3.2142 \(\int \frac{a+b x}{\sqrt{d+e x} (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=225 \[ -\frac{5 e^2 \sqrt{d+e x}}{8 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac{5 e^3 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 \sqrt{b} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}}+\frac{5 e \sqrt{d+e x}}{12 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{\sqrt{d+e x}}{3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.141333, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {770, 21, 51, 63, 208} \[ -\frac{5 e^2 \sqrt{d+e x}}{8 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac{5 e^3 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 \sqrt{b} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}}+\frac{5 e \sqrt{d+e x}}{12 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{\sqrt{d+e x}}{3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 51

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

Mathematica [C]  time = 0.0247045, size = 66, normalized size = 0.29 \[ \frac{2 e^3 (a+b x) \sqrt{d+e x} \, _2F_1\left (\frac{1}{2},4;\frac{3}{2};-\frac{b (d+e x)}{a e-b d}\right )}{\sqrt{(a+b x)^2} (a e-b d)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^3*(a + b*x)*Sqrt[d + e*x]*Hypergeometric2F1[1/2, 4, 3/2, -((b*(d + e*x))/(-(b*d) + a*e))])/((-(b*d) + a*e
)^4*Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.013, size = 334, normalized size = 1.5 \begin{align*}{\frac{ \left ( bx+a \right ) ^{2}}{24\, \left ( ae-bd \right ) ^{3}} \left ( 15\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{3}{b}^{3}{e}^{3}+45\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}a{b}^{2}{e}^{3}+15\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{x}^{2}{b}^{2}{e}^{2}+45\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{2}b{e}^{3}+40\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}xab{e}^{2}-10\,\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}+33\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}{e}^{2}-26\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}abde+8\,\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{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)/(b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^(1/2),x)

[Out]

1/24*(15*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^3*b^3*e^3+45*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2)
)*x^2*a*b^2*e^3+15*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^2*b^2*e^2+45*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2)
)*x*a^2*b*e^3+40*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*a*b*e^2-10*((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+33*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*e^2-26*((a*e-b*d
)*b)^(1/2)*(e*x+d)^(1/2)*a*b*d*e+8*((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)/(a
*e-b*d)^3/((b*x+a)^2)^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.12744, size = 1805, normalized size = 8.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^(1/2),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 - 34*a*b^3*d^2*e + 59*a^2*b^2*d*e^2 - 33*
a^3*b*e^3 + 15*(b^4*d*e^2 - a*b^3*e^3)*x^2 - 10*(b^4*d^2*e - 5*a*b^3*d*e^2 + 4*a^2*b^2*e^3)*x)*sqrt(e*x + d))/
(a^3*b^5*d^4 - 4*a^4*b^4*d^3*e + 6*a^5*b^3*d^2*e^2 - 4*a^6*b^2*d*e^3 + a^7*b*e^4 + (b^8*d^4 - 4*a*b^7*d^3*e +
6*a^2*b^6*d^2*e^2 - 4*a^3*b^5*d*e^3 + a^4*b^4*e^4)*x^3 + 3*(a*b^7*d^4 - 4*a^2*b^6*d^3*e + 6*a^3*b^5*d^2*e^2 -
4*a^4*b^4*d*e^3 + a^5*b^3*e^4)*x^2 + 3*(a^2*b^6*d^4 - 4*a^3*b^5*d^3*e + 6*a^4*b^4*d^2*e^2 - 4*a^5*b^3*d*e^3 +
a^6*b^2*e^4)*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)*arct
an(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (8*b^4*d^3 - 34*a*b^3*d^2*e + 59*a^2*b^2*d*e^2 - 33*a^3
*b*e^3 + 15*(b^4*d*e^2 - a*b^3*e^3)*x^2 - 10*(b^4*d^2*e - 5*a*b^3*d*e^2 + 4*a^2*b^2*e^3)*x)*sqrt(e*x + d))/(a^
3*b^5*d^4 - 4*a^4*b^4*d^3*e + 6*a^5*b^3*d^2*e^2 - 4*a^6*b^2*d*e^3 + a^7*b*e^4 + (b^8*d^4 - 4*a*b^7*d^3*e + 6*a
^2*b^6*d^2*e^2 - 4*a^3*b^5*d*e^3 + a^4*b^4*e^4)*x^3 + 3*(a*b^7*d^4 - 4*a^2*b^6*d^3*e + 6*a^3*b^5*d^2*e^2 - 4*a
^4*b^4*d*e^3 + a^5*b^3*e^4)*x^2 + 3*(a^2*b^6*d^4 - 4*a^3*b^5*d^3*e + 6*a^4*b^4*d^2*e^2 - 4*a^5*b^3*d*e^3 + a^6
*b^2*e^4)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(b**2*x**2+2*a*b*x+a**2)**(5/2)/(e*x+d)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.27293, size = 552, normalized size = 2.45 \begin{align*} -\frac{5 \, \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{3}}{8 \,{\left (b^{3} d^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt{-b^{2} d + a b e}} - \frac{15 \,{\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} + 33 \, \sqrt{x e + d} b^{2} d^{2} e^{3} + 40 \,{\left (x e + d\right )}^{\frac{3}{2}} a b e^{4} - 66 \, \sqrt{x e + d} a b d e^{4} + 33 \, \sqrt{x e + d} a^{2} e^{5}}{24 \,{\left (b^{3} d^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^3/((b^3*d^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 3*a*b^2*d
^2*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 3*a^2*b*d*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a^3*e^3*sgn((x*e
+ d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) - 1/24*(15*(x*e + d)^(5/2)*b^2*e^3 - 40*(x*e + d)^(3/2)*b^2*d
*e^3 + 33*sqrt(x*e + d)*b^2*d^2*e^3 + 40*(x*e + d)^(3/2)*a*b*e^4 - 66*sqrt(x*e + d)*a*b*d*e^4 + 33*sqrt(x*e +
d)*a^2*e^5)/((b^3*d^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 3*a*b^2*d^2*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) +
3*a^2*b*d*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a^3*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2))*((x*e + d)*b -
b*d + a*e)^3)