3.2087 \(\int \frac {(a+b x) (d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=172 \[ -\frac {3 e^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{5/2} (b d-a e)^{5/2}}+\frac {3 e^3 \sqrt {d+e x}}{64 b^2 (a+b x) (b d-a e)^2}-\frac {e^2 \sqrt {d+e x}}{32 b^2 (a+b x)^2 (b d-a e)}-\frac {e \sqrt {d+e x}}{8 b^2 (a+b x)^3}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4} \]

[Out]

-1/4*(e*x+d)^(3/2)/b/(b*x+a)^4-3/64*e^4*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(5/2)/(-a*e+b*d)^(5/
2)-1/8*e*(e*x+d)^(1/2)/b^2/(b*x+a)^3-1/32*e^2*(e*x+d)^(1/2)/b^2/(-a*e+b*d)/(b*x+a)^2+3/64*e^3*(e*x+d)^(1/2)/b^
2/(-a*e+b*d)^2/(b*x+a)

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Rubi [A]  time = 0.09, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 47, 51, 63, 208} \[ \frac {3 e^3 \sqrt {d+e x}}{64 b^2 (a+b x) (b d-a e)^2}-\frac {e^2 \sqrt {d+e x}}{32 b^2 (a+b x)^2 (b d-a e)}-\frac {3 e^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{5/2} (b d-a e)^{5/2}}-\frac {e \sqrt {d+e x}}{8 b^2 (a+b x)^3}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e*Sqrt[d + e*x])/(8*b^2*(a + b*x)^3) - (e^2*Sqrt[d + e*x])/(32*b^2*(b*d - a*e)*(a + b*x)^2) + (3*e^3*Sqrt[d
+ e*x])/(64*b^2*(b*d - a*e)^2*(a + b*x)) - (d + e*x)^(3/2)/(4*b*(a + b*x)^4) - (3*e^4*ArcTanh[(Sqrt[b]*Sqrt[d
+ e*x])/Sqrt[b*d - a*e]])/(64*b^(5/2)*(b*d - a*e)^(5/2))

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, 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)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac {(d+e x)^{3/2}}{(a+b x)^5} \, dx\\ &=-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}+\frac {(3 e) \int \frac {\sqrt {d+e x}}{(a+b x)^4} \, dx}{8 b}\\ &=-\frac {e \sqrt {d+e x}}{8 b^2 (a+b x)^3}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}+\frac {e^2 \int \frac {1}{(a+b x)^3 \sqrt {d+e x}} \, dx}{16 b^2}\\ &=-\frac {e \sqrt {d+e x}}{8 b^2 (a+b x)^3}-\frac {e^2 \sqrt {d+e x}}{32 b^2 (b d-a e) (a+b x)^2}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}-\frac {\left (3 e^3\right ) \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{64 b^2 (b d-a e)}\\ &=-\frac {e \sqrt {d+e x}}{8 b^2 (a+b x)^3}-\frac {e^2 \sqrt {d+e x}}{32 b^2 (b d-a e) (a+b x)^2}+\frac {3 e^3 \sqrt {d+e x}}{64 b^2 (b d-a e)^2 (a+b x)}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}+\frac {\left (3 e^4\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{128 b^2 (b d-a e)^2}\\ &=-\frac {e \sqrt {d+e x}}{8 b^2 (a+b x)^3}-\frac {e^2 \sqrt {d+e x}}{32 b^2 (b d-a e) (a+b x)^2}+\frac {3 e^3 \sqrt {d+e x}}{64 b^2 (b d-a e)^2 (a+b x)}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}+\frac {\left (3 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^2 (b d-a e)^2}\\ &=-\frac {e \sqrt {d+e x}}{8 b^2 (a+b x)^3}-\frac {e^2 \sqrt {d+e x}}{32 b^2 (b d-a e) (a+b x)^2}+\frac {3 e^3 \sqrt {d+e x}}{64 b^2 (b d-a e)^2 (a+b x)}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}-\frac {3 e^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{5/2} (b d-a e)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.30 \[ \frac {2 e^4 (d+e x)^{5/2} \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};-\frac {b (d+e x)}{a e-b d}\right )}{5 (a e-b d)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^4*(d + e*x)^(5/2)*Hypergeometric2F1[5/2, 5, 7/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(5*(-(b*d) + a*e)^5)

