### 3.532 $$\int \frac{\sqrt{a+c x^2}}{(d+e x)^4} \, dx$$

Optimal. Leaf size=144 $-\frac{a c^2 d \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{5/2}}-\frac{c d \sqrt{a+c x^2} (a e-c d x)}{2 (d+e x)^2 \left (a e^2+c d^2\right )^2}-\frac{e \left (a+c x^2\right )^{3/2}}{3 (d+e x)^3 \left (a e^2+c d^2\right )}$

[Out]

-(c*d*(a*e - c*d*x)*Sqrt[a + c*x^2])/(2*(c*d^2 + a*e^2)^2*(d + e*x)^2) - (e*(a + c*x^2)^(3/2))/(3*(c*d^2 + a*e
^2)*(d + e*x)^3) - (a*c^2*d*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 = 0.0863029, antiderivative size = 144, 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 = {731, 721, 725, 206} $-\frac{a c^2 d \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{5/2}}-\frac{c d \sqrt{a+c x^2} (a e-c d x)}{2 (d+e x)^2 \left (a e^2+c d^2\right )^2}-\frac{e \left (a+c x^2\right )^{3/2}}{3 (d+e x)^3 \left (a e^2+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[a + c*x^2]/(d + e*x)^4,x]

[Out]

-(c*d*(a*e - c*d*x)*Sqrt[a + c*x^2])/(2*(c*d^2 + a*e^2)^2*(d + e*x)^2) - (e*(a + c*x^2)^(3/2))/(3*(c*d^2 + a*e
^2)*(d + e*x)^3) - (a*c^2*d*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 731

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*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0]

Rule 721

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(-2*a*e + (2*c*
d)*x)*(a + c*x^2)^p)/(2*(m + 1)*(c*d^2 + a*e^2)), x] - Dist[(4*a*c*p)/(2*(m + 1)*(c*d^2 + a*e^2)), Int[(d + e*
x)^(m + 2)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2,
0] && GtQ[p, 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{\sqrt{a+c x^2}}{(d+e x)^4} \, dx &=-\frac{e \left (a+c x^2\right )^{3/2}}{3 \left (c d^2+a e^2\right ) (d+e x)^3}+\frac{(c d) \int \frac{\sqrt{a+c x^2}}{(d+e x)^3} \, dx}{c d^2+a e^2}\\ &=-\frac{c d (a e-c d x) \sqrt{a+c x^2}}{2 \left (c d^2+a e^2\right )^2 (d+e x)^2}-\frac{e \left (a+c x^2\right )^{3/2}}{3 \left (c d^2+a e^2\right ) (d+e x)^3}+\frac{\left (a c^2 d\right ) \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{2 \left (c d^2+a e^2\right )^2}\\ &=-\frac{c d (a e-c d x) \sqrt{a+c x^2}}{2 \left (c d^2+a e^2\right )^2 (d+e x)^2}-\frac{e \left (a+c x^2\right )^{3/2}}{3 \left (c d^2+a e^2\right ) (d+e x)^3}-\frac{\left (a c^2 d\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{c d (a e-c d x) \sqrt{a+c x^2}}{2 \left (c d^2+a e^2\right )^2 (d+e x)^2}-\frac{e \left (a+c x^2\right )^{3/2}}{3 \left (c d^2+a e^2\right ) (d+e x)^3}-\frac{a c^2 d \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 = 0.169825, size = 173, normalized size = 1.2 $\frac{\sqrt{a+c x^2} \sqrt{a e^2+c d^2} \left (-2 a^2 e^3-a c e \left (5 d^2+3 d e x+2 e^2 x^2\right )+c^2 d^2 x (3 d+e x)\right )-3 a c^2 d (d+e x)^3 \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )+3 a c^2 d (d+e x)^3 \log (d+e x)}{6 (d+e x)^3 \left (a e^2+c d^2\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[a + c*x^2]/(d + e*x)^4,x]

[Out]

(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]*(-2*a^2*e^3 + c^2*d^2*x*(3*d + e*x) - a*c*e*(5*d^2 + 3*d*e*x + 2*e^2*x^2)
) + 3*a*c^2*d*(d + e*x)^3*Log[d + e*x] - 3*a*c^2*d*(d + e*x)^3*Log[a*e - c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt[a +
c*x^2]])/(6*(c*d^2 + a*e^2)^(5/2)*(d + e*x)^3)

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Maple [B]  time = 0.207, size = 1262, normalized size = 8.8 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)^(1/2)/(e*x+d)^4,x)

[Out]

-1/3/e^2/(a*e^2+c*d^2)/(d/e+x)^3*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)-1/2/e*c*d/(a*e^2+c*d^2)
^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)^(3/2)-1/2*c^2*d^2/(a*e^2+c*d^2)^3/(d/e+x)*(c*(d/e
+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)-1/2/e*c^3*d^3/(a*e^2+c*d^2)^3*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e
^2+c*d^2)/e^2)^(1/2)+1/2/e^2*c^(7/2)*d^4/(a*e^2+c*d^2)^3*ln((-c*d/e+(d/e+x)*c)/c^(1/2)+(c*(d/e+x)^2-2*c*d/e*(d
/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))+1/2/e*c^3*d^3/(a*e^2+c*d^2)^3/((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))*
a+1/2/e^3*c^4*d^5/(a*e^2+c*d^2)^3/((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*c^3*d^2/(a*e^2+c*d^2)^3*
(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*x+1/2*c^(5/2)*d^2/(a*e^2+c*d^2)^3*ln((-c*d/e+(d/e+x)*c)/
c^(1/2)+(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))*a+1/2/e*c^2*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)-1/2/e^2*c^(5/2)*d^2/(a*e^2+c*d^2)^2*ln((-c*d/e+(d/e+x)*c)/c^(1/2)+(c*(d/e
+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))-1/2/e*c^2*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))*a-1/2/e^3*c^3*d^3/(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))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.25644, size = 1719, normalized size = 11.94 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

