3.2144 \(\int \frac{a+b x}{(d+e x)^{5/2} (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=328 \[ -\frac{105 b e^3 (a+b x)}{8 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^5}-\frac{35 e^3 (a+b x)}{8 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^4}-\frac{21 e^2}{8 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^3}+\frac{105 b^{3/2} e^3 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{11/2}}+\frac{3 e}{4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac{1}{3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)} \]

[Out]

(-21*e^2)/(8*(b*d - a*e)^3*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(3*(b*d - a*e)*(a + b*x)^2*(d +
e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*e)/(4*(b*d - a*e)^2*(a + b*x)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*
x + b^2*x^2]) - (35*e^3*(a + b*x))/(8*(b*d - a*e)^4*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (105*b*e^
3*(a + b*x))/(8*(b*d - a*e)^5*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (105*b^(3/2)*e^3*(a + b*x)*ArcTan
h[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*(b*d - a*e)^(11/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.243993, antiderivative size = 328, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{105 b e^3 (a+b x)}{8 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^5}-\frac{35 e^3 (a+b x)}{8 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^4}-\frac{21 e^2}{8 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^3}+\frac{105 b^{3/2} e^3 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{11/2}}+\frac{3 e}{4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac{1}{3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-21*e^2)/(8*(b*d - a*e)^3*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(3*(b*d - a*e)*(a + b*x)^2*(d +
e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*e)/(4*(b*d - a*e)^2*(a + b*x)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*
x + b^2*x^2]) - (35*e^3*(a + b*x))/(8*(b*d - a*e)^4*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (105*b*e^
3*(a + b*x))/(8*(b*d - a*e)^5*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (105*b^(3/2)*e^3*(a + b*x)*ArcTan
h[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*(b*d - a*e)^(11/2)*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}{(d+e x)^{5/2} \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 (d+e x)^{5/2}} \, 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 (d+e x)^{5/2}} \, dx}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{1}{3 (b d-a e) (a+b x)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (3 e \left (a b+b^2 x\right )\right ) \int \frac{1}{(a+b x)^3 (d+e x)^{5/2}} \, dx}{2 b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{1}{3 (b d-a e) (a+b x)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (21 e^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{(a+b x)^2 (d+e x)^{5/2}} \, dx}{8 b (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{21 e^2}{8 (b d-a e)^3 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{3 (b d-a e) (a+b x)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (105 