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fricas [B]  time = 0.89, size = 1043, normalized size = 6.06 \[ \left [\frac {3 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (16 \, b^{5} d^{4} - 40 \, a b^{4} d^{3} e + 26 \, a^{2} b^{3} d^{2} e^{2} + a^{3} b^{2} d e^{3} - 3 \, a^{4} b e^{4} - 3 \, {\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} + {\left (2 \, b^{5} d^{2} e^{2} - 13 \, a b^{4} d e^{3} + 11 \, a^{2} b^{3} e^{4}\right )} x^{2} + {\left (24 \, b^{5} d^{3} e - 68 \, a b^{4} d^{2} e^{2} + 55 \, a^{2} b^{3} d e^{3} - 11 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{128 \, {\left (a^{4} b^{6} d^{3} - 3 \, a^{5} b^{5} d^{2} e + 3 \, a^{6} b^{4} d e^{2} - a^{7} b^{3} e^{3} + {\left (b^{10} d^{3} - 3 \, a b^{9} d^{2} e + 3 \, a^{2} b^{8} d e^{2} - a^{3} b^{7} e^{3}\right )} x^{4} + 4 \, {\left (a b^{9} d^{3} - 3 \, a^{2} b^{8} d^{2} e + 3 \, a^{3} b^{7} d e^{2} - a^{4} b^{6} e^{3}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} d^{3} - 3 \, a^{3} b^{7} d^{2} e + 3 \, a^{4} b^{6} d e^{2} - a^{5} b^{5} e^{3}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} d^{3} - 3 \, a^{4} b^{6} d^{2} e + 3 \, a^{5} b^{5} d e^{2} - a^{6} b^{4} e^{3}\right )} x\right )}}, \frac {3 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (16 \, b^{5} d^{4} - 40 \, a b^{4} d^{3} e + 26 \, a^{2} b^{3} d^{2} e^{2} + a^{3} b^{2} d e^{3} - 3 \, a^{4} b e^{4} - 3 \, {\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} + {\left (2 \, b^{5} d^{2} e^{2} - 13 \, a b^{4} d e^{3} + 11 \, a^{2} b^{3} e^{4}\right )} x^{2} + {\left (24 \, b^{5} d^{3} e - 68 \, a b^{4} d^{2} e^{2} + 55 \, a^{2} b^{3} d e^{3} - 11 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{64 \, {\left (a^{4} b^{6} d^{3} - 3 \, a^{5} b^{5} d^{2} e + 3 \, a^{6} b^{4} d e^{2} - a^{7} b^{3} e^{3} + {\left (b^{10} d^{3} - 3 \, a b^{9} d^{2} e + 3 \, a^{2} b^{8} d e^{2} - a^{3} b^{7} e^{3}\right )} x^{4} + 4 \, {\left (a b^{9} d^{3} - 3 \, a^{2} b^{8} d^{2} e + 3 \, a^{3} b^{7} d e^{2} - a^{4} b^{6} e^{3}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} d^{3} - 3 \, a^{3} b^{7} d^{2} e + 3 \, a^{4} b^{6} d e^{2} - a^{5} b^{5} e^{3}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} d^{3} - 3 \, a^{4} b^{6} d^{2} e + 3 \, a^{5} b^{5} d e^{2} - a^{6} b^{4} e^{3}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

