3.2091 \(\int \frac{a+b x}{(d+e x)^{5/2} (a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=233 \[ -\frac{1155 b^{3/2} e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 (b d-a e)^{13/2}}+\frac{1155 b e^4}{64 \sqrt{d+e x} (b d-a e)^6}+\frac{385 e^4}{64 (d+e x)^{3/2} (b d-a e)^5}+\frac{231 e^3}{64 (a+b x) (d+e x)^{3/2} (b d-a e)^4}-\frac{33 e^2}{32 (a+b x)^2 (d+e x)^{3/2} (b d-a e)^3}+\frac{11 e}{24 (a+b x)^3 (d+e x)^{3/2} (b d-a e)^2}-\frac{1}{4 (a+b x)^4 (d+e x)^{3/2} (b d-a e)} \]

[Out]

(385*e^4)/(64*(b*d - a*e)^5*(d + e*x)^(3/2)) - 1/(4*(b*d - a*e)*(a + b*x)^4*(d + e*x)^(3/2)) + (11*e)/(24*(b*d
- a*e)^2*(a + b*x)^3*(d + e*x)^(3/2)) - (33*e^2)/(32*(b*d - a*e)^3*(a + b*x)^2*(d + e*x)^(3/2)) + (231*e^3)/(
64*(b*d - a*e)^4*(a + b*x)*(d + e*x)^(3/2)) + (1155*b*e^4)/(64*(b*d - a*e)^6*Sqrt[d + e*x]) - (1155*b^(3/2)*e^
4*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*(b*d - a*e)^(13/2))

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Rubi [A]  time = 0.20486, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {27, 51, 63, 208} \[ -\frac{1155 b^{3/2} e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 (b d-a e)^{13/2}}+\frac{1155 b e^4}{64 \sqrt{d+e x} (b d-a e)^6}+\frac{385 e^4}{64 (d+e x)^{3/2} (b d-a e)^5}+\frac{231 e^3}{64 (a+b x) (d+e x)^{3/2} (b d-a e)^4}-\frac{33 e^2}{32 (a+b x)^2 (d+e x)^{3/2} (b d-a e)^3}+\frac{11 e}{24 (a+b x)^3 (d+e x)^{3/2} (b d-a e)^2}-\frac{1}{4 (a+b x)^4 (d+e x)^{3/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(385*e^4)/(64*(b*d - a*e)^5*(d + e*x)^(3/2)) - 1/(4*(b*d - a*e)*(a + b*x)^4*(d + e*x)^(3/2)) + (11*e)/(24*(b*d
- a*e)^2*(a + b*x)^3*(d + e*x)^(3/2)) - (33*e^2)/(32*(b*d - a*e)^3*(a + b*x)^2*(d + e*x)^(3/2)) + (231*e^3)/(
64*(b*d - a*e)^4*(a + b*x)*(d + e*x)^(3/2)) + (1155*b*e^4)/(64*(b*d - a*e)^6*Sqrt[d + e*x]) - (1155*b^(3/2)*e^
4*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*(b*d - a*e)^(13/2))

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 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 )^3} \, dx &=\int \frac{1}{(a+b x)^5 (d+e x)^{5/2}} \, dx\\ &=-\frac{1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}-\frac{(11 e) \int \frac{1}{(a+b x)^4 (d+e x)^{5/2}} \, dx}{8 (b d-a e)}\\ &=-\frac{1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac{11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}+\frac{\left (33 e^2\right ) \int \frac{1}{(a+b x)^3 (d+e x)^{5/2}} \, dx}{16 (b d-a e)^2}\\ &=-\frac{1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac{11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac{33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}-\frac{\left (231 e^3\right ) \int \frac{1}{(a+b x)^2 (d+e x)^{5/2}} \, dx}{64 (b d-a e)^3}\\ &=-\frac{1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac{11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac{33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}+\frac{231 e^3}{64 (b d-a e)^4 (a+b x) (d+e x)^{3/2}}+\frac{\left (1155 e^4\right ) \int \frac{1}{(a+b x) (d+e x)^{5/2}} \, dx}{128 (b d-a e)^4}\\ &=\frac{385 e^4}{64 (b d-a e)^5 (d+e x)^{3/2}}-\frac{1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac{11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac{33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}+\frac{231 e^3}{64 (b d-a e)^4 (a+b x) (d+e x)^{3/2}}+\frac{\left (1155 b e^4\right ) \int \frac{1}{(a+b x) (d+e x)^{3/2}} \, dx}{128 (b d-a e)^5}\\ &=\frac{385 e^4}{64 (b d-a e)^5 (d+e x)^{3/2}}-\frac{1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac{11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac{33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}+\frac{231 e^3}{64 (b d-a e)^4 (a+b x) (d+e x)^{3/2}}+\frac{1155 b e^4}{64 (b d-a e)^6 \sqrt{d+e x}}+\frac{\left (1155 b^2 e^4\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{128 (b d-a e)^6}\\ &=\frac{385 e^4}{64 (b d-a e)^5 (d+e x)^{3/2}}-\frac{1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac{11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac{33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}+\frac{231 e^3}{64 (b d-a e)^4 (a+b x) (d+e x)^{3/2}}+\frac{1155 b e^4}{64 (b d-a e)^6 \sqrt{d+e x}}+\frac{\left (1155 b^2 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 )}{64 (b d-a e)^6}\\ &=\frac{385 e^4}{64 (b d-a e)^5 (d+e x)^{3/2}}-\frac{1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac{11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac{33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}+\frac{231 e^3}{64 (b d-a e)^4 (a+b x) (d+e x)^{3/2}}+\frac{1155 b e^4}{64 (b d-a e)^6 \sqrt{d+e x}}-\frac{1155 b^{3/2} e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 (b d-a e)^{13/2}}\\ \end{align*}