e^3 \left (a b+b^2 x\right )\right ) \int \frac{1}{(a+b x) (d+e x)^{5/2}} \, dx}{16 b (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{21 e^2}{8 (b d-a e)^3 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{3 (b d-a e) (a+b x)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 e^3 (a+b x)}{8 (b d-a e)^4 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (105 e^3 \left (a b+b^2 x\right )\right ) \int \frac{1}{(a+b x) (d+e x)^{3/2}} \, dx}{16 (b d-a e)^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{21 e^2}{8 (b d-a e)^3 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{3 (b d-a e) (a+b x)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 e^3 (a+b x)}{8 (b d-a e)^4 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{105 b e^3 (a+b x)}{8 (b d-a e)^5 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (105 b e^3 \left (a b+b^2 x\right )\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{16 (b d-a e)^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{21 e^2}{8 (b d-a e)^3 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{3 (b d-a e) (a+b x)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 e^3 (a+b x)}{8 (b d-a e)^4 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{105 b e^3 (a+b x)}{8 (b d-a e)^5 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (105 b 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 d-a e)^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{21 e^2}{8 (b d-a e)^3 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{3 (b d-a e) (a+b x)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 e^3 (a+b x)}{8 (b d-a e)^4 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{105 b e^3 (a+b x)}{8 (b d-a e)^5 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{105 b^{3/2} e^3 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 (b d-a e)^{11/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.034832, size = 68, normalized size = 0.21 \[ -\frac{2 e^3 (a+b x) \, _2F_1\left (-\frac{3}{2},4;-\frac{1}{2};-\frac{b (d+e x)}{a e-b d}\right )}{3 \sqrt{(a+b x)^2} (d+e x)^{3/2} (a e-b d)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.023, size = 563, normalized size = 1.7 \begin{align*}{\frac{ \left ( bx+a \right ) ^{2}}{24\, \left ( ae-bd \right ) ^{5}} \left ( 315\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{3/2}{x}^{3}{b}^{5}{e}^{3}+945\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{3/2}{x}^{2}a{b}^{4}{e}^{3}+315\,\sqrt{ \left ( ae-bd \right ) b}{x}^{4}{b}^{4}{e}^{4}+945\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{3/2}x{a}^{2}{b}^{3}{e}^{3}+840\,\sqrt{ \left ( ae-bd \right ) b}{x}^{3}a{b}^{3}{e}^{4}+420\,\sqrt{ \left ( ae-bd \right ) b}{x}^{3}{b}^{4}d{e}^{3}+315\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{3/2}{a}^{3}{b}^{2}{e}^{3}+693\,\sqrt{ \left ( ae-bd \right ) b}{x}^{2}{a}^{2}{b}^{2}{e}^{4}+1134\,\sqrt{ \left ( ae-bd \right ) b}{x}^{2}a{b}^{3}d{e}^{3}+63\,\sqrt{ \left ( ae-bd \right ) b}{x}^{2}{b}^{4}{d}^{2}{e}^{2}+144\,\sqrt{ \left ( ae-bd \right ) b}x{a}^{3}b{e}^{4}+954\,\sqrt{ \left ( ae-bd \right ) b}x{a}^{2}{b}^{2}d{e}^{3}+180\,\sqrt{ \left ( ae-bd \right ) b}xa{b}^{3}{d}^{2}{e}^{2}-18\,\sqrt{ \left ( ae-bd \right ) b}x{b}^{4}{d}^{3}e-16\,\sqrt{ \left ( ae-bd \right ) b}{a}^{4}{e}^{4}+208\,\sqrt{ \left ( ae-bd \right ) b}{a}^{3}bd{e}^{3}+165\,\sqrt{ \left ( ae-bd \right ) b}{a}^{2}{b}^{2}{d}^{2}{e}^{2}-50\,\sqrt{ \left ( ae-bd \right ) b}a{b}^{3}{d}^{3}e+8\,\sqrt{ \left ( ae-bd \right ) b}{b}^{4}{d}^{4} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( ex+d \right ) ^{-{\frac{3}{2}}} \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)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/24*(315*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*(e*x+d)^(3/2)*x^3*b^5*e^3+945*arctan((e*x+d)^(1/2)*b/((a
*e-b*d)*b)^(1/2))*(e*x+d)^(3/2)*x^2*a*b^4*e^3+315*((a*e-b*d)*b)^(1/2)*x^4*b^4*e^4+945*arctan((e*x+d)^(1/2)*b/(
(a*e-b*d)*b)^(1/2))*(e*x+d)^(3/2)*x*a^2*b^3*e^3+840*((a*e-b*d)*b)^(1/2)*x^3*a*b^3*e^4+420*((a*e-b*d)*b)^(1/2)*
x^3*b^4*d*e^3+315*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*(e*x+d)^(3/2)*a^3*b^2*e^3+693*((a*e-b*d)*b)^(1/2
)*x^2*a^2*b^2*e^4+1134*((a*e-b*d)*b)^(1/2)*x^2*a*b^3*d*e^3+63*((a*e-b*d)*b)^(1/2)*x^2*b^4*d^2*e^2+144*((a*e-b*
d)*b)^(1/2)*x*a^3*b*e^4+954*((a*e-b*d)*b)^(1/2)*x*a^2*b^2*d*e^3+180*((a*e-b*d)*b)^(1/2)*x*a*b^3*d^2*e^2-18*((a
*e-b*d)*b)^(1/2)*x*b^4*d^3*e-16*((a*e-b*d)*b)^(1/2)*a^4*e^4+208*((a*e-b*d)*b)^(1/2)*a^3*b*d*e^3+165*((a*e-b*d)
*b)^(1/2)*a^2*b^2*d^2*e^2-50*((a*e-b*d)*b)^(1/2)*a*b^3*d^3*e+8*((a*e-b*d)*b)^(1/2)*b^4*d^4)*(b*x+a)^2/((a*e-b*
d)*b)^(1/2)/(e*x+d)^(3/2)/(a*e-b*d)^5/((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}}{\left (e x + d\right )}^{\frac{5}{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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.28824, size = 3717, normalized size = 11.33 \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)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/48*(315*(b^4*e^5*x^5 + a^3*b*d^2*e^3 + (2*b^4*d*e^4 + 3*a*b^3*e^5)*x^4 + (b^4*d^2*e^3 + 6*a*b^3*d*e^4 + 3*
a^2*b^2*e^5)*x^3 + (3*a*b^3*d^2*e^3 + 6*a^2*b^2*d*e^4 + a^3*b*e^5)*x^2 + (3*a^2*b^2*d^2*e^3 + 2*a^3*b*d*e^4)*x
)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e - 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) +
2*(315*b^4*e^4*x^4 + 8*b^4*d^4 - 50*a*b^3*d^3*e + 165*a^2*b^2*d^2*e^2 + 208*a^3*b*d*e^3 - 16*a^4*e^4 + 420*(b
^4*d*e^3 + 2*a*b^3*e^4)*x^3 + 63*(b^4*d^2*e^2 + 18*a*b^3*d*e^3 + 11*a^2*b^2*e^4)*x^2 - 18*(b^4*d^3*e - 10*a*b^
3*d^2*e^2 - 53*a^2*b^2*d*e^3 - 8*a^3*b*e^4)*x)*sqrt(e*x + d))/(a^3*b^5*d^7 - 5*a^4*b^4*d^6*e + 10*a^5*b^3*d^5*
e^2 - 10*a^6*b^2*d^4*e^3 + 5*a^7*b*d^3*e^4 - a^8*d^2*e^5 + (b^8*d^5*e^2 - 5*a*b^7*d^4*e^3 + 10*a^2*b^6*d^3*e^4
- 10*a^3*b^5*d^2*e^5 + 5*a^4*b^4*d*e^6 - a^5*b^3*e^7)*x^5 + (2*b^8*d^6*e - 7*a*b^7*d^5*e^2 + 5*a^2*b^6*d^4*e^
3 + 10*a^3*b^5*d^3*e^4 - 20*a^4*b^4*d^2*e^5 + 13*a^5*b^3*d*e^6 - 3*a^6*b^2*e^7)*x^4 + (b^8*d^7 + a*b^7*d^6*e -
17*a^2*b^6*d^5*e^2 + 35*a^3*b^5*d^4*e^3 - 25*a^4*b^4*d^3*e^4 - a^5*b^3*d^2*e^5 + 9*a^6*b^2*d*e^6 - 3*a^7*b*e^
7)*x^3 + (3*a*b^7*d^7 - 9*a^2*b^6*d^6*e + a^3*b^5*d^5*e^2 + 25*a^4*b^4*d^4*e^3 - 35*a^5*b^3*d^3*e^4 + 17*a^6*b
^2*d^2*e^5 - a^7*b*d*e^6 - a^8*e^7)*x^2 + (3*a^2*b^6*d^7 - 13*a^3*b^5*d^6*e + 20*a^4*b^4*d^5*e^2 - 10*a^5*b^3*
