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

Optimal. Leaf size=313 \[ -\frac {3 e^4 (-a B e-A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{7/2} (b d-a e)^{7/2}}+\frac {3 e^3 \sqrt {d+e x} (-a B e-A b e+2 b B d)}{128 b^3 (a+b x) (b d-a e)^3}-\frac {e^2 \sqrt {d+e x} (-a B e-A b e+2 b B d)}{64 b^3 (a+b x)^2 (b d-a e)^2}-\frac {e \sqrt {d+e x} (-a B e-A b e+2 b B d)}{16 b^3 (a+b x)^3 (b d-a e)}-\frac {(d+e x)^{3/2} (-a B e-A b e+2 b B d)}{8 b^2 (a+b x)^4 (b d-a e)}-\frac {(d+e x)^{5/2} (A b-a B)}{5 b (a+b x)^5 (b d-a e)} \]

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Rubi [A]  time = 0.27, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {27, 78, 47, 51, 63, 208} \begin {gather*} \frac {3 e^3 \sqrt {d+e x} (-a B e-A b e+2 b B d)}{128 b^3 (a+b x) (b d-a e)^3}-\frac {e^2 \sqrt {d+e x} (-a B e-A b e+2 b B d)}{64 b^3 (a+b x)^2 (b d-a e)^2}-\frac {3 e^4 (-a B e-A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{7/2} (b d-a e)^{7/2}}-\frac {e \sqrt {d+e x} (-a B e-A b e+2 b B d)}{16 b^3 (a+b x)^3 (b d-a e)}-\frac {(d+e x)^{3/2} (-a B e-A b e+2 b B d)}{8 b^2 (a+b x)^4 (b d-a e)}-\frac {(d+e x)^{5/2} (A b-a B)}{5 b (a+b x)^5 (b d-a e)} \end {gather*}

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*(2*b*B*d - A*b*e - a*B*e)*Sqrt[d + e*x])/(16*b^3*(b*d - a*e)*(a + b*x)^3) - (e^2*(2*b*B*d - A*b*e - a*B*e)
*Sqrt[d + e*x])/(64*b^3*(b*d - a*e)^2*(a + b*x)^2) + (3*e^3*(2*b*B*d - A*b*e - a*B*e)*Sqrt[d + e*x])/(128*b^3*
(b*d - a*e)^3*(a + b*x)) - ((2*b*B*d - A*b*e - a*B*e)*(d + e*x)^(3/2))/(8*b^2*(b*d - a*e)*(a + b*x)^4) - ((A*b
- a*B)*(d + e*x)^(5/2))/(5*b*(b*d - a*e)*(a + b*x)^5) - (3*e^4*(2*b*B*d - A*b*e - a*B*e)*ArcTanh[(Sqrt[b]*Sqr
t[d + e*x])/Sqrt[b*d - a*e]])/(128*b^(7/2)*(b*d - a*e)^(7/2))