Mathematica [C]  time = 0.0181044, size = 52, normalized size = 0.22 \[ -\frac{2 e^4 \, _2F_1\left (-\frac{3}{2},5;-\frac{1}{2};-\frac{b (d+e x)}{a e-b d}\right )}{3 (d+e x)^{3/2} (a e-b d)^5} \]

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),x]

[Out]

(-2*e^4*Hypergeometric2F1[-3/2, 5, -1/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(3*(-(b*d) + a*e)^5*(d + e*x)^(3/2)
)

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Maple [B]  time = 0.023, size = 473, normalized size = 2. \begin{align*} -{\frac{2\,{e}^{4}}{3\, \left ( ae-bd \right ) ^{5}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}+10\,{\frac{{e}^{4}b}{ \left ( ae-bd \right ) ^{6}\sqrt{ex+d}}}+{\frac{515\,{e}^{4}{b}^{5}}{64\, \left ( ae-bd \right ) ^{6} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{7}{2}}}}+{\frac{5153\,{b}^{4}{e}^{5}a}{192\, \left ( ae-bd \right ) ^{6} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{5153\,{e}^{4}{b}^{5}d}{192\, \left ( ae-bd \right ) ^{6} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{5855\,{e}^{6}{b}^{3}{a}^{2}}{192\, \left ( ae-bd \right ) ^{6} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{5855\,{b}^{4}{e}^{5}ad}{96\, \left ( ae-bd \right ) ^{6} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{5855\,{e}^{4}{b}^{5}{d}^{2}}{192\, \left ( ae-bd \right ) ^{6} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{765\,{e}^{7}{b}^{2}{a}^{3}}{64\, \left ( ae-bd \right ) ^{6} \left ( bex+ae \right ) ^{4}}\sqrt{ex+d}}-{\frac{2295\,{e}^{6}{b}^{3}d{a}^{2}}{64\, \left ( ae-bd \right ) ^{6} \left ( bex+ae \right ) ^{4}}\sqrt{ex+d}}+{\frac{2295\,{b}^{4}{e}^{5}a{d}^{2}}{64\, \left ( ae-bd \right ) ^{6} \left ( bex+ae \right ) ^{4}}\sqrt{ex+d}}-{\frac{765\,{e}^{4}{b}^{5}{d}^{3}}{64\, \left ( ae-bd \right ) ^{6} \left ( bex+ae \right ) ^{4}}\sqrt{ex+d}}+{\frac{1155\,{e}^{4}{b}^{2}}{64\, \left ( ae-bd \right ) ^{6}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \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,x)

[Out]

