4*e^3 + 3*a^3*c^2*d^2*e^5 + a^4*c*e^7)*x^3 + (a*c^4*d^
7 + 3*a^2*c^3*d^5*e^2 + 3*a^3*c^2*d^3*e^4 + a^4*c*d*e^6)*x^2 + (a^2*c^3*d^6*e + 3*a^3*c^2*d^4*e^3 + 3*a^4*c*d^
2*e^5 + a^5*e^7)*x), -(3*(a*c^2*d*e^3*x^3 + a*c^2*d^2*e^2*x^2 + a^2*c*d*e^3*x + a^2*c*d^2*e^2)*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))
 - (2*a*c^2*d^4*e + a^2*c*d^2*e^3 - a^3*e^5 + (c^3*d^4*e - a*c^2*d^2*e^3 - 2*a^2*c*e^5)*x^2 + (c^3*d^5 + 2*a*c
^2*d^3*e^2 + a^2*c*d*e^4)*x)*sqrt(c*x^2 + a))/(a^2*c^3*d^7 + 3*a^3*c^2*d^5*e^2 + 3*a^4*c*d^3*e^4 + a^5*d*e^6 +
 (a*c^4*d^6*e + 3*a^2*c^3*d^4*e^3 + 3*a^3*c^2*d^2*e^5 + a^4*c*e^7)*x^3 + (a*c^4*d^7 + 3*a^2*c^3*d^5*e^2 + 3*a^
3*c^2*d^3*e^4 + a^4*c*d*e^6)*x^2 + (a^2*c^3*d^6*e + 3*a^3*c^2*d^4*e^3 + 3*a^4*c*d^2*e^5 + a^5*e^7)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**2/(c*x**2+a)**(3/2),x)

[Out]

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

[Out]

Timed out