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 {(A+B x) (d+e x)^{3/2}}{(a+b x)^6} \, dx\\ &=-\frac {(A b-a B) (d+e x)^{5/2}}{5 b (b d-a e) (a+b x)^5}+\frac {(2 b B d-A b e-a B e) \int \frac {(d+e x)^{3/2}}{(a+b x)^5} \, dx}{2 b (b d-a e)}\\ &=-\frac {(2 b B d-A b e-a B e) (d+e x)^{3/2}}{8 b^2 (b d-a e) (a+b x)^4}-\frac {(A b-a B) (d+e x)^{5/2}}{5 b (b d-a e) (a+b x)^5}+\frac {(3 e (2 b B d-A b e-a B e)) \int \frac {\sqrt {d+e x}}{(a+b x)^4} \, dx}{16 b^2 (b d-a e)}\\ &=-\frac {e (2 b B d-A b e-a B e) \sqrt {d+e x}}{16 b^3 (b d-a e) (a+b x)^3}-\frac {(2 b B d-A b e-a B e) (d+e x)^{3/2}}{8 b^2 (b d-a e) (a+b x)^4}-\frac {(A b-a B) (d+e x)^{5/2}}{5 b (b d-a e) (a+b x)^5}+\frac {\left (e^2 (2 b B d-A b e-a B e)\right ) \int \frac {1}{(a+b x)^3 \sqrt {d+e x}} \, dx}{32 b^3 (b d-a e)}\\ &=-\frac {e (2 b B d-A b e-a B e) \sqrt {d+e x}}{16 b^3 (b d-a e) (a+b x)^3}-\frac {e^2 (2 b B d-A b e-a B e) \sqrt {d+e x}}{64 b^3 (b d-a e)^2 (a+b x)^2}-\frac {(2 b B d-A b e-a B e) (d+e x)^{3/2}}{8 b^2 (b d-a e) (a+b x)^4}-\frac {(A b-a B) (d+e x)^{5/2}}{5 b (b d-a e) (a+b x)^5}-\frac {\left (3 e^3 (2 b B d-A b e-a B e)\right ) \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{128 b^3 (b d-a e)^2}\\ &=-\frac {e (2 b B d-A b e-a B e) \sqrt {d+e x}}{16 b^3 (b d-a e) (a+b x)^3}-\frac {e^2 (2 b B d-A b e-a B e) \sqrt {d+e x}}{64 b^3 (b d-a e)^2 (a+b x)^2}+\frac {3 e^3 (2 b B d-A b e-a B e) \sqrt {d+e x}}{128 b^3 (b d-a e)^3 (a+b x)}-\frac {(2 b B d-A b e-a B e) (d+e x)^{3/2}}{8 b^2 (b d-a e) (a+b x)^4}-\frac {(A b-a B) (d+e x)^{5/2}}{5 b (b d-a e) (a+b x)^5}+\frac {\left (3 e^4 (2 b B d-A b e-a B e)\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{256 b^3 (b d-a e)^3}\\ &=-\frac {e (2 b B d-A b e-a B e) \sqrt {d+e x}}{16 b^3 (b d-a e) (a+b x)^3}-\frac {e^2 (2 b B d-A b e-a B e) \sqrt {d+e x}}{64 b^3 (b d-a e)^2 (a+b x)^2}+\frac {3 e^3 (2 b B d-A b e-a B e) \sqrt {d+e x}}{128 b^3 (b d-a e)^3 (a+b x)}-\frac {(2 b B d-A b e-a B e) (d+e x)^{3/2}}{8 b^2 (b d-a e) (a+b x)^4}-\frac {(A b-a B) (d+e x)^{5/2}}{5 b (b d-a e) (a+b x)^5}+\frac {\left (3 e^3 (2 b B d-A b e-a B e)\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{128 b^3 (b d-a e)^3}\\ &=-\frac {e (2 b B d-A b e-a B e) \sqrt {d+e x}}{16 b^3 (b d-a e) (a+b x)^3}-\frac {e^2 (2 b B d-A b e-a B e) \sqrt {d+e x}}{64 b^3 (b d-a e)^2 (a+b x)^2}+\frac {3 e^3 (2 b B d-A b e-a B e) \sqrt {d+e x}}{128 b^3 (b d-a e)^3 (a+b x)}-\frac {(2 b B d-A b e-a B e) (d+e x)^{3/2}}{8 b^2 (b d-a e) (a+b x)^4}-\frac {(A b-a B) (d+e x)^{5/2}}{5 b (b d-a e) (a+b x)^5}-\frac {3 e^4 (2 b B d-A b e-a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{7/2} (b d-a e)^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 98, normalized size = 0.31 \begin {gather*} \frac {(d+e x)^{5/2} \left (\frac {5 e^4 (a B e+A b e-2 b B d) \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^5}+\frac {5 a B-5 A b}{(a+b x)^5}\right )}{25 b (b d-a e)} \end {gather*}

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]

((d + e*x)^(5/2)*((-5*A*b + 5*a*B)/(a + b*x)^5 + (5*e^4*(-2*b*B*d + A*b*e + a*B*e)*Hypergeometric2F1[5/2, 5, 7
/2, (b*(d + e*x))/(b*d - a*e)])/(b*d - a*e)^5))/(25*b*(b*d - a*e))

