### 3.1647 $$\int \frac{(d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx$$

Optimal. Leaf size=110 $\frac{5 e \sqrt{d+e x} (b d-a e)}{b^3}-\frac{5 e (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2}}-\frac{(d+e x)^{5/2}}{b (a+b x)}+\frac{5 e (d+e x)^{3/2}}{3 b^2}$

[Out]

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

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Rubi [A]  time = 0.0575072, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.179, Rules used = {27, 47, 50, 63, 208} $\frac{5 e \sqrt{d+e x} (b d-a e)}{b^3}-\frac{5 e (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2}}-\frac{(d+e x)^{5/2}}{b (a+b x)}+\frac{5 e (d+e x)^{3/2}}{3 b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 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 50

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

Mathematica [C]  time = 0.0166566, size = 50, normalized size = 0.45 $\frac{2 e (d+e x)^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};-\frac{b (d+e x)}{a e-b d}\right )}{7 (a e-b d)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*e*(d + e*x)^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(7*(-(b*d) + a*e)^2)

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Maple [B]  time = 0.202, size = 258, normalized size = 2.4 \begin{align*}{\frac{2\,e}{3\,{b}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-4\,{\frac{a{e}^{2}\sqrt{ex+d}}{{b}^{3}}}+4\,{\frac{e\sqrt{ex+d}d}{{b}^{2}}}-{\frac{{a}^{2}{e}^{3}}{{b}^{3} \left ( bxe+ae \right ) }\sqrt{ex+d}}+2\,{\frac{\sqrt{ex+d}ad{e}^{2}}{{b}^{2} \left ( bxe+ae \right ) }}-{\frac{e{d}^{2}}{b \left ( bxe+ae \right ) }\sqrt{ex+d}}+5\,{\frac{{a}^{2}{e}^{3}}{{b}^{3}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }-10\,{\frac{ad{e}^{2}}{{b}^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+5\,{\frac{e{d}^{2}}{b\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \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),x)

[Out]

2/3*e*(e*x+d)^(3/2)/b^2-4/b^3*a*e^2*(e*x+d)^(1/2)+4*e/b^2*(e*x+d)^(1/2)*d-1/b^3*(e*x+d)^(1/2)/(b*e*x+a*e)*a^2*
e^3+2/b^2*(e*x+d)^(1/2)/(b*e*x+a*e)*a*d*e^2-e/b*(e*x+d)^(1/2)/(b*e*x+a*e)*d^2+5/b^3/((a*e-b*d)*b)^(1/2)*arctan
(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^2*e^3-10/b^2/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^
(1/2))*a*d*e^2+5*e/b/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*d^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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.84785, size = 707, normalized size = 6.43 \begin{align*} \left [-\frac{15 \,{\left (a b d e - a^{2} e^{2} +{\left (b^{2} d e - a b e^{2}\right )} x\right )} \sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e + 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) - 2 \,{\left (2 \, b^{2} e^{2} x^{2} - 3 \, b^{2} d^{2} + 20 \, a b d e - 15 \, a^{2} e^{2} + 2 \,{\left (7 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}}{6 \,{\left (b^{4} x + a b^{3}\right )}}, -\frac{15 \,{\left (a b d e - a^{2} e^{2} +{\left (b^{2} d e - a b e^{2}\right )} x\right )} \sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) -{\left (2 \, b^{2} e^{2} x^{2} - 3 \, b^{2} d^{2} + 20 \, a b d e - 15 \, a^{2} e^{2} + 2 \,{\left (7 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}}{3 \,{\left (b^{4} x + a b^{3}\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),x, algorithm="fricas")

[Out]

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

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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),x)

[Out]

Timed out

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Giac [B]  time = 1.18858, size = 258, normalized size = 2.35 \begin{align*} \frac{5 \,{\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{\sqrt{-b^{2} d + a b e} b^{3}} - \frac{\sqrt{x e + d} b^{2} d^{2} e - 2 \, \sqrt{x e + d} a b d e^{2} + \sqrt{x e + d} a^{2} e^{3}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{3}} + \frac{2 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} b^{4} e + 6 \, \sqrt{x e + d} b^{4} d e - 6 \, \sqrt{x e + d} a b^{3} e^{2}\right )}}{3 \, b^{6}} \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),x, algorithm="giac")

[Out]

5*(b^2*d^2*e - 2*a*b*d*e^2 + a^2*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^3)
- (sqrt(x*e + d)*b^2*d^2*e - 2*sqrt(x*e + d)*a*b*d*e^2 + sqrt(x*e + d)*a^2*e^3)/(((x*e + d)*b - b*d + a*e)*b^3
) + 2/3*((x*e + d)^(3/2)*b^4*e + 6*sqrt(x*e + d)*b^4*d*e - 6*sqrt(x*e + d)*a*b^3*e^2)/b^6