[1/12*(3*(a*c^2*d*e^3*x^3 + 3*a*c^2*d^2*e^2*x^2 + 3*a*c^2*d^3*e*x + a*c^2*d^4)*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*(5*a*c^2*d^4*e + 7*a^2*c*d^2*e^3 + 2*a^3*e^5 - (c^3*d^4*e - a*c^2*d^2*e^3 - 2*
a^2*c*e^5)*x^2 - 3*(c^3*d^5 - a^2*c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^9 + 3*a*c^2*d^7*e^2 + 3*a^2*c*d^5*e^4 +
a^3*d^3*e^6 + (c^3*d^6*e^3 + 3*a*c^2*d^4*e^5 + 3*a^2*c*d^2*e^7 + a^3*e^9)*x^3 + 3*(c^3*d^7*e^2 + 3*a*c^2*d^5*e
^4 + 3*a^2*c*d^3*e^6 + a^3*d*e^8)*x^2 + 3*(c^3*d^8*e + 3*a*c^2*d^6*e^3 + 3*a^2*c*d^4*e^5 + a^3*d^2*e^7)*x), -1
/6*(3*(a*c^2*d*e^3*x^3 + 3*a*c^2*d^2*e^2*x^2 + 3*a*c^2*d^3*e*x + a*c^2*d^4)*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)) + (5*a*c^2*d^4*e +
7*a^2*c*d^2*e^3 + 2*a^3*e^5 - (c^3*d^4*e - a*c^2*d^2*e^3 - 2*a^2*c*e^5)*x^2 - 3*(c^3*d^5 - a^2*c*d*e^4)*x)*sq
rt(c*x^2 + a))/(c^3*d^9 + 3*a*c^2*d^7*e^2 + 3*a^2*c*d^5*e^4 + a^3*d^3*e^6 + (c^3*d^6*e^3 + 3*a*c^2*d^4*e^5 + 3
*a^2*c*d^2*e^7 + a^3*e^9)*x^3 + 3*(c^3*d^7*e^2 + 3*a*c^2*d^5*e^4 + 3*a^2*c*d^3*e^6 + a^3*d*e^8)*x^2 + 3*(c^3*d
^8*e + 3*a*c^2*d^6*e^3 + 3*a^2*c*d^4*e^5 + a^3*d^2*e^7)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + c x^{2}}}{\left (d + e x\right )^{4}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+a)**(1/2)/(e*x+d)**4,x)

[Out]

Integral(sqrt(a + c*x**2)/(d + e*x)**4, x)

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Giac [B]  time = 1.41163, size = 699, normalized size = 4.85 \begin{align*} -\frac{a c^{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 )}{{\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{6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{4} c^{\frac{7}{2}} d^{4} e + 4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} c^{4} d^{5} - 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} a c^{\frac{7}{2}} d^{4} e - 14 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} a c^{3} d^{3} e^{2} - 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{4} a c^{\frac{5}{2}} d^{2} e^{3} - 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{5} a c^{2} d e^{4} + 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a^{2} c^{3} d^{3} e^{2} + 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} a^{2} c^{\frac{5}{2}} d^{2} e^{3} + 12 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} a^{2} c^{2} d e^{4} + 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{4} a^{2} c^{\frac{3}{2}} e^{5} - a^{3} c^{\frac{5}{2}} d^{2} e^{3} - 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a^{3} c^{2} d e^{4} + 2 \, a^{4} c^{\frac{3}{2}} e^{5}}{3 \,{\left (c^{2} d^{4} e^{2} + 2 \, a c d^{2} e^{4} + a^{2} e^{6}\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 )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

-a*c^2*d*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)) + 1/3*(6*(sqrt(c)*x - sqrt(c*x^2 + a))^4*c^(7/2)*d^4*e + 4*(sqrt(c)*x - sqrt(
c*x^2 + a))^3*c^4*d^5 - 6*(sqrt(c)*x - sqrt(c*x^2 + a))^2*a*c^(7/2)*d^4*e - 14*(sqrt(c)*x - sqrt(c*x^2 + a))^3
*a*c^3*d^3*e^2 - 3*(sqrt(c)*x - sqrt(c*x^2 + a))^4*a*c^(5/2)*d^2*e^3 - 3*(sqrt(c)*x - sqrt(c*x^2 + a))^5*a*c^2
*d*e^4 + 6*(sqrt(c)*x - sqrt(c*x^2 + a))*a^2*c^3*d^3*e^2 + 24*(sqrt(c)*x - sqrt(c*x^2 + a))^2*a^2*c^(5/2)*d^2*
e^3 + 12*(sqrt(c)*x - sqrt(c*x^2 + a))^3*a^2*c^2*d*e^4 + 6*(sqrt(c)*x - sqrt(c*x^2 + a))^4*a^2*c^(3/2)*e^5 - a
^3*c^(5/2)*d^2*e^3 - 9*(sqrt(c)*x - sqrt(c*x^2 + a))*a^3*c^2*d*e^4 + 2*a^4*c^(3/2)*e^5)/((c^2*d^4*e^2 + 2*a*c*
d^2*e^4 + a^2*e^6)*((sqrt(c)*x - sqrt(c*x^2 + a))^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 + a))*sqrt(c)*d - a*e)^3)