### 3.1657 $$\int \frac{(d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx$$

Optimal. Leaf size=126 $-\frac{5 e^2 \sqrt{d+e x}}{8 b^3 (a+b x)}-\frac{5 e^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{7/2} \sqrt{b d-a e}}-\frac{5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{5/2}}{3 b (a+b x)^3}$

[Out]

(-5*e^2*Sqrt[d + e*x])/(8*b^3*(a + b*x)) - (5*e*(d + e*x)^(3/2))/(12*b^2*(a + b*x)^2) - (d + e*x)^(5/2)/(3*b*(
a + b*x)^3) - (5*e^3*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(7/2)*Sqrt[b*d - a*e])

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Rubi [A]  time = 0.0593683, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {27, 47, 63, 208} $-\frac{5 e^2 \sqrt{d+e x}}{8 b^3 (a+b x)}-\frac{5 e^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{7/2} \sqrt{b d-a e}}-\frac{5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{5/2}}{3 b (a+b x)^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(5/2)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(-5*e^2*Sqrt[d + e*x])/(8*b^3*(a + b*x)) - (5*e*(d + e*x)^(3/2))/(12*b^2*(a + b*x)^2) - (d + e*x)^(5/2)/(3*b*(
a + b*x)^3) - (5*e^3*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(7/2)*Sqrt[b*d - a*e])

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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{(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{(d+e x)^{5/2}}{(a+b x)^4} \, dx\\ &=-\frac{(d+e x)^{5/2}}{3 b (a+b x)^3}+\frac{(5 e) \int \frac{(d+e x)^{3/2}}{(a+b x)^3} \, dx}{6 b}\\ &=-\frac{5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{5/2}}{3 b (a+b x)^3}+\frac{\left (5 e^2\right ) \int \frac{\sqrt{d+e x}}{(a+b x)^2} \, dx}{8 b^2}\\ &=-\frac{5 e^2 \sqrt{d+e x}}{8 b^3 (a+b x)}-\frac{5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{5/2}}{3 b (a+b x)^3}+\frac{\left (5 e^3\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{16 b^3}\\ &=-\frac{5 e^2 \sqrt{d+e x}}{8 b^3 (a+b x)}-\frac{5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{5/2}}{3 b (a+b x)^3}+\frac{\left (5 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^3}\\ &=-\frac{5 e^2 \sqrt{d+e x}}{8 b^3 (a+b x)}-\frac{5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{5/2}}{3 b (a+b x)^3}-\frac{5 e^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{7/2} \sqrt{b d-a e}}\\ \end{align*}

Mathematica [A]  time = 0.144229, size = 119, normalized size = 0.94 $\frac{5 e^3 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{a e-b d}}\right )}{8 b^{7/2} \sqrt{a e-b d}}-\frac{\sqrt{d+e x} \left (15 a^2 e^2+10 a b e (d+4 e x)+b^2 \left (8 d^2+26 d e x+33 e^2 x^2\right )\right )}{24 b^3 (a+b x)^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^(5/2)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

-(Sqrt[d + e*x]*(15*a^2*e^2 + 10*a*b*e*(d + 4*e*x) + b^2*(8*d^2 + 26*d*e*x + 33*e^2*x^2)))/(24*b^3*(a + b*x)^3
) + (5*e^3*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(8*b^(7/2)*Sqrt[-(b*d) + a*e])

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Maple [A]  time = 0.2, size = 204, normalized size = 1.6 \begin{align*} -{\frac{11\,{e}^{3}}{8\, \left ( bxe+ae \right ) ^{3}b} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{e}^{4}a}{3\, \left ( bxe+ae \right ) ^{3}{b}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{e}^{3}d}{3\, \left ( bxe+ae \right ) ^{3}b} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{a}^{2}{e}^{5}}{8\, \left ( bxe+ae \right ) ^{3}{b}^{3}}\sqrt{ex+d}}+{\frac{5\,{e}^{4}ad}{4\, \left ( bxe+ae \right ) ^{3}{b}^{2}}\sqrt{ex+d}}-{\frac{5\,{e}^{3}{d}^{2}}{8\, \left ( bxe+ae \right ) ^{3}b}\sqrt{ex+d}}+{\frac{5\,{e}^{3}}{8\,{b}^{3}}\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((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

-11/8*e^3/(b*e*x+a*e)^3/b*(e*x+d)^(5/2)-5/3*e^4/(b*e*x+a*e)^3/b^2*(e*x+d)^(3/2)*a+5/3*e^3/(b*e*x+a*e)^3/b*(e*x
+d)^(3/2)*d-5/8*e^5/(b*e*x+a*e)^3/b^3*(e*x+d)^(1/2)*a^2+5/4*e^4/(b*e*x+a*e)^3/b^2*(e*x+d)^(1/2)*a*d-5/8*e^3/(b
*e*x+a*e)^3/b*(e*x+d)^(1/2)*d^2+5/8*e^3/b^3/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))

