3.1704 \(\int \frac{(d+e x)^{5/2}}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=212 \[ \frac{2 (a+b x) \sqrt{d+e x} (b d-a e)^2}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (d+e x)^{3/2} (b d-a e)}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (a+b x) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.109214, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {646, 50, 63, 208} \[ \frac{2 (a+b x) \sqrt{d+e x} (b d-a e)^2}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (d+e x)^{3/2} (b d-a e)}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (a+b x) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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}}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{(d+e x)^{5/2}}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (\left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{a b+b^2 x} \, dx}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (\left (b^2 d-a b e\right )^2 \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{a b+b^2 x} \, dx}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (b d-a e)^2 (a+b x) \sqrt{d+e x}}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (\left (b^2 d-a b e\right )^3 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{b^6 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (b d-a e)^2 (a+b x) \sqrt{d+e x}}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (b^2 d-a b e\right )^3 \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b-\frac{b^2 d}{e}+\frac{b^2 x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{b^6 e \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (b d-a e)^2 (a+b x) \sqrt{d+e x}}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (b d-a e)^{5/2} (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.105951, size = 127, normalized size = 0.6 \[ \frac{2 (a+b x) \left (\sqrt{b} \sqrt{d+e x} \left (15 a^2 e^2-5 a b e (7 d+e x)+b^2 \left (23 d^2+11 d e x+3 e^2 x^2\right )\right )-15 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )\right )}{15 b^{7/2} \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*(Sqrt[b]*Sqrt[d + e*x]*(15*a^2*e^2 - 5*a*b*e*(7*d + e*x) + b^2*(23*d^2 + 11*d*e*x + 3*e^2*x^2)) -
15*(b*d - a*e)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]]))/(15*b^(7/2)*Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.232, size = 309, normalized size = 1.5 \begin{align*}{\frac{2\,bx+2\,a}{15\,{b}^{3}} \left ( 3\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}{b}^{2}-5\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}abe+5\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{b}^{2}d-15\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{3}{e}^{3}+45\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}bd{e}^{2}-45\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) a{b}^{2}{d}^{2}e+15\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){b}^{3}{d}^{3}+15\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}{e}^{2}-30\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}abde+15\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{2}{d}^{2} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}{\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*x+a)^2)^(1/2),x)

[Out]

2/15*(b*x+a)*(3*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*b^2-5*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b*e+5*((a*e-b*d)*b
)^(1/2)*(e*x+d)^(3/2)*b^2*d-15*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^3*e^3+45*arctan(b*(e*x+d)^(1/2)/(
(a*e-b*d)*b)^(1/2))*a^2*b*d*e^2-45*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*b^2*d^2*e+15*arctan(b*(e*x+d)
^(1/2)/((a*e-b*d)*b)^(1/2))*b^3*d^3+15*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*e^2-30*((a*e-b*d)*b)^(1/2)*(e*x+d
)^(1/2)*a*b*d*e+15*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^2*d^2)/((b*x+a)^2)^(1/2)/b^3/((a*e-b*d)*b)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{5}{2}}}{\sqrt{{\left (b x + a\right )}^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.71745, size = 644, normalized size = 3.04 \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((e*x+d)^(5/2)/((b*x+a)^2)^(1/2),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]

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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*x+a)**2)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.19106, size = 324, normalized size = 1.53 \begin{align*} \frac{2 \,{\left (b^{3} d^{3} \mathrm{sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm{sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm{sgn}\left (b x + a\right )\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} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{4} d \mathrm{sgn}\left (b x + a\right ) + 15 \, \sqrt{x e + d} b^{4} d^{2} \mathrm{sgn}\left (b x + a\right ) - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{3} e \mathrm{sgn}\left (b x + a\right ) - 30 \, \sqrt{x e + d} a b^{3} d e \mathrm{sgn}\left (b x + a\right ) + 15 \, \sqrt{x e + d} a^{2} b^{2} e^{2} \mathrm{sgn}\left (b x + a\right )\right )}}{15 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2*(b^3*d^3*sgn(b*x + a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) - a^3*e^3*sgn(b*x + a))*arct
an(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*sgn(b*x + a)
+ 5*(x*e + d)^(3/2)*b^4*d*sgn(b*x + a) + 15*sqrt(x*e + d)*b^4*d^2*sgn(b*x + a) - 5*(x*e + d)^(3/2)*a*b^3*e*sg
n(b*x + a) - 30*sqrt(x*e + d)*a*b^3*d*e*sgn(b*x + a) + 15*sqrt(x*e + d)*a^2*b^2*e^2*sgn(b*x + a))/b^5