### 3.1661 $$\int \frac{1}{(d+e x)^{3/2} (a^2+2 a b x+b^2 x^2)^2} \, dx$$

Optimal. Leaf size=173 $-\frac{35 e^3}{8 \sqrt{d+e x} (b d-a e)^4}-\frac{35 e^2}{24 (a+b x) \sqrt{d+e x} (b d-a e)^3}+\frac{35 \sqrt{b} e^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 (b d-a e)^{9/2}}+\frac{7 e}{12 (a+b x)^2 \sqrt{d+e x} (b d-a e)^2}-\frac{1}{3 (a+b x)^3 \sqrt{d+e x} (b d-a e)}$

[Out]

(-35*e^3)/(8*(b*d - a*e)^4*Sqrt[d + e*x]) - 1/(3*(b*d - a*e)*(a + b*x)^3*Sqrt[d + e*x]) + (7*e)/(12*(b*d - a*e
)^2*(a + b*x)^2*Sqrt[d + e*x]) - (35*e^2)/(24*(b*d - a*e)^3*(a + b*x)*Sqrt[d + e*x]) + (35*Sqrt[b]*e^3*ArcTanh
[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*(b*d - a*e)^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0835521, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {27, 51, 63, 208} $-\frac{35 e^3}{8 \sqrt{d+e x} (b d-a e)^4}-\frac{35 e^2}{24 (a+b x) \sqrt{d+e x} (b d-a e)^3}+\frac{35 \sqrt{b} e^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 (b d-a e)^{9/2}}+\frac{7 e}{12 (a+b x)^2 \sqrt{d+e x} (b d-a e)^2}-\frac{1}{3 (a+b x)^3 \sqrt{d+e x} (b d-a e)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-35*e^3)/(8*(b*d - a*e)^4*Sqrt[d + e*x]) - 1/(3*(b*d - a*e)*(a + b*x)^3*Sqrt[d + e*x]) + (7*e)/(12*(b*d - a*e
)^2*(a + b*x)^2*Sqrt[d + e*x]) - (35*e^2)/(24*(b*d - a*e)^3*(a + b*x)*Sqrt[d + e*x]) + (35*Sqrt[b]*e^3*ArcTanh
[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*(b*d - a*e)^(9/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{1}{(a+b x)^4 (d+e x)^{3/2}} \, dx\\ &=-\frac{1}{3 (b d-a e) (a+b x)^3 \sqrt{d+e x}}-\frac{(7 e) \int \frac{1}{(a+b x)^3 (d+e x)^{3/2}} \, dx}{6 (b d-a e)}\\ &=-\frac{1}{3 (b d-a e) (a+b x)^3 \sqrt{d+e x}}+\frac{7 e}{12 (b d-a e)^2 (a+b x)^2 \sqrt{d+e x}}+\frac{\left (35 e^2\right ) \int \frac{1}{(a+b x)^2 (d+e x)^{3/2}} \, dx}{24 (b d-a e)^2}\\ &=-\frac{1}{3 (b d-a e) (a+b x)^3 \sqrt{d+e x}}+\frac{7 e}{12 (b d-a e)^2 (a+b x)^2 \sqrt{d+e x}}-\frac{35 e^2}{24 (b d-a e)^3 (a+b x) \sqrt{d+e x}}-\frac{\left (35 e^3\right ) \int \frac{1}{(a+b x) (d+e x)^{3/2}} \, dx}{16 (b d-a e)^3}\\ &=-\frac{35 e^3}{8 (b d-a e)^4 \sqrt{d+e x}}-\frac{1}{3 (b d-a e) (a+b x)^3 \sqrt{d+e x}}+\frac{7 e}{12 (b d-a e)^2 (a+b x)^2 \sqrt{d+e x}}-\frac{35 e^2}{24 (b d-a e)^3 (a+b x) \sqrt{d+e x}}-\frac{\left (35 b e^3\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{16 (b d-a e)^4}\\ &=-\frac{35 e^3}{8 (b d-a e)^4 \sqrt{d+e x}}-\frac{1}{3 (b d-a e) (a+b x)^3 \sqrt{d+e x}}+\frac{7 e}{12 (b d-a e)^2 (a+b x)^2 \sqrt{d+e x}}-\frac{35 e^2}{24 (b d-a e)^3 (a+b x) \sqrt{d+e x}}-\frac{\left (35 b e^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{8 (b d-a e)^4}\\ &=-\frac{35 e^3}{8 (b d-a e)^4 \sqrt{d+e x}}-\frac{1}{3 (b d-a e) (a+b x)^3 \sqrt{d+e x}}+\frac{7 e}{12 (b d-a e)^2 (a+b x)^2 \sqrt{d+e x}}-\frac{35 e^2}{24 (b d-a e)^3 (a+b x) \sqrt{d+e x}}+\frac{35 \sqrt{b} e^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 (b d-a e)^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0147443, size = 50, normalized size = 0.29 $-\frac{2 e^3 \, _2F_1\left (-\frac{1}{2},4;\frac{1}{2};-\frac{b (d+e x)}{a e-b d}\right )}{\sqrt{d+e x} (a e-b d)^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*e^3*Hypergeometric2F1[-1/2, 4, 1/2, -((b*(d + e*x))/(-(b*d) + a*e))])/((-(b*d) + a*e)^4*Sqrt[d + e*x])