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IntegrateAlgebraic [B]  time = 3.19, size = 660, normalized size = 2.11 \begin {gather*} -\frac {e^4 \sqrt {d+e x} \left (15 a^5 B e^5+15 a^4 A b e^5+70 a^4 b B e^4 (d+e x)-90 a^4 b B d e^4+70 a^3 A b^2 e^4 (d+e x)-60 a^3 A b^2 d e^4+210 a^3 b^2 B d^2 e^3+128 a^3 b^2 B e^3 (d+e x)^2-350 a^3 b^2 B d e^3 (d+e x)+90 a^2 A b^3 d^2 e^3-128 a^2 A b^3 e^3 (d+e x)^2-210 a^2 A b^3 d e^3 (d+e x)-240 a^2 b^3 B d^3 e^2+630 a^2 b^3 B d^2 e^2 (d+e x)-70 a^2 b^3 B e^2 (d+e x)^3-256 a^2 b^3 B d e^2 (d+e x)^2-60 a A b^4 d^3 e^2+210 a A b^4 d^2 e^2 (d+e x)-70 a A b^4 e^2 (d+e x)^3+256 a A b^4 d e^2 (d+e x)^2+135 a b^4 B d^4 e-490 a b^4 B d^3 e (d+e x)+128 a b^4 B d^2 e (d+e x)^2-15 a b^4 B e (d+e x)^4+210 a b^4 B d e (d+e x)^3+15 A b^5 d^4 e-70 A b^5 d^3 e (d+e x)-128 A b^5 d^2 e (d+e x)^2-15 A b^5 e (d+e x)^4+70 A b^5 d e (d+e x)^3-30 b^5 B d^5+140 b^5 B d^4 (d+e x)-140 b^5 B d^2 (d+e x)^3+30 b^5 B d (d+e x)^4\right )}{640 b^3 (b d-a e)^3 (-a e-b (d+e x)+b d)^5}-\frac {3 \left (-a B e^5-A b e^5+2 b B d e^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{128 b^{7/2} (b d-a e)^3 \sqrt {a e-b d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/640*(e^4*Sqrt[d + e*x]*(-30*b^5*B*d^5 + 15*A*b^5*d^4*e + 135*a*b^4*B*d^4*e - 60*a*A*b^4*d^3*e^2 - 240*a^2*b
^3*B*d^3*e^2 + 90*a^2*A*b^3*d^2*e^3 + 210*a^3*b^2*B*d^2*e^3 - 60*a^3*A*b^2*d*e^4 - 90*a^4*b*B*d*e^4 + 15*a^4*A
*b*e^5 + 15*a^5*B*e^5 + 140*b^5*B*d^4*(d + e*x) - 70*A*b^5*d^3*e*(d + e*x) - 490*a*b^4*B*d^3*e*(d + e*x) + 210
*a*A*b^4*d^2*e^2*(d + e*x) + 630*a^2*b^3*B*d^2*e^2*(d + e*x) - 210*a^2*A*b^3*d*e^3*(d + e*x) - 350*a^3*b^2*B*d
*e^3*(d + e*x) + 70*a^3*A*b^2*e^4*(d + e*x) + 70*a^4*b*B*e^4*(d + e*x) - 128*A*b^5*d^2*e*(d + e*x)^2 + 128*a*b
^4*B*d^2*e*(d + e*x)^2 + 256*a*A*b^4*d*e^2*(d + e*x)^2 - 256*a^2*b^3*B*d*e^2*(d + e*x)^2 - 128*a^2*A*b^3*e^3*(
d + e*x)^2 + 128*a^3*b^2*B*e^3*(d + e*x)^2 - 140*b^5*B*d^2*(d + e*x)^3 + 70*A*b^5*d*e*(d + e*x)^3 + 210*a*b^4*
B*d*e*(d + e*x)^3 - 70*a*A*b^4*e^2*(d + e*x)^3 - 70*a^2*b^3*B*e^2*(d + e*x)^3 + 30*b^5*B*d*(d + e*x)^4 - 15*A*
b^5*e*(d + e*x)^4 - 15*a*b^4*B*e*(d + e*x)^4))/(b^3*(b*d - a*e)^3*(b*d - a*e - b*(d + e*x))^5) - (3*(2*b*B*d*e
^4 - A*b*e^5 - a*B*e^5)*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(128*b^(7/2)*(b*d - a*
e)^3*Sqrt[-(b*d) + a*e])

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fricas [B]  time = 0.51, size = 2349, normalized size = 7.50

result too large to display

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]