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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((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.02755, size = 1162, normalized size = 9.22 \begin{align*} \left [\frac{15 \,{\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt{b^{2} d - a b e} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{b^{2} d - a b e} \sqrt{e x + d}}{b x + a}\right ) - 2 \,{\left (8 \, b^{4} d^{3} + 2 \, a b^{3} d^{2} e + 5 \, a^{2} b^{2} d e^{2} - 15 \, a^{3} b e^{3} + 33 \,{\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \,{\left (13 \, b^{4} d^{2} e + 7 \, a b^{3} d e^{2} - 20 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt{e x + d}}{48 \,{\left (a^{3} b^{5} d - a^{4} b^{4} e +{\left (b^{8} d - a b^{7} e\right )} x^{3} + 3 \,{\left (a b^{7} d - a^{2} b^{6} e\right )} x^{2} + 3 \,{\left (a^{2} b^{6} d - a^{3} b^{5} e\right )} x\right )}}, \frac{15 \,{\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt{-b^{2} d + a b e} \arctan \left (\frac{\sqrt{-b^{2} d + a b e} \sqrt{e x + d}}{b e x + b d}\right ) -{\left (8 \, b^{4} d^{3} + 2 \, a b^{3} d^{2} e + 5 \, a^{2} b^{2} d e^{2} - 15 \, a^{3} b e^{3} + 33 \,{\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \,{\left (13 \, b^{4} d^{2} e + 7 \, a b^{3} d e^{2} - 20 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt{e x + d}}{24 \,{\left (a^{3} b^{5} d - a^{4} b^{4} e +{\left (b^{8} d - a b^{7} e\right )} x^{3} + 3 \,{\left (a b^{7} d - a^{2} b^{6} e\right )} x^{2} + 3 \,{\left (a^{2} b^{6} d - a^{3} b^{5} e\right )} x\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[1/48*(15*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a
*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(8*b^4*d^3 + 2*a*b^3*d^2*e + 5*a^2*b^2*d*e^2 - 15*a^3
*b*e^3 + 33*(b^4*d*e^2 - a*b^3*e^3)*x^2 + 2*(13*b^4*d^2*e + 7*a*b^3*d*e^2 - 20*a^2*b^2*e^3)*x)*sqrt(e*x + d))/
(a^3*b^5*d - a^4*b^4*e + (b^8*d - a*b^7*e)*x^3 + 3*(a*b^7*d - a^2*b^6*e)*x^2 + 3*(a^2*b^6*d - a^3*b^5*e)*x), 1
/24*(15*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*
b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (8*b^4*d^3 + 2*a*b^3*d^2*e + 5*a^2*b^2*d*e^2 - 15*a^3*b*e^3 + 33*(b^4*d*e^
2 - a*b^3*e^3)*x^2 + 2*(13*b^4*d^2*e + 7*a*b^3*d*e^2 - 20*a^2*b^2*e^3)*x)*sqrt(e*x + d))/(a^3*b^5*d - a^4*b^4*
e + (b^8*d - a*b^7*e)*x^3 + 3*(a*b^7*d - a^2*b^6*e)*x^2 + 3*(a^2*b^6*d - a^3*b^5*e)*x)]

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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((e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.25548, size = 223, normalized size = 1.77 \begin{align*} \frac{5 \, \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{3}}{8 \, \sqrt{-b^{2} d + a b e} b^{3}} - \frac{33 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{2} e^{3} - 40 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{2} d e^{3} + 15 \, \sqrt{x e + d} b^{2} d^{2} e^{3} + 40 \,{\left (x e + d\right )}^{\frac{3}{2}} a b e^{4} - 30 \, \sqrt{x e + d} a b d e^{4} + 15 \, \sqrt{x e + d} a^{2} e^{5}}{24 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

5/8*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^3/(sqrt(-b^2*d + a*b*e)*b^3) - 1/24*(33*(x*e + d)^(5/2)*b^2
*e^3 - 40*(x*e + d)^(3/2)*b^2*d*e^3 + 15*sqrt(x*e + d)*b^2*d^2*e^3 + 40*(x*e + d)^(3/2)*a*b*e^4 - 30*sqrt(x*e
+ d)*a*b*d*e^4 + 15*sqrt(x*e + d)*a^2*e^5)/(((x*e + d)*b - b*d + a*e)^3*b^3)