3.2067 \(\int \frac{(a+b x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=112 \[ \frac{2 \sqrt{d+e x} (b d-a e)^2}{b^3}+\frac{2 (d+e x)^{3/2} (b d-a e)}{3 b^2}-\frac{2 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2}}+\frac{2 (d+e x)^{5/2}}{5 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0585831, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {27, 50, 63, 208} \[ \frac{2 \sqrt{d+e x} (b d-a e)^2}{b^3}+\frac{2 (d+e x)^{3/2} (b d-a e)}{3 b^2}-\frac{2 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2}}+\frac{2 (d+e x)^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*d - a*e)^2*Sqrt[d + e*x])/b^3 + (2*(b*d - a*e)*(d + e*x)^(3/2))/(3*b^2) + (2*(d + e*x)^(5/2))/(5*b) - (2
*(b*d - a*e)^(5/2)*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 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{(a+b x) (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} \, dx\\ &=\frac{2 (d+e x)^{5/2}}{5 b}+\frac{(b d-a e) \int \frac{(d+e x)^{3/2}}{a+b x} \, dx}{b}\\ &=\frac{2 (b d-a e) (d+e x)^{3/2}}{3 b^2}+\frac{2 (d+e x)^{5/2}}{5 b}+\frac{(b d-a e)^2 \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{b^2}\\ &=\frac{2 (b d-a e)^2 \sqrt{d+e x}}{b^3}+\frac{2 (b d-a e) (d+e x)^{3/2}}{3 b^2}+\frac{2 (d+e x)^{5/2}}{5 b}+\frac{(b d-a e)^3 \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{b^3}\\ &=\frac{2 (b d-a e)^2 \sqrt{d+e x}}{b^3}+\frac{2 (b d-a e) (d+e x)^{3/2}}{3 b^2}+\frac{2 (d+e x)^{5/2}}{5 b}+\frac{\left (2 (b d-a e)^3\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 e}\\ &=\frac{2 (b d-a e)^2 \sqrt{d+e x}}{b^3}+\frac{2 (b d-a e) (d+e x)^{3/2}}{3 b^2}+\frac{2 (d+e x)^{5/2}}{5 b}-\frac{2 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.120746, size = 105, normalized size = 0.94 \[ \frac{2 (b d-a e) \left (\sqrt{b} \sqrt{d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )\right )}{3 b^{7/2}}+\frac{2 (d+e x)^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^(5/2))/(5*b) + (2*(b*d - a*e)*(Sqrt[b]*Sqrt[d + e*x]*(4*b*d - 3*a*e + b*e*x) - 3*(b*d - a*e)^(3/2
)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]]))/(3*b^(7/2))

________________________________________________________________________________________

Maple [B]  time = 0.007, size = 263, normalized size = 2.4 \begin{align*}{\frac{2}{5\,b} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{2\,ae}{3\,{b}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{2\,d}{3\,b} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+2\,{\frac{{a}^{2}{e}^{2}\sqrt{ex+d}}{{b}^{3}}}-4\,{\frac{ade\sqrt{ex+d}}{{b}^{2}}}+2\,{\frac{{d}^{2}\sqrt{ex+d}}{b}}-2\,{\frac{{e}^{3}{a}^{3}}{{b}^{3}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+6\,{\frac{d{e}^{2}{a}^{2}}{{b}^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }-6\,{\frac{a{d}^{2}e}{b\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+2\,{\frac{{d}^{3}}{\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(e*x+d)^(5/2)/b-2/3/b^2*(e*x+d)^(3/2)*a*e+2/3/b*(e*x+d)^(3/2)*d+2/b^3*a^2*e^2*(e*x+d)^(1/2)-4/b^2*a*d*e*(e
*x+d)^(1/2)+2/b*d^2*(e*x+d)^(1/2)-2/b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*e^3*a^
3+6/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*d*e^2*a^2-6/b/((a*e-b*d)*b)^(1/2)*arct
an((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a*d^2*e+2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/
2))*d^3

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.05439, size = 644, normalized size = 5.75 \begin{align*} \left [\frac{15 \,{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\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 (3 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 35 \, a b d e + 15 \, a^{2} e^{2} +{\left (11 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}}{15 \, b^{3}}, -\frac{2 \,{\left (15 \,{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\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 (3 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 35 \, a b d e + 15 \, a^{2} e^{2} +{\left (11 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}\right )}}{15 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*(b^2*d^2 - 2*a*b*d*e + a^2*e^2)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqr
t((b*d - a*e)/b))/(b*x + a)) + 2*(3*b^2*e^2*x^2 + 23*b^2*d^2 - 35*a*b*d*e + 15*a^2*e^2 + (11*b^2*d*e - 5*a*b*e
^2)*x)*sqrt(e*x + d))/b^3, -2/15*(15*(b^2*d^2 - 2*a*b*d*e + a^2*e^2)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d
)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (3*b^2*e^2*x^2 + 23*b^2*d^2 - 35*a*b*d*e + 15*a^2*e^2 + (11*b^2*d*e -
5*a*b*e^2)*x)*sqrt(e*x + d))/b^3]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.21615, size = 243, normalized size = 2.17 \begin{align*} \frac{2 \,{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} 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{2 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{4} + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{4} d + 15 \, \sqrt{x e + d} b^{4} d^{2} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{3} e - 30 \, \sqrt{x e + d} a b^{3} d e + 15 \, \sqrt{x e + d} a^{2} b^{2} e^{2}\right )}}{15 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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