d^4*e^3 - 5*a^6*b^2*d^3*e^4 + 7*a^7*b*d^2*e^5 - 2*a^8*d*e^6)*x), 1/24*(315*(b^4*e^5*x^5 + a^3*b*d^2*e^3 + (2*b
^4*d*e^4 + 3*a*b^3*e^5)*x^4 + (b^4*d^2*e^3 + 6*a*b^3*d*e^4 + 3*a^2*b^2*e^5)*x^3 + (3*a*b^3*d^2*e^3 + 6*a^2*b^2
*d*e^4 + a^3*b*e^5)*x^2 + (3*a^2*b^2*d^2*e^3 + 2*a^3*b*d*e^4)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt
(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) - (315*b^4*e^4*x^4 + 8*b^4*d^4 - 50*a*b^3*d^3*e + 165*a^2*b^2*d^
2*e^2 + 208*a^3*b*d*e^3 - 16*a^4*e^4 + 420*(b^4*d*e^3 + 2*a*b^3*e^4)*x^3 + 63*(b^4*d^2*e^2 + 18*a*b^3*d*e^3 +
11*a^2*b^2*e^4)*x^2 - 18*(b^4*d^3*e - 10*a*b^3*d^2*e^2 - 53*a^2*b^2*d*e^3 - 8*a^3*b*e^4)*x)*sqrt(e*x + d))/(a^
3*b^5*d^7 - 5*a^4*b^4*d^6*e + 10*a^5*b^3*d^5*e^2 - 10*a^6*b^2*d^4*e^3 + 5*a^7*b*d^3*e^4 - a^8*d^2*e^5 + (b^8*d
^5*e^2 - 5*a*b^7*d^4*e^3 + 10*a^2*b^6*d^3*e^4 - 10*a^3*b^5*d^2*e^5 + 5*a^4*b^4*d*e^6 - a^5*b^3*e^7)*x^5 + (2*b
^8*d^6*e - 7*a*b^7*d^5*e^2 + 5*a^2*b^6*d^4*e^3 + 10*a^3*b^5*d^3*e^4 - 20*a^4*b^4*d^2*e^5 + 13*a^5*b^3*d*e^6 -
3*a^6*b^2*e^7)*x^4 + (b^8*d^7 + a*b^7*d^6*e - 17*a^2*b^6*d^5*e^2 + 35*a^3*b^5*d^4*e^3 - 25*a^4*b^4*d^3*e^4 - a
^5*b^3*d^2*e^5 + 9*a^6*b^2*d*e^6 - 3*a^7*b*e^7)*x^3 + (3*a*b^7*d^7 - 9*a^2*b^6*d^6*e + a^3*b^5*d^5*e^2 + 25*a^
4*b^4*d^4*e^3 - 35*a^5*b^3*d^3*e^4 + 17*a^6*b^2*d^2*e^5 - a^7*b*d*e^6 - a^8*e^7)*x^2 + (3*a^2*b^6*d^7 - 13*a^3
*b^5*d^6*e + 20*a^4*b^4*d^5*e^2 - 10*a^5*b^3*d^4*e^3 - 5*a^6*b^2*d^3*e^4 + 7*a^7*b*d^2*e^5 - 2*a^8*d*e^6)*x)]

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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)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.3224, size = 933, normalized size = 2.84 \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)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

-105/8*b^2*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^3/((b^5*d^5*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 5*a
*b^4*d^4*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 10*a^2*b^3*d^3*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 10*a^3
*b^2*d^2*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 5*a^4*b*d*e^4*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a^5*e^5*s
gn((x*e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) - 1/24*(315*(x*e + d)^4*b^4*e^3 - 840*(x*e + d)^3*b^4
*d*e^3 + 693*(x*e + d)^2*b^4*d^2*e^3 - 144*(x*e + d)*b^4*d^3*e^3 - 16*b^4*d^4*e^3 + 840*(x*e + d)^3*a*b^3*e^4
- 1386*(x*e + d)^2*a*b^3*d*e^4 + 432*(x*e + d)*a*b^3*d^2*e^4 + 64*a*b^3*d^3*e^4 + 693*(x*e + d)^2*a^2*b^2*e^5
- 432*(x*e + d)*a^2*b^2*d*e^5 - 96*a^2*b^2*d^2*e^5 + 144*(x*e + d)*a^3*b*e^6 + 64*a^3*b*d*e^6 - 16*a^4*e^7)/((
b^5*d^5*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 5*a*b^4*d^4*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 10*a^2*b^3*d^3
*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 10*a^3*b^2*d^2*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 5*a^4*b*d*e^
4*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a^5*e^5*sgn((x*e + d)*b*e - b*d*e + a*e^2))*((x*e + d)^(3/2)*b - sqrt(x
*e + d)*b*d + sqrt(x*e + d)*a*e)^3)