________________________________________________________________________________________

Maple [B]  time = 0.21, size = 292, normalized size = 1.7 \begin{align*} -2\,{\frac{{e}^{3}}{ \left ( ae-bd \right ) ^{4}\sqrt{ex+d}}}-{\frac{19\,{b}^{3}{e}^{3}}{8\, \left ( ae-bd \right ) ^{4} \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{17\,{e}^{4}{b}^{2}a}{3\, \left ( ae-bd \right ) ^{4} \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{17\,{b}^{3}{e}^{3}d}{3\, \left ( ae-bd \right ) ^{4} \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{29\,{e}^{5}b{a}^{2}}{8\, \left ( ae-bd \right ) ^{4} \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}+{\frac{29\,{e}^{4}{b}^{2}ad}{4\, \left ( ae-bd \right ) ^{4} \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}-{\frac{29\,{b}^{3}{e}^{3}{d}^{2}}{8\, \left ( ae-bd \right ) ^{4} \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}-{\frac{35\,{e}^{3}b}{8\, \left ( ae-bd \right ) ^{4}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*e^3/(a*e-b*d)^4/(e*x+d)^(1/2)-19/8*e^3*b^3/(a*e-b*d)^4/(b*e*x+a*e)^3*(e*x+d)^(5/2)-17/3*e^4*b^2/(a*e-b*d)^4
/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a+17/3*e^3*b^3/(a*e-b*d)^4/(b*e*x+a*e)^3*(e*x+d)^(3/2)*d-29/8*e^5*b/(a*e-b*d)^4/(
b*e*x+a*e)^3*(e*x+d)^(1/2)*a^2+29/4*e^4*b^2/(a*e-b*d)^4/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a*d-29/8*e^3*b^3/(a*e-b*d)
^4/(b*e*x+a*e)^3*(e*x+d)^(1/2)*d^2-35/8*e^3*b/(a*e-b*d)^4/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d
)*b)^(1/2))