[1/1280*(15*(2*B*a^5*b*d*e^4 - (B*a^6 + A*a^5*b)*e^5 + (2*B*b^6*d*e^4 - (B*a*b^5 + A*b^6)*e^5)*x^5 + 5*(2*B*a*
b^5*d*e^4 - (B*a^2*b^4 + A*a*b^5)*e^5)*x^4 + 10*(2*B*a^2*b^4*d*e^4 - (B*a^3*b^3 + A*a^2*b^4)*e^5)*x^3 + 10*(2*
B*a^3*b^3*d*e^4 - (B*a^4*b^2 + A*a^3*b^3)*e^5)*x^2 + 5*(2*B*a^4*b^2*d*e^4 - (B*a^5*b + A*a^4*b^2)*e^5)*x)*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*(32*(B*a*b^6 +
4*A*b^7)*d^5 - 16*(6*B*a^2*b^5 + 29*A*a*b^6)*d^4*e + 4*(19*B*a^3*b^4 + 146*A*a^2*b^5)*d^3*e^2 + 2*(4*B*a^4*b^3
- 129*A*a^3*b^4)*d^2*e^3 - 5*(7*B*a^5*b^2 + A*a^4*b^3)*d*e^4 + 15*(B*a^6*b + A*a^5*b^2)*e^5 - 15*(2*B*b^7*d^2
*e^3 - (3*B*a*b^6 + A*b^7)*d*e^4 + (B*a^2*b^5 + A*a*b^6)*e^5)*x^4 + 10*(2*B*b^7*d^3*e^2 - (17*B*a*b^6 + A*b^7)
*d^2*e^3 + 2*(11*B*a^2*b^5 + 4*A*a*b^6)*d*e^4 - 7*(B*a^3*b^4 + A*a^2*b^5)*e^5)*x^3 + 2*(120*B*b^7*d^4*e - 2*(2
27*B*a*b^6 - 2*A*b^7)*d^3*e^2 + 27*(21*B*a^2*b^5 - A*a*b^6)*d^2*e^3 - 3*(99*B*a^3*b^4 - 29*A*a^2*b^5)*d*e^4 +
64*(B*a^4*b^3 - A*a^3*b^4)*e^5)*x^2 + 2*(80*B*b^7*d^5 - 8*(31*B*a*b^6 - 11*A*b^7)*d^4*e + 2*(107*B*a^2*b^5 - 1
72*A*a*b^6)*d^3*e^2 + (B*a^3*b^4 + 489*A*a^2*b^5)*d^2*e^3 - 2*(41*B*a^4*b^3 + 134*A*a^3*b^4)*d*e^4 + 35*(B*a^5
*b^2 + A*a^4*b^3)*e^5)*x)*sqrt(e*x + d))/(a^5*b^8*d^4 - 4*a^6*b^7*d^3*e + 6*a^7*b^6*d^2*e^2 - 4*a^8*b^5*d*e^3
+ a^9*b^4*e^4 + (b^13*d^4 - 4*a*b^12*d^3*e + 6*a^2*b^11*d^2*e^2 - 4*a^3*b^10*d*e^3 + a^4*b^9*e^4)*x^5 + 5*(a*b
^12*d^4 - 4*a^2*b^11*d^3*e + 6*a^3*b^10*d^2*e^2 - 4*a^4*b^9*d*e^3 + a^5*b^8*e^4)*x^4 + 10*(a^2*b^11*d^4 - 4*a^
3*b^10*d^3*e + 6*a^4*b^9*d^2*e^2 - 4*a^5*b^8*d*e^3 + a^6*b^7*e^4)*x^3 + 10*(a^3*b^10*d^4 - 4*a^4*b^9*d^3*e + 6
*a^5*b^8*d^2*e^2 - 4*a^6*b^7*d*e^3 + a^7*b^6*e^4)*x^2 + 5*(a^4*b^9*d^4 - 4*a^5*b^8*d^3*e + 6*a^6*b^7*d^2*e^2 -
4*a^7*b^6*d*e^3 + a^8*b^5*e^4)*x), 1/640*(15*(2*B*a^5*b*d*e^4 - (B*a^6 + A*a^5*b)*e^5 + (2*B*b^6*d*e^4 - (B*a
*b^5 + A*b^6)*e^5)*x^5 + 5*(2*B*a*b^5*d*e^4 - (B*a^2*b^4 + A*a*b^5)*e^5)*x^4 + 10*(2*B*a^2*b^4*d*e^4 - (B*a^3*
b^3 + A*a^2*b^4)*e^5)*x^3 + 10*(2*B*a^3*b^3*d*e^4 - (B*a^4*b^2 + A*a^3*b^3)*e^5)*x^2 + 5*(2*B*a^4*b^2*d*e^4 -
(B*a^5*b + A*a^4*b^2)*e^5)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) -
(32*(B*a*b^6 + 4*A*b^7)*d^5 - 16*(6*B*a^2*b^5 + 29*A*a*b^6)*d^4*e + 4*(19*B*a^3*b^4 + 146*A*a^2*b^5)*d^3*e^2 +
2*(4*B*a^4*b^3 - 129*A*a^3*b^4)*d^2*e^3 - 5*(7*B*a^5*b^2 + A*a^4*b^3)*d*e^4 + 15*(B*a^6*b + A*a^5*b^2)*e^5 -
15*(2*B*b^7*d^2*e^3 - (3*B*a*b^6 + A*b^7)*d*e^4 + (B*a^2*b^5 + A*a*b^6)*e^5)*x^4 + 10*(2*B*b^7*d^3*e^2 - (17*B
*a*b^6 + A*b^7)*d^2*e^3 + 2*(11*B*a^2*b^5 + 4*A*a*b^6)*d*e^4 - 7*(B*a^3*b^4 + A*a^2*b^5)*e^5)*x^3 + 2*(120*B*b
^7*d^4*e - 2*(227*B*a*b^6 - 2*A*b^7)*d^3*e^2 + 27*(21*B*a^2*b^5 - A*a*b^6)*d^2*e^3 - 3*(99*B*a^3*b^4 - 29*A*a^
2*b^5)*d*e^4 + 64*(B*a^4*b^3 - A*a^3*b^4)*e^5)*x^2 + 2*(80*B*b^7*d^5 - 8*(31*B*a*b^6 - 11*A*b^7)*d^4*e + 2*(10
7*B*a^2*b^5 - 172*A*a*b^6)*d^3*e^2 + (B*a^3*b^4 + 489*A*a^2*b^5)*d^2*e^3 - 2*(41*B*a^4*b^3 + 134*A*a^3*b^4)*d*
e^4 + 35*(B*a^5*b^2 + A*a^4*b^3)*e^5)*x)*sqrt(e*x + d))/(a^5*b^8*d^4 - 4*a^6*b^7*d^3*e + 6*a^7*b^6*d^2*e^2 - 4
*a^8*b^5*d*e^3 + a^9*b^4*e^4 + (b^13*d^4 - 4*a*b^12*d^3*e + 6*a^2*b^11*d^2*e^2 - 4*a^3*b^10*d*e^3 + a^4*b^9*e^
4)*x^5 + 5*(a*b^12*d^4 - 4*a^2*b^11*d^3*e + 6*a^3*b^10*d^2*e^2 - 4*a^4*b^9*d*e^3 + a^5*b^8*e^4)*x^4 + 10*(a^2*
b^11*d^4 - 4*a^3*b^10*d^3*e + 6*a^4*b^9*d^2*e^2 - 4*a^5*b^8*d*e^3 + a^6*b^7*e^4)*x^3 + 10*(a^3*b^10*d^4 - 4*a^
4*b^9*d^3*e + 6*a^5*b^8*d^2*e^2 - 4*a^6*b^7*d*e^3 + a^7*b^6*e^4)*x^2 + 5*(a^4*b^9*d^4 - 4*a^5*b^8*d^3*e + 6*a^
6*b^7*d^2*e^2 - 4*a^7*b^6*d*e^3 + a^8*b^5*e^4)*x)]