-2/3*e^4/(a*e-b*d)^5/(e*x+d)^(3/2)+10*e^4/(a*e-b*d)^6*b/(e*x+d)^(1/2)+515/64*e^4/(a*e-b*d)^6*b^5/(b*e*x+a*e)^4
*(e*x+d)^(7/2)+5153/192*e^5/(a*e-b*d)^6*b^4/(b*e*x+a*e)^4*(e*x+d)^(5/2)*a-5153/192*e^4/(a*e-b*d)^6*b^5/(b*e*x+
a*e)^4*(e*x+d)^(5/2)*d+5855/192*e^6/(a*e-b*d)^6*b^3/(b*e*x+a*e)^4*(e*x+d)^(3/2)*a^2-5855/96*e^5/(a*e-b*d)^6*b^
4/(b*e*x+a*e)^4*(e*x+d)^(3/2)*a*d+5855/192*e^4/(a*e-b*d)^6*b^5/(b*e*x+a*e)^4*(e*x+d)^(3/2)*d^2+765/64*e^7/(a*e
-b*d)^6*b^2/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a^3-2295/64*e^6/(a*e-b*d)^6*b^3/(b*e*x+a*e)^4*(e*x+d)^(1/2)*d*a^2+2295
/64*e^5/(a*e-b*d)^6*b^4/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a*d^2-765/64*e^4/(a*e-b*d)^6*b^5/(b*e*x+a*e)^4*(e*x+d)^(1/
2)*d^3+1155/64*e^4/(a*e-b*d)^6*b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.42769, size = 5126, normalized size = 22. \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)^3,x, algorithm="fricas")

[Out]

