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

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

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Rubi [A]  time = 0.08, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {741, 807, 725, 206} \begin {gather*} \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}} \end {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 0.12, size = 139, normalized size = 0.92 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.85, size = 209, normalized size = 1.38 \begin {gather*} \frac {-a^2 e^3+2 a c d^2 e+a c d e^2 x-2 a c e^3 x^2+c^2 d^3 x+c^2 d^2 e x^2}{a \sqrt {a+c x^2} (d+e x) \left (a e^2+c d^2\right )^2}+\frac {6 c d e^2 \sqrt {-a e^2-c d^2} \tan ^{-1}\left (-\frac {e \sqrt {a+c x^2}}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} d}{\sqrt {-a e^2-c d^2}}\right )}{\left (a e^2+c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.54, size = 900, normalized size = 5.96 \begin {gather*} \left [\frac {3 \, {\left (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}\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 (2 \, a c^{2} d^{4} e + a^{2} c d^{2} e^{3} - a^{3} e^{5} + {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3} - 2 \, a^{2} c e^{5}\right )} x^{2} + {\left (c^{3} d^{5} + 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (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} + {\left (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}\right )} x^{3} + {\left (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}\right )} x^{2} + {\left (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}\right )} x\right )}}, -\frac {3 \, {\left (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}\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 (2 \, a c^{2} d^{4} e + a^{2} c d^{2} e^{3} - a^{3} e^{5} + {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3} - 2 \, a^{2} c e^{5}\right )} x^{2} + {\left (c^{3} d^{5} + 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} x\right )} \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} + {\left (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}\right )} x^{3} + {\left (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}\right )} x^{2} + {\left (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}\right )} x}\right ] \end {gather*}

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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maple [B]  time = 0.06, size = 400, normalized size = 2.65 \begin {gather*} \frac {3 c^{2} d^{2} x}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, a}-\frac {3 c d e \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}+\frac {3 c d e}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}-\frac {2 c x}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, a}-\frac {1}{\left (a \,e^{2}+c \,d^{2}\right ) \left (x +\frac {d}{e}\right ) \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}} \end {gather*}

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

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maxima [A]  time = 1.82, size = 284, normalized size = 1.88 \begin {gather*} \frac {3 \, c^{2} d^{2} x}{\sqrt {c x^{2} + a} a c^{2} d^{4} + 2 \, \sqrt {c x^{2} + a} a^{2} c d^{2} e^{2} + \sqrt {c x^{2} + a} a^{3} e^{4}} + \frac {3 \, c d}{\frac {\sqrt {c x^{2} + a} c^{2} d^{4}}{e} + 2 \, \sqrt {c x^{2} + a} a c d^{2} e + \sqrt {c x^{2} + a} a^{2} e^{3}} - \frac {2 \, c x}{\sqrt {c x^{2} + a} a c d^{2} + \sqrt {c x^{2} + a} a^{2} e^{2}} - \frac {1}{\sqrt {c x^{2} + a} c d^{2} x + \sqrt {c x^{2} + a} a e^{2} x + \frac {\sqrt {c x^{2} + a} c d^{3}}{e} + \sqrt {c x^{2} + a} a d e} + \frac {3 \, c d \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{{\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {5}{2}} e^{3}} \end {gather*}

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]

3*c^2*d^2*x/(sqrt(c*x^2 + a)*a*c^2*d^4 + 2*sqrt(c*x^2 + a)*a^2*c*d^2*e^2 + sqrt(c*x^2 + a)*a^3*e^4) + 3*c*d/(s
qrt(c*x^2 + a)*c^2*d^4/e + 2*sqrt(c*x^2 + a)*a*c*d^2*e + sqrt(c*x^2 + a)*a^2*e^3) - 2*c*x/(sqrt(c*x^2 + a)*a*c
*d^2 + sqrt(c*x^2 + a)*a^2*e^2) - 1/(sqrt(c*x^2 + a)*c*d^2*x + sqrt(c*x^2 + a)*a*e^2*x + sqrt(c*x^2 + a)*c*d^3
/e + sqrt(c*x^2 + a)*a*d*e) + 3*c*d*arcsinh(c*d*x/(sqrt(a*c)*abs(e*x + d)) - a*e/(sqrt(a*c)*abs(e*x + d)))/((a
+ c*d^2/e^2)^(5/2)*e^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (c\,x^2+a\right )}^{3/2}\,{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + c x^{2}\right )^{\frac {3}{2}} \left (d + e x\right )^{2}}\, dx \end {gather*}

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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