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giac [B]  time = 0.27, size = 814, normalized size = 2.60 \begin {gather*} \frac {3 \, {\left (2 \, B b d e^{4} - B a e^{5} - A b e^{5}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{128 \, {\left (b^{6} d^{3} - 3 \, a b^{5} d^{2} e + 3 \, a^{2} b^{4} d e^{2} - a^{3} b^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} + \frac {30 \, {\left (x e + d\right )}^{\frac {9}{2}} B b^{5} d e^{4} - 140 \, {\left (x e + d\right )}^{\frac {7}{2}} B b^{5} d^{2} e^{4} + 140 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{5} d^{4} e^{4} - 30 \, \sqrt {x e + d} B b^{5} d^{5} e^{4} - 15 \, {\left (x e + d\right )}^{\frac {9}{2}} B a b^{4} e^{5} - 15 \, {\left (x e + d\right )}^{\frac {9}{2}} A b^{5} e^{5} + 210 \, {\left (x e + d\right )}^{\frac {7}{2}} B a b^{4} d e^{5} + 70 \, {\left (x e + d\right )}^{\frac {7}{2}} A b^{5} d e^{5} + 128 \, {\left (x e + d\right )}^{\frac {5}{2}} B a b^{4} d^{2} e^{5} - 128 \, {\left (x e + d\right )}^{\frac {5}{2}} A b^{5} d^{2} e^{5} - 490 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{4} d^{3} e^{5} - 70 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{5} d^{3} e^{5} + 135 \, \sqrt {x e + d} B a b^{4} d^{4} e^{5} + 15 \, \sqrt {x e + d} A b^{5} d^{4} e^{5} - 70 \, {\left (x e + d\right )}^{\frac {7}{2}} B a^{2} b^{3} e^{6} - 70 \, {\left (x e + d\right )}^{\frac {7}{2}} A a b^{4} e^{6} - 256 \, {\left (x e + d\right )}^{\frac {5}{2}} B a^{2} b^{3} d e^{6} + 256 \, {\left (x e + d\right )}^{\frac {5}{2}} A a b^{4} d e^{6} + 630 \, {\left (x e + d\right )}^{\frac {3}{2}} B a^{2} b^{3} d^{2} e^{6} + 210 \, {\left (x e + d\right )}^{\frac {3}{2}} A a b^{4} d^{2} e^{6} - 240 \, \sqrt {x e + d} B a^{2} b^{3} d^{3} e^{6} - 60 \, \sqrt {x e + d} A a b^{4} d^{3} e^{6} + 128 \, {\left (x e + d\right )}^{\frac {5}{2}} B a^{3} b^{2} e^{7} - 128 \, {\left (x e + d\right )}^{\frac {5}{2}} A a^{2} b^{3} e^{7} - 350 \, {\left (x e + d\right )}^{\frac {3}{2}} B a^{3} b^{2} d e^{7} - 210 \, {\left (x e + d\right )}^{\frac {3}{2}} A a^{2} b^{3} d e^{7} + 210 \, \sqrt {x e + d} B a^{3} b^{2} d^{2} e^{7} + 90 \, \sqrt {x e + d} A a^{2} b^{3} d^{2} e^{7} + 70 \, {\left (x e + d\right )}^{\frac {3}{2}} B a^{4} b e^{8} + 70 \, {\left (x e + d\right )}^{\frac {3}{2}} A a^{3} b^{2} e^{8} - 90 \, \sqrt {x e + d} B a^{4} b d e^{8} - 60 \, \sqrt {x e + d} A a^{3} b^{2} d e^{8} + 15 \, \sqrt {x e + d} B a^{5} e^{9} + 15 \, \sqrt {x e + d} A a^{4} b e^{9}}{640 \, {\left (b^{6} d^{3} - 3 \, a b^{5} d^{2} e + 3 \, a^{2} b^{4} d e^{2} - a^{3} b^{3} e^{3}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{5}} \end {gather*}