[1/128*(3*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(b^2*d - a*b*e)*lo
g((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(16*b^5*d^4 - 40*a*b^4*d^3*e + 26
*a^2*b^3*d^2*e^2 + a^3*b^2*d*e^3 - 3*a^4*b*e^4 - 3*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (2*b^5*d^2*e^2 - 13*a*b^4*d*e
^3 + 11*a^2*b^3*e^4)*x^2 + (24*b^5*d^3*e - 68*a*b^4*d^2*e^2 + 55*a^2*b^3*d*e^3 - 11*a^3*b^2*e^4)*x)*sqrt(e*x +
d))/(a^4*b^6*d^3 - 3*a^5*b^5*d^2*e + 3*a^6*b^4*d*e^2 - a^7*b^3*e^3 + (b^10*d^3 - 3*a*b^9*d^2*e + 3*a^2*b^8*d*
e^2 - a^3*b^7*e^3)*x^4 + 4*(a*b^9*d^3 - 3*a^2*b^8*d^2*e + 3*a^3*b^7*d*e^2 - a^4*b^6*e^3)*x^3 + 6*(a^2*b^8*d^3
- 3*a^3*b^7*d^2*e + 3*a^4*b^6*d*e^2 - a^5*b^5*e^3)*x^2 + 4*(a^3*b^7*d^3 - 3*a^4*b^6*d^2*e + 3*a^5*b^5*d*e^2 -
a^6*b^4*e^3)*x), 1/64*(3*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(-b
^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (16*b^5*d^4 - 40*a*b^4*d^3*e + 26*a^2
*b^3*d^2*e^2 + a^3*b^2*d*e^3 - 3*a^4*b*e^4 - 3*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (2*b^5*d^2*e^2 - 13*a*b^4*d*e^3 +
11*a^2*b^3*e^4)*x^2 + (24*b^5*d^3*e - 68*a*b^4*d^2*e^2 + 55*a^2*b^3*d*e^3 - 11*a^3*b^2*e^4)*x)*sqrt(e*x + d))
/(a^4*b^6*d^3 - 3*a^5*b^5*d^2*e + 3*a^6*b^4*d*e^2 - a^7*b^3*e^3 + (b^10*d^3 - 3*a*b^9*d^2*e + 3*a^2*b^8*d*e^2
- a^3*b^7*e^3)*x^4 + 4*(a*b^9*d^3 - 3*a^2*b^8*d^2*e + 3*a^3*b^7*d*e^2 - a^4*b^6*e^3)*x^3 + 6*(a^2*b^8*d^3 - 3*
a^3*b^7*d^2*e + 3*a^4*b^6*d*e^2 - a^5*b^5*e^3)*x^2 + 4*(a^3*b^7*d^3 - 3*a^4*b^6*d^2*e + 3*a^5*b^5*d*e^2 - a^6*
b^4*e^3)*x)]

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giac [B]  time = 0.20, size = 289, normalized size = 1.68 \[ \frac {3 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{4}}{64 \, {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e}} + \frac {3 \, {\left (x e + d\right )}^{\frac {7}{2}} b^{3} e^{4} - 11 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{3} d e^{4} - 11 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{4} + 3 \, \sqrt {x e + d} b^{3} d^{3} e^{4} + 11 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{2} e^{5} + 22 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{2} d e^{5} - 9 \, \sqrt {x e + d} a b^{2} d^{2} e^{5} - 11 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b e^{6} + 9 \, \sqrt {x e + d} a^{2} b d e^{6} - 3 \, \sqrt {x e + d} a^{3} e^{7}}{64 \, {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")

[Out]

3/64*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^4/((b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)*sqrt(-b^2*d + a*b
*e)) + 1/64*(3*(x*e + d)^(7/2)*b^3*e^4 - 11*(x*e + d)^(5/2)*b^3*d*e^4 - 11*(x*e + d)^(3/2)*b^3*d^2*e^4 + 3*sqr
t(x*e + d)*b^3*d^3*e^4 + 11*(x*e + d)^(5/2)*a*b^2*e^5 + 22*(x*e + d)^(3/2)*a*b^2*d*e^5 - 9*sqrt(x*e + d)*a*b^2
*d^2*e^5 - 11*(x*e + d)^(3/2)*a^2*b*e^6 + 9*sqrt(x*e + d)*a^2*b*d*e^6 - 3*sqrt(x*e + d)*a^3*e^7)/((b^4*d^2 - 2
*a*b^3*d*e + a^2*b^2*e^2)*((x*e + d)*b - b*d + a*e)^4)

