### 3.567 $$\int \frac{1}{(d+e x)^3 \sqrt{a+c x^2}} \, dx$$

Optimal. Leaf size=145 $-\frac{3 c d e \sqrt{a+c x^2}}{2 (d+e x) \left (a e^2+c d^2\right )^2}-\frac{e \sqrt{a+c x^2}}{2 (d+e x)^2 \left (a e^2+c d^2\right )}-\frac{c \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{5/2}}$

[Out]

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

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Rubi [A]  time = 0.0707614, antiderivative size = 145, 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 = {745, 807, 725, 206} $-\frac{3 c d e \sqrt{a+c x^2}}{2 (d+e x) \left (a e^2+c d^2\right )^2}-\frac{e \sqrt{a+c x^2}}{2 (d+e x)^2 \left (a e^2+c d^2\right )}-\frac{c \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 745

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

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

Mathematica [A]  time = 0.139823, size = 161, normalized size = 1.11 $\frac{-e \sqrt{a+c x^2} \sqrt{a e^2+c d^2} \left (a e^2+c d (4 d+3 e x)\right )-c (d+e x)^2 \left (2 c d^2-a e^2\right ) \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )+c (d+e x)^2 \left (2 c d^2-a e^2\right ) \log (d+e x)}{2 (d+e x)^2 \left (a e^2+c d^2\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-(e*Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]*(a*e^2 + c*d*(4*d + 3*e*x))) + c*(2*c*d^2 - a*e^2)*(d + e*x)^2*Log[d
+ e*x] - c*(2*c*d^2 - a*e^2)*(d + e*x)^2*Log[a*e - c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]])/(2*(c*d^2 + a
*e^2)^(5/2)*(d + e*x)^2)

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Maple [B]  time = 0.193, size = 426, normalized size = 2.9 \begin{align*} -{\frac{1}{2\,e \left ( a{e}^{2}+c{d}^{2} \right ) }\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}}}} \left ({\frac{d}{e}}+x \right ) ^{-2}}-{\frac{3\,cd}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{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}}}} \left ({\frac{d}{e}}+x \right ) ^{-1}}-{\frac{3\,{c}^{2}{d}^{2}}{2\,e \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}}}}}}}+{\frac{c}{2\,e \left ( a{e}^{2}+c{d}^{2} \right ) }\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}}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.10872, size = 1432, normalized size = 9.88 \begin{align*} \left [-\frac{{\left (2 \, c^{2} d^{4} - a c d^{2} e^{2} +{\left (2 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x^{2} + 2 \,{\left (2 \, c^{2} d^{3} e - a c d e^{3}\right )} x\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 (4 \, c^{2} d^{4} e + 5 \, a c d^{2} e^{3} + a^{2} e^{5} + 3 \,{\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt{c x^{2} + a}}{4 \,{\left (c^{3} d^{8} + 3 \, a c^{2} d^{6} e^{2} + 3 \, a^{2} c d^{4} e^{4} + a^{3} d^{2} e^{6} +{\left (c^{3} d^{6} e^{2} + 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} + a^{3} e^{8}\right )} x^{2} + 2 \,{\left (c^{3} d^{7} e + 3 \, a c^{2} d^{5} e^{3} + 3 \, a^{2} c d^{3} e^{5} + a^{3} d e^{7}\right )} x\right )}}, -\frac{{\left (2 \, c^{2} d^{4} - a c d^{2} e^{2} +{\left (2 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x^{2} + 2 \,{\left (2 \, c^{2} d^{3} e - a c d e^{3}\right )} x\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 (4 \, c^{2} d^{4} e + 5 \, a c d^{2} e^{3} + a^{2} e^{5} + 3 \,{\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt{c x^{2} + a}}{2 \,{\left (c^{3} d^{8} + 3 \, a c^{2} d^{6} e^{2} + 3 \, a^{2} c d^{4} e^{4} + a^{3} d^{2} e^{6} +{\left (c^{3} d^{6} e^{2} + 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} + a^{3} e^{8}\right )} x^{2} + 2 \,{\left (c^{3} d^{7} e + 3 \, a c^{2} d^{5} e^{3} + 3 \, a^{2} c d^{3} e^{5} + a^{3} d e^{7}\right )} x\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((2*c^2*d^4 - a*c*d^2*e^2 + (2*c^2*d^2*e^2 - a*c*e^4)*x^2 + 2*(2*c^2*d^3*e - a*c*d*e^3)*x)*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*(4*c^2*d^4*e + 5*a*c*d^2*e^3 + a^2*e^5 + 3*(c^2*d^3*e^2 + a*
c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^8 + 3*a*c^2*d^6*e^2 + 3*a^2*c*d^4*e^4 + a^3*d^2*e^6 + (c^3*d^6*e^2 + 3*a*c
^2*d^4*e^4 + 3*a^2*c*d^2*e^6 + a^3*e^8)*x^2 + 2*(c^3*d^7*e + 3*a*c^2*d^5*e^3 + 3*a^2*c*d^3*e^5 + a^3*d*e^7)*x)
, -1/2*((2*c^2*d^4 - a*c*d^2*e^2 + (2*c^2*d^2*e^2 - a*c*e^4)*x^2 + 2*(2*c^2*d^3*e - a*c*d*e^3)*x)*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)) + (4*c^2*d^4*e + 5*a*c*d^2*e^3 + a^2*e^5 + 3*(c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^8 + 3*a*
c^2*d^6*e^2 + 3*a^2*c*d^4*e^4 + a^3*d^2*e^6 + (c^3*d^6*e^2 + 3*a*c^2*d^4*e^4 + 3*a^2*c*d^2*e^6 + a^3*e^8)*x^2
+ 2*(c^3*d^7*e + 3*a*c^2*d^5*e^3 + 3*a^2*c*d^3*e^5 + a^3*d*e^7)*x)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.42993, size = 466, normalized size = 3.21 \begin{align*} -c{\left (\frac{{\left (2 \, c d^{2} - a e^{2}\right )} \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 )}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt{-c d^{2} - a e^{2}}} + \frac{2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} c d^{2} e + 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} c^{\frac{3}{2}} d^{3} - 10 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a c d^{2} e - 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} a \sqrt{c} d e^{2} -{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} a e^{3} + 3 \, a^{2} \sqrt{c} d e^{2} -{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a^{2} e^{3}}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} e + 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} \sqrt{c} d - a e\right )}^{2}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*((2*c*d^2 - a*e^2)*arctan(((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))/((c^2*d^4 + 2
*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c*d^2 - a*e^2)) + (2*(sqrt(c)*x - sqrt(c*x^2 + a))^3*c*d^2*e + 6*(sqrt(c)*x - sq
rt(c*x^2 + a))^2*c^(3/2)*d^3 - 10*(sqrt(c)*x - sqrt(c*x^2 + a))*a*c*d^2*e - 3*(sqrt(c)*x - sqrt(c*x^2 + a))^2*
a*sqrt(c)*d*e^2 - (sqrt(c)*x - sqrt(c*x^2 + a))^3*a*e^3 + 3*a^2*sqrt(c)*d*e^2 - (sqrt(c)*x - sqrt(c*x^2 + a))*
a^2*e^3)/((c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*((sqrt(c)*x - sqrt(c*x^2 + a))^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 +
a))*sqrt(c)*d - a*e)^2))