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]

3/128*(2*B*b*d*e^4 - B*a*e^5 - A*b*e^5)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^6*d^3 - 3*a*b^5*d^2*e
+ 3*a^2*b^4*d*e^2 - a^3*b^3*e^3)*sqrt(-b^2*d + a*b*e)) + 1/640*(30*(x*e + d)^(9/2)*B*b^5*d*e^4 - 140*(x*e + d
)^(7/2)*B*b^5*d^2*e^4 + 140*(x*e + d)^(3/2)*B*b^5*d^4*e^4 - 30*sqrt(x*e + d)*B*b^5*d^5*e^4 - 15*(x*e + d)^(9/2
)*B*a*b^4*e^5 - 15*(x*e + d)^(9/2)*A*b^5*e^5 + 210*(x*e + d)^(7/2)*B*a*b^4*d*e^5 + 70*(x*e + d)^(7/2)*A*b^5*d*
e^5 + 128*(x*e + d)^(5/2)*B*a*b^4*d^2*e^5 - 128*(x*e + d)^(5/2)*A*b^5*d^2*e^5 - 490*(x*e + d)^(3/2)*B*a*b^4*d^
3*e^5 - 70*(x*e + d)^(3/2)*A*b^5*d^3*e^5 + 135*sqrt(x*e + d)*B*a*b^4*d^4*e^5 + 15*sqrt(x*e + d)*A*b^5*d^4*e^5
- 70*(x*e + d)^(7/2)*B*a^2*b^3*e^6 - 70*(x*e + d)^(7/2)*A*a*b^4*e^6 - 256*(x*e + d)^(5/2)*B*a^2*b^3*d*e^6 + 25
6*(x*e + d)^(5/2)*A*a*b^4*d*e^6 + 630*(x*e + d)^(3/2)*B*a^2*b^3*d^2*e^6 + 210*(x*e + d)^(3/2)*A*a*b^4*d^2*e^6
- 240*sqrt(x*e + d)*B*a^2*b^3*d^3*e^6 - 60*sqrt(x*e + d)*A*a*b^4*d^3*e^6 + 128*(x*e + d)^(5/2)*B*a^3*b^2*e^7 -
128*(x*e + d)^(5/2)*A*a^2*b^3*e^7 - 350*(x*e + d)^(3/2)*B*a^3*b^2*d*e^7 - 210*(x*e + d)^(3/2)*A*a^2*b^3*d*e^7
+ 210*sqrt(x*e + d)*B*a^3*b^2*d^2*e^7 + 90*sqrt(x*e + d)*A*a^2*b^3*d^2*e^7 + 70*(x*e + d)^(3/2)*B*a^4*b*e^8 +
70*(x*e + d)^(3/2)*A*a^3*b^2*e^8 - 90*sqrt(x*e + d)*B*a^4*b*d*e^8 - 60*sqrt(x*e + d)*A*a^3*b^2*d*e^8 + 15*sqr
t(x*e + d)*B*a^5*e^9 + 15*sqrt(x*e + d)*A*a^4*b*e^9)/((b^6*d^3 - 3*a*b^5*d^2*e + 3*a^2*b^4*d*e^2 - a^3*b^3*e^3
)*((x*e + d)*b - b*d + a*e)^5)

