3.2131 \(\int \frac{(a+b x) (d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=198 \[ -\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) \sqrt{d+e x} (b d-a e)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (a+b x) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(5*e*(b*d - a*e)*(a + b*x)*Sqrt[d + e*x])/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*e*(a + b*x)*(d + e*x)^(3/2)
)/(3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (d + e*x)^(5/2)/(b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e*(b*d - a*e)
^(3/2)*(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 = 0.121972, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {768, 646, 50, 63, 208} \[ -\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) \sqrt{d+e x} (b d-a e)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (a+b x) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^(5/2))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(5*e*(b*d - a*e)*(a + b*x)*Sqrt[d + e*x])/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*e*(a + b*x)*(d + e*x)^(3/2)
)/(3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (d + e*x)^(5/2)/(b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e*(b*d - a*e)
^(3/2)*(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 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

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{(a+b x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(5 e) \int \frac{(d+e x)^{3/2}}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx}{2 b}\\ &=-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{a b+b^2 x} \, dx}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{a b+b^2 x} \, dx}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e (b d-a e) (a+b x) \sqrt{d+e x}}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e \left (b^2 d-a b e\right )^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{2 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e (b d-a e) (a+b x) \sqrt{d+e x}}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 \left (b^2 d-a b e\right )^2 \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^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e (b d-a e) (a+b x) \sqrt{d+e x}}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (b d-a e)^{3/2} (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 [C]  time = 0.0325365, size = 66, normalized size = 0.33 \[ \frac{2 e (a+b x) (d+e x)^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};-\frac{b (d+e x)}{a e-b d}\right )}{7 \sqrt{(a+b x)^2} (a e-b d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^(5/2))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(2*e*(a + b*x)*(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*Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.015, size = 409, normalized size = 2.1 \begin{align*}{\frac{ \left ( bx+a \right ) ^{2}}{3\,{b}^{3}} \left ( 2\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}x{b}^{2}e+15\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{2}b{e}^{3}-30\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xa{b}^{2}d{e}^{2}+15\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{b}^{3}{d}^{2}e+2\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}abe-12\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}xab{e}^{2}+12\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}x{b}^{2}de+15\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{3}{e}^{3}-30\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}bd{e}^{2}+15\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) a{b}^{2}{d}^{2}e-15\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}{e}^{2}+18\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}abde-3\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{2}{d}^{2} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \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)^(3/2),x)

[Out]

1/3*(2*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*x*b^2*e+15*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*a^2*b*e^3-30
*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*a*b^2*d*e^2+15*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*b^
3*d^2*e+2*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b*e-12*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*a*b*e^2+12*((a*e-b*d)
*b)^(1/2)*(e*x+d)^(1/2)*x*b^2*d*e+15*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^3*e^3-30*arctan((e*x+d)^(1/
2)*b/((a*e-b*d)*b)^(1/2))*a^2*b*d*e^2+15*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a*b^2*d^2*e-15*((a*e-b*d)
*b)^(1/2)*(e*x+d)^(1/2)*a^2*e^2+18*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b*d*e-3*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/
2)*b^2*d^2)*(b*x+a)^2/((a*e-b*d)*b)^(1/2)/b^3/((b*x+a)^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}{\left (e x + d\right )}^{\frac{5}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}}\,{d x} \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)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(e*x + d)^(5/2)/(b^2*x^2 + 2*a*b*x + a^2)^(3/2), x)

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Fricas [A]  time = 1.01768, size = 707, normalized size = 3.57 \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((b*x+a)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),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((b*x+a)*(e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.21331, size = 363, normalized size = 1.83 \begin{align*} \frac{5 \,{\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{\left (-1\right )}}{\sqrt{-b^{2} d + a b e} b^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac{{\left (\sqrt{x e + d} b^{2} d^{2} e^{2} - 2 \, \sqrt{x e + d} a b d e^{3} + \sqrt{x e + d} a^{2} e^{4}\right )} e^{\left (-1\right )}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac{2 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} b^{4} e^{4} + 6 \, \sqrt{x e + d} b^{4} d e^{4} - 6 \, \sqrt{x e + d} a b^{3} e^{5}\right )} e^{\left (-3\right )}}{3 \, b^{6} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\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)^(3/2),x, algorithm="giac")

[Out]

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