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.46893, size = 2441, normalized size = 14.11 \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)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[1/48*(105*(b^3*e^4*x^4 + a^3*d*e^3 + (b^3*d*e^3 + 3*a*b^2*e^4)*x^3 + 3*(a*b^2*d*e^3 + a^2*b*e^4)*x^2 + (3*a^2
*b*d*e^3 + a^3*e^4)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d
- a*e)))/(b*x + a)) - 2*(105*b^3*e^3*x^3 + 8*b^3*d^3 - 38*a*b^2*d^2*e + 87*a^2*b*d*e^2 + 48*a^3*e^3 + 35*(b^3*
d*e^2 + 8*a*b^2*e^3)*x^2 - 7*(2*b^3*d^2*e - 14*a*b^2*d*e^2 - 33*a^2*b*e^3)*x)*sqrt(e*x + d))/(a^3*b^4*d^5 - 4*
a^4*b^3*d^4*e + 6*a^5*b^2*d^3*e^2 - 4*a^6*b*d^2*e^3 + a^7*d*e^4 + (b^7*d^4*e - 4*a*b^6*d^3*e^2 + 6*a^2*b^5*d^2
*e^3 - 4*a^3*b^4*d*e^4 + a^4*b^3*e^5)*x^4 + (b^7*d^5 - a*b^6*d^4*e - 6*a^2*b^5*d^3*e^2 + 14*a^3*b^4*d^2*e^3 -
11*a^4*b^3*d*e^4 + 3*a^5*b^2*e^5)*x^3 + 3*(a*b^6*d^5 - 3*a^2*b^5*d^4*e + 2*a^3*b^4*d^3*e^2 + 2*a^4*b^3*d^2*e^3
- 3*a^5*b^2*d*e^4 + a^6*b*e^5)*x^2 + (3*a^2*b^5*d^5 - 11*a^3*b^4*d^4*e + 14*a^4*b^3*d^3*e^2 - 6*a^5*b^2*d^2*e
^3 - a^6*b*d*e^4 + a^7*e^5)*x), 1/24*(105*(b^3*e^4*x^4 + a^3*d*e^3 + (b^3*d*e^3 + 3*a*b^2*e^4)*x^3 + 3*(a*b^2*
d*e^3 + a^2*b*e^4)*x^2 + (3*a^2*b*d*e^3 + a^3*e^4)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*s
qrt(-b/(b*d - a*e))/(b*e*x + b*d)) - (105*b^3*e^3*x^3 + 8*b^3*d^3 - 38*a*b^2*d^2*e + 87*a^2*b*d*e^2 + 48*a^3*e
^3 + 35*(b^3*d*e^2 + 8*a*b^2*e^3)*x^2 - 7*(2*b^3*d^2*e - 14*a*b^2*d*e^2 - 33*a^2*b*e^3)*x)*sqrt(e*x + d))/(a^3
*b^4*d^5 - 4*a^4*b^3*d^4*e + 6*a^5*b^2*d^3*e^2 - 4*a^6*b*d^2*e^3 + a^7*d*e^4 + (b^7*d^4*e - 4*a*b^6*d^3*e^2 +
6*a^2*b^5*d^2*e^3 - 4*a^3*b^4*d*e^4 + a^4*b^3*e^5)*x^4 + (b^7*d^5 - a*b^6*d^4*e - 6*a^2*b^5*d^3*e^2 + 14*a^3*b
^4*d^2*e^3 - 11*a^4*b^3*d*e^4 + 3*a^5*b^2*e^5)*x^3 + 3*(a*b^6*d^5 - 3*a^2*b^5*d^4*e + 2*a^3*b^4*d^3*e^2 + 2*a^
4*b^3*d^2*e^3 - 3*a^5*b^2*d*e^4 + a^6*b*e^5)*x^2 + (3*a^2*b^5*d^5 - 11*a^3*b^4*d^4*e + 14*a^4*b^3*d^3*e^2 - 6*
a^5*b^2*d^2*e^3 - a^6*b*d*e^4 + a^7*e^5)*x)]

________________________________________________________________________________________

Sympy [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)**(3/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.22019, size = 437, normalized size = 2.53 \begin{align*} -\frac{35 \, b \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{3}}{8 \,{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt{-b^{2} d + a b e}} - \frac{2 \, e^{3}}{{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt{x e + d}} - \frac{57 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{3} e^{3} - 136 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} d e^{3} + 87 \, \sqrt{x e + d} b^{3} d^{2} e^{3} + 136 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{2} e^{4} - 174 \, \sqrt{x e + d} a b^{2} d e^{4} + 87 \, \sqrt{x e + d} a^{2} b e^{5}}{24 \,{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-35/8*b*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^3/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3
*b*d*e^3 + a^4*e^4)*sqrt(-b^2*d + a*b*e)) - 2*e^3/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^
3 + a^4*e^4)*sqrt(x*e + d)) - 1/24*(57*(x*e + d)^(5/2)*b^3*e^3 - 136*(x*e + d)^(3/2)*b^3*d*e^3 + 87*sqrt(x*e +
d)*b^3*d^2*e^3 + 136*(x*e + d)^(3/2)*a*b^2*e^4 - 174*sqrt(x*e + d)*a*b^2*d*e^4 + 87*sqrt(x*e + d)*a^2*b*e^5)/
((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*((x*e + d)*b - b*d + a*e)^3)