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maple [B]  time = 0.09, size = 871, normalized size = 2.78 \begin {gather*} \frac {3 \left (e x +d \right )^{\frac {9}{2}} A \,b^{2} e^{5}}{128 \left (b e x +a e \right )^{5} \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}+\frac {3 \left (e x +d \right )^{\frac {9}{2}} B a b \,e^{5}}{128 \left (b e x +a e \right )^{5} \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}-\frac {3 \left (e x +d \right )^{\frac {9}{2}} B \,b^{2} d \,e^{4}}{64 \left (b e x +a e \right )^{5} \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}+\frac {7 \left (e x +d \right )^{\frac {7}{2}} A b \,e^{5}}{64 \left (b e x +a e \right )^{5} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {7 \left (e x +d \right )^{\frac {7}{2}} B a \,e^{5}}{64 \left (b e x +a e \right )^{5} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}-\frac {7 \left (e x +d \right )^{\frac {7}{2}} B b d \,e^{4}}{32 \left (b e x +a e \right )^{5} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}-\frac {3 \sqrt {e x +d}\, A a \,e^{6}}{128 \left (b e x +a e \right )^{5} b^{2}}+\frac {3 \sqrt {e x +d}\, A d \,e^{5}}{128 \left (b e x +a e \right )^{5} b}+\frac {\left (e x +d \right )^{\frac {5}{2}} A \,e^{5}}{5 \left (b e x +a e \right )^{5} \left (a e -b d \right )}-\frac {3 \sqrt {e x +d}\, B \,a^{2} e^{6}}{128 \left (b e x +a e \right )^{5} b^{3}}-\frac {\left (e x +d \right )^{\frac {5}{2}} B a \,e^{5}}{5 \left (b e x +a e \right )^{5} \left (a e -b d \right ) b}+\frac {9 \sqrt {e x +d}\, B a d \,e^{5}}{128 \left (b e x +a e \right )^{5} b^{2}}-\frac {3 \sqrt {e x +d}\, B \,d^{2} e^{4}}{64 \left (b e x +a e \right )^{5} b}-\frac {7 \left (e x +d \right )^{\frac {3}{2}} A \,e^{5}}{64 \left (b e x +a e \right )^{5} b}+\frac {3 A \,e^{5} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{128 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {7 \left (e x +d \right )^{\frac {3}{2}} B a \,e^{5}}{64 \left (b e x +a e \right )^{5} b^{2}}+\frac {3 B a \,e^{5} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{128 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {\left (a e -b d \right ) b}\, b^{3}}+\frac {7 \left (e x +d \right )^{\frac {3}{2}} B d \,e^{4}}{32 \left (b e x +a e \right )^{5} b}-\frac {3 B d \,e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{64 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {\left (a e -b d \right ) b}\, b^{2}} \end {gather*}

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]

3/128*e^5/(b*e*x+a*e)^5*b^2/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)*(e*x+d)^(9/2)*A+3/128*e^5/(b*e*x+a*e
)^5*b/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)*(e*x+d)^(9/2)*a*B-3/64*e^4/(b*e*x+a*e)^5*b^2/(a^3*e^3-3*a^
2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)*(e*x+d)^(9/2)*B*d+7/64*e^5/(b*e*x+a*e)^5/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^
(7/2)*A*b+7/64*e^5/(b*e*x+a*e)^5/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(7/2)*a*B-7/32*e^4/(b*e*x+a*e)^5/(a^2*e^2
-2*a*b*d*e+b^2*d^2)*(e*x+d)^(7/2)*B*b*d+1/5*e^5/(b*e*x+a*e)^5/(a*e-b*d)*(e*x+d)^(5/2)*A-1/5*e^5/(b*e*x+a*e)^5/
(a*e-b*d)/b*(e*x+d)^(5/2)*B*a-7/64*e^5/(b*e*x+a*e)^5/b*(e*x+d)^(3/2)*A-7/64*e^5/(b*e*x+a*e)^5/b^2*(e*x+d)^(3/2
)*a*B+7/32*e^4/(b*e*x+a*e)^5/b*(e*x+d)^(3/2)*B*d-3/128*e^6/(b*e*x+a*e)^5/b^2*(e*x+d)^(1/2)*A*a+3/128*e^5/(b*e*
x+a*e)^5/b*(e*x+d)^(1/2)*A*d-3/128*e^6/(b*e*x+a*e)^5/b^3*(e*x+d)^(1/2)*B*a^2+9/128*e^5/(b*e*x+a*e)^5/b^2*(e*x+
d)^(1/2)*B*a*d-3/64*e^4/(b*e*x+a*e)^5/b*(e*x+d)^(1/2)*B*d^2+3/128*e^5/b^2/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e
-b^3*d^3)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A+3/128*e^5/b^3/(a^3*e^3-3*a^2*b*d*e
^2+3*a*b^2*d^2*e-b^3*d^3)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*B-3/64*e^4/b^2/(a^
3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*d