[1/384*(3465*(b^5*e^6*x^6 + a^4*b*d^2*e^4 + 2*(b^5*d*e^5 + 2*a*b^4*e^6)*x^5 + (b^5*d^2*e^4 + 8*a*b^4*d*e^5 + 6
*a^2*b^3*e^6)*x^4 + 4*(a*b^4*d^2*e^4 + 3*a^2*b^3*d*e^5 + a^3*b^2*e^6)*x^3 + (6*a^2*b^3*d^2*e^4 + 8*a^3*b^2*d*e
^5 + a^4*b*e^6)*x^2 + 2*(2*a^3*b^2*d^2*e^4 + a^4*b*d*e^5)*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*(3465*b^5*e^5*x^5 - 48*b^5*d^5 + 328*a*b^4*d^4*e
- 1030*a^2*b^3*d^3*e^2 + 2295*a^3*b^2*d^2*e^3 + 2048*a^4*b*d*e^4 - 128*a^5*e^5 + 1155*(4*b^5*d*e^4 + 11*a*b^4
*e^5)*x^4 + 231*(3*b^5*d^2*e^3 + 74*a*b^4*d*e^4 + 73*a^2*b^3*e^5)*x^3 - 99*(2*b^5*d^3*e^2 - 27*a*b^4*d^2*e^3 -
232*a^2*b^3*d*e^4 - 93*a^3*b^2*e^5)*x^2 + 11*(8*b^5*d^4*e - 68*a*b^4*d^3*e^2 + 345*a^2*b^3*d^2*e^3 + 1162*a^3
*b^2*d*e^4 + 128*a^4*b*e^5)*x)*sqrt(e*x + d))/(a^4*b^6*d^8 - 6*a^5*b^5*d^7*e + 15*a^6*b^4*d^6*e^2 - 20*a^7*b^3
*d^5*e^3 + 15*a^8*b^2*d^4*e^4 - 6*a^9*b*d^3*e^5 + a^10*d^2*e^6 + (b^10*d^6*e^2 - 6*a*b^9*d^5*e^3 + 15*a^2*b^8*
d^4*e^4 - 20*a^3*b^7*d^3*e^5 + 15*a^4*b^6*d^2*e^6 - 6*a^5*b^5*d*e^7 + a^6*b^4*e^8)*x^6 + 2*(b^10*d^7*e - 4*a*b
^9*d^6*e^2 + 3*a^2*b^8*d^5*e^3 + 10*a^3*b^7*d^4*e^4 - 25*a^4*b^6*d^3*e^5 + 24*a^5*b^5*d^2*e^6 - 11*a^6*b^4*d*e
^7 + 2*a^7*b^3*e^8)*x^5 + (b^10*d^8 + 2*a*b^9*d^7*e - 27*a^2*b^8*d^6*e^2 + 64*a^3*b^7*d^5*e^3 - 55*a^4*b^6*d^4
*e^4 - 6*a^5*b^5*d^3*e^5 + 43*a^6*b^4*d^2*e^6 - 28*a^7*b^3*d*e^7 + 6*a^8*b^2*e^8)*x^4 + 4*(a*b^9*d^8 - 3*a^2*b
^8*d^7*e - 2*a^3*b^7*d^6*e^2 + 19*a^4*b^6*d^5*e^3 - 30*a^5*b^5*d^4*e^4 + 19*a^6*b^4*d^3*e^5 - 2*a^7*b^3*d^2*e^
6 - 3*a^8*b^2*d*e^7 + a^9*b*e^8)*x^3 + (6*a^2*b^8*d^8 - 28*a^3*b^7*d^7*e + 43*a^4*b^6*d^6*e^2 - 6*a^5*b^5*d^5*
e^3 - 55*a^6*b^4*d^4*e^4 + 64*a^7*b^3*d^3*e^5 - 27*a^8*b^2*d^2*e^6 + 2*a^9*b*d*e^7 + a^10*e^8)*x^2 + 2*(2*a^3*
b^7*d^8 - 11*a^4*b^6*d^7*e + 24*a^5*b^5*d^6*e^2 - 25*a^6*b^4*d^5*e^3 + 10*a^7*b^3*d^4*e^4 + 3*a^8*b^2*d^3*e^5
- 4*a^9*b*d^2*e^6 + a^10*d*e^7)*x), -1/192*(3465*(b^5*e^6*x^6 + a^4*b*d^2*e^4 + 2*(b^5*d*e^5 + 2*a*b^4*e^6)*x^
5 + (b^5*d^2*e^4 + 8*a*b^4*d*e^5 + 6*a^2*b^3*e^6)*x^4 + 4*(a*b^4*d^2*e^4 + 3*a^2*b^3*d*e^5 + a^3*b^2*e^6)*x^3
+ (6*a^2*b^3*d^2*e^4 + 8*a^3*b^2*d*e^5 + a^4*b*e^6)*x^2 + 2*(2*a^3*b^2*d^2*e^4 + a^4*b*d*e^5)*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)) - (3465*b^5*e^5*x^5 - 48*b^5*d^5
+ 328*a*b^4*d^4*e - 1030*a^2*b^3*d^3*e^2 + 2295*a^3*b^2*d^2*e^3 + 2048*a^4*b*d*e^4 - 128*a^5*e^5 + 1155*(4*b^
5*d*e^4 + 11*a*b^4*e^5)*x^4 + 231*(3*b^5*d^2*e^3 + 74*a*b^4*d*e^4 + 73*a^2*b^3*e^5)*x^3 - 99*(2*b^5*d^3*e^2 -
27*a*b^4*d^2*e^3 - 232*a^2*b^3*d*e^4 - 93*a^3*b^2*e^5)*x^2 + 11*(8*b^5*d^4*e - 68*a*b^4*d^3*e^2 + 345*a^2*b^3*
d^2*e^3 + 1162*a^3*b^2*d*e^4 + 128*a^4*b*e^5)*x)*sqrt(e*x + d))/(a^4*b^6*d^8 - 6*a^5*b^5*d^7*e + 15*a^6*b^4*d^
6*e^2 - 20*a^7*b^3*d^5*e^3 + 15*a^8*b^2*d^4*e^4 - 6*a^9*b*d^3*e^5 + a^10*d^2*e^6 + (b^10*d^6*e^2 - 6*a*b^9*d^5
*e^3 + 15*a^2*b^8*d^4*e^4 - 20*a^3*b^7*d^3*e^5 + 15*a^4*b^6*d^2*e^6 - 6*a^5*b^5*d*e^7 + a^6*b^4*e^8)*x^6 + 2*(
b^10*d^7*e - 4*a*b^9*d^6*e^2 + 3*a^2*b^8*d^5*e^3 + 10*a^3*b^7*d^4*e^4 - 25*a^4*b^6*d^3*e^5 + 24*a^5*b^5*d^2*e^
6 - 11*a^6*b^4*d*e^7 + 2*a^7*b^3*e^8)*x^5 + (b^10*d^8 + 2*a*b^9*d^7*e - 27*a^2*b^8*d^6*e^2 + 64*a^3*b^7*d^5*e^
3 - 55*a^4*b^6*d^4*e^4 - 6*a^5*b^5*d^3*e^5 + 43*a^6*b^4*d^2*e^6 - 28*a^7*b^3*d*e^7 + 6*a^8*b^2*e^8)*x^4 + 4*(a
*b^9*d^8 - 3*a^2*b^8*d^7*e - 2*a^3*b^7*d^6*e^2 + 19*a^4*b^6*d^5*e^3 - 30*a^5*b^5*d^4*e^4 + 19*a^6*b^4*d^3*e^5
- 2*a^7*b^3*d^2*e^6 - 3*a^8*b^2*d*e^7 + a^9*b*e^8)*x^3 + (6*a^2*b^8*d^8 - 28*a^3*b^7*d^7*e + 43*a^4*b^6*d^6*e^
2 - 6*a^5*b^5*d^5*e^3 - 55*a^6*b^4*d^4*e^4 + 64*a^7*b^3*d^3*e^5 - 27*a^8*b^2*d^2*e^6 + 2*a^9*b*d*e^7 + a^10*e^
8)*x^2 + 2*(2*a^3*b^7*d^8 - 11*a^4*b^6*d^7*e + 24*a^5*b^5*d^6*e^2 - 25*a^6*b^4*d^5*e^3 + 10*a^7*b^3*d^4*e^4 +
3*a^8*b^2*d^3*e^5 - 4*a^9*b*d^2*e^6 + a^10*d*e^7)*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)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.25989, size = 675, normalized size = 2.9 \begin{align*} \frac{1155 \, b^{2} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{4}}{64 \,{\left (b^{6} d^{6} - 6 \, a b^{5} d^{5} e + 15 \, a^{2} b^{4} d^{4} e^{2} - 20 \, a^{3} b^{3} d^{3} e^{3} + 15 \, a^{4} b^{2} d^{2} e^{4} - 6 \, a^{5} b d e^{5} + a^{6} e^{6}\right )} \sqrt{-b^{2} d + a b e}} + \frac{2 \,{\left (15 \,{\left (x e + d\right )} b e^{4} + b d e^{4} - a e^{5}\right )}}{3 \,{\left (b^{6} d^{6} - 6 \, a b^{5} d^{5} e + 15 \, a^{2} b^{4} d^{4} e^{2} - 20 \, a^{3} b^{3} d^{3} e^{3} + 15 \, a^{4} b^{2} d^{2} e^{4} - 6 \, a^{5} b d e^{5} + a^{6} e^{6}\right )}{\left (x e + d\right )}^{\frac{3}{2}}} + \frac{1545 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{5} e^{4} - 5153 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{5} d e^{4} + 5855 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{5} d^{2} e^{4} - 2295 \, \sqrt{x e + d} b^{5} d^{3} e^{4} + 5153 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{4} e^{5} - 11710 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{4} d e^{5} + 6885 \, \sqrt{x e + d} a b^{4} d^{2} e^{5} + 5855 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b^{3} e^{6} - 6885 \, \sqrt{x e + d} a^{2} b^{3} d e^{6} + 2295 \, \sqrt{x e + d} a^{3} b^{2} e^{7}}{192 \,{\left (b^{6} d^{6} - 6 \, a b^{5} d^{5} e + 15 \, a^{2} b^{4} d^{4} e^{2} - 20 \, a^{3} b^{3} d^{3} e^{3} + 15 \, a^{4} b^{2} d^{2} e^{4} - 6 \, a^{5} b d e^{5} + a^{6} e^{6}\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{4}} \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,x, algorithm="giac")