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maple [A]  time = 0.07, size = 222, normalized size = 1.29 \[ \frac {3 \left (e x +d \right )^{\frac {7}{2}} b \,e^{4}}{64 \left (b e x +a e \right )^{4} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}-\frac {3 \sqrt {e x +d}\, a \,e^{5}}{64 \left (b e x +a e \right )^{4} b^{2}}+\frac {3 \sqrt {e x +d}\, d \,e^{4}}{64 \left (b e x +a e \right )^{4} b}+\frac {11 \left (e x +d \right )^{\frac {5}{2}} e^{4}}{64 \left (b e x +a e \right )^{4} \left (a e -b d \right )}-\frac {11 \left (e x +d \right )^{\frac {3}{2}} e^{4}}{64 \left (b e x +a e \right )^{4} b}+\frac {3 e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{64 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (a e -b d \right ) b}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/64*e^4/(b*e*x+a*e)^4*b/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(7/2)+11/64*e^4/(b*e*x+a*e)^4/(a*e-b*d)*(e*x+d)^(
5/2)-11/64*e^4/(b*e*x+a*e)^4/b*(e*x+d)^(3/2)-3/64*e^5/(b*e*x+a*e)^4/b^2*(e*x+d)^(1/2)*a+3/64*e^4/(b*e*x+a*e)^4
/b*(e*x+d)^(1/2)*d+3/64*e^4/b^2/(a^2*e^2-2*a*b*d*e+b^2*d^2)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d
)*b)^(1/2)*b)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 2.14, size = 296, normalized size = 1.72 \[ \frac {3\,e^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{64\,b^{5/2}\,{\left (a\,e-b\,d\right )}^{5/2}}-\frac {\frac {11\,e^4\,{\left (d+e\,x\right )}^{3/2}}{64\,b}-\frac {11\,e^4\,{\left (d+e\,x\right )}^{5/2}}{64\,\left (a\,e-b\,d\right )}+\frac {3\,e^4\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}}{64\,b^2}-\frac {3\,b\,e^4\,{\left (d+e\,x\right )}^{7/2}}{64\,{\left (a\,e-b\,d\right )}^2}}{b^4\,{\left (d+e\,x\right )}^4-\left (4\,b^4\,d-4\,a\,b^3\,e\right )\,{\left (d+e\,x\right )}^3-\left (d+e\,x\right )\,\left (-4\,a^3\,b\,e^3+12\,a^2\,b^2\,d\,e^2-12\,a\,b^3\,d^2\,e+4\,b^4\,d^3\right )+a^4\,e^4+b^4\,d^4+{\left (d+e\,x\right )}^2\,\left (6\,a^2\,b^2\,e^2-12\,a\,b^3\,d\,e+6\,b^4\,d^2\right )+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e-4\,a^3\,b\,d\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*e^4*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(64*b^(5/2)*(a*e - b*d)^(5/2)) - ((11*e^4*(d + e*x)^
(3/2))/(64*b) - (11*e^4*(d + e*x)^(5/2))/(64*(a*e - b*d)) + (3*e^4*(a*e - b*d)*(d + e*x)^(1/2))/(64*b^2) - (3*
b*e^4*(d + e*x)^(7/2))/(64*(a*e - b*d)^2))/(b^4*(d + e*x)^4 - (4*b^4*d - 4*a*b^3*e)*(d + e*x)^3 - (d + e*x)*(4
*b^4*d^3 - 4*a^3*b*e^3 + 12*a^2*b^2*d*e^2 - 12*a*b^3*d^2*e) + a^4*e^4 + b^4*d^4 + (d + e*x)^2*(6*b^4*d^2 + 6*a
^2*b^2*e^2 - 12*a*b^3*d*e) + 6*a^2*b^2*d^2*e^2 - 4*a*b^3*d^3*e - 4*a^3*b*d*e^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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