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

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 = 2.23, size = 535, normalized size = 1.71 \begin {gather*} \frac {\frac {7\,{\left (d+e\,x\right )}^{7/2}\,\left (A\,b\,e^5+B\,a\,e^5-2\,B\,b\,d\,e^4\right )}{64\,{\left (a\,e-b\,d\right )}^2}-\frac {7\,{\left (d+e\,x\right )}^{3/2}\,\left (A\,b\,e^5+B\,a\,e^5-2\,B\,b\,d\,e^4\right )}{64\,b^2}-\frac {3\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}\,\left (A\,b\,e^5+B\,a\,e^5-2\,B\,b\,d\,e^4\right )}{128\,b^3}+\frac {3\,b\,{\left (d+e\,x\right )}^{9/2}\,\left (A\,b\,e^5+B\,a\,e^5-2\,B\,b\,d\,e^4\right )}{128\,{\left (a\,e-b\,d\right )}^3}+\frac {\left (A\,b\,e^5-B\,a\,e^5\right )\,{\left (d+e\,x\right )}^{5/2}}{5\,b\,\left (a\,e-b\,d\right )}}{\left (d+e\,x\right )\,\left (5\,a^4\,b\,e^4-20\,a^3\,b^2\,d\,e^3+30\,a^2\,b^3\,d^2\,e^2-20\,a\,b^4\,d^3\,e+5\,b^5\,d^4\right )-{\left (d+e\,x\right )}^2\,\left (-10\,a^3\,b^2\,e^3+30\,a^2\,b^3\,d\,e^2-30\,a\,b^4\,d^2\,e+10\,b^5\,d^3\right )+b^5\,{\left (d+e\,x\right )}^5-\left (5\,b^5\,d-5\,a\,b^4\,e\right )\,{\left (d+e\,x\right )}^4+a^5\,e^5-b^5\,d^5+{\left (d+e\,x\right )}^3\,\left (10\,a^2\,b^3\,e^2-20\,a\,b^4\,d\,e+10\,b^5\,d^2\right )-10\,a^2\,b^3\,d^3\,e^2+10\,a^3\,b^2\,d^2\,e^3+5\,a\,b^4\,d^4\,e-5\,a^4\,b\,d\,e^4}+\frac {3\,e^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^4\,\sqrt {d+e\,x}\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{\sqrt {a\,e-b\,d}\,\left (A\,b\,e^5+B\,a\,e^5-2\,B\,b\,d\,e^4\right )}\right )\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{128\,b^{7/2}\,{\left (a\,e-b\,d\right )}^{7/2}} \end {gather*}

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]

((7*(d + e*x)^(7/2)*(A*b*e^5 + B*a*e^5 - 2*B*b*d*e^4))/(64*(a*e - b*d)^2) - (7*(d + e*x)^(3/2)*(A*b*e^5 + B*a*
e^5 - 2*B*b*d*e^4))/(64*b^2) - (3*(a*e - b*d)*(d + e*x)^(1/2)*(A*b*e^5 + B*a*e^5 - 2*B*b*d*e^4))/(128*b^3) + (
3*b*(d + e*x)^(9/2)*(A*b*e^5 + B*a*e^5 - 2*B*b*d*e^4))/(128*(a*e - b*d)^3) + ((A*b*e^5 - B*a*e^5)*(d + e*x)^(5
/2))/(5*b*(a*e - b*d)))/((d + e*x)*(5*b^5*d^4 + 5*a^4*b*e^4 - 20*a^3*b^2*d*e^3 + 30*a^2*b^3*d^2*e^2 - 20*a*b^4
*d^3*e) - (d + e*x)^2*(10*b^5*d^3 - 10*a^3*b^2*e^3 + 30*a^2*b^3*d*e^2 - 30*a*b^4*d^2*e) + b^5*(d + e*x)^5 - (5
*b^5*d - 5*a*b^4*e)*(d + e*x)^4 + a^5*e^5 - b^5*d^5 + (d + e*x)^3*(10*b^5*d^2 + 10*a^2*b^3*e^2 - 20*a*b^4*d*e)
- 10*a^2*b^3*d^3*e^2 + 10*a^3*b^2*d^2*e^3 + 5*a*b^4*d^4*e - 5*a^4*b*d*e^4) + (3*e^4*atan((b^(1/2)*e^4*(d + e*
x)^(1/2)*(A*b*e + B*a*e - 2*B*b*d))/((a*e - b*d)^(1/2)*(A*b*e^5 + B*a*e^5 - 2*B*b*d*e^4)))*(A*b*e + B*a*e - 2*
B*b*d))/(128*b^(7/2)*(a*e - b*d)^(7/2))

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

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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