[Out]

1155/64*b^2*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^4/((b^6*d^6 - 6*a*b^5*d^5*e + 15*a^2*b^4*d^4*e^2 -
20*a^3*b^3*d^3*e^3 + 15*a^4*b^2*d^2*e^4 - 6*a^5*b*d*e^5 + a^6*e^6)*sqrt(-b^2*d + a*b*e)) + 2/3*(15*(x*e + d)*b
*e^4 + b*d*e^4 - a*e^5)/((b^6*d^6 - 6*a*b^5*d^5*e + 15*a^2*b^4*d^4*e^2 - 20*a^3*b^3*d^3*e^3 + 15*a^4*b^2*d^2*e
^4 - 6*a^5*b*d*e^5 + a^6*e^6)*(x*e + d)^(3/2)) + 1/192*(1545*(x*e + d)^(7/2)*b^5*e^4 - 5153*(x*e + d)^(5/2)*b^
5*d*e^4 + 5855*(x*e + d)^(3/2)*b^5*d^2*e^4 - 2295*sqrt(x*e + d)*b^5*d^3*e^4 + 5153*(x*e + d)^(5/2)*a*b^4*e^5 -
11710*(x*e + d)^(3/2)*a*b^4*d*e^5 + 6885*sqrt(x*e + d)*a*b^4*d^2*e^5 + 5855*(x*e + d)^(3/2)*a^2*b^3*e^6 - 688
5*sqrt(x*e + d)*a^2*b^3*d*e^6 + 2295*sqrt(x*e + d)*a^3*b^2*e^7)/((b^6*d^6 - 6*a*b^5*d^5*e + 15*a^2*b^4*d^4*e^2
- 20*a^3*b^3*d^3*e^3 + 15*a^4*b^2*d^2*e^4 - 6*a^5*b*d*e^5 + a^6*e^6)*((x*e + d)*b - b*d + a*e)^4)