3.19.49 \(\int \frac {(a+b x) (d+e x)^{9/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=175 \[ -\frac {63 e^2 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{11/2}}+\frac {63 e^2 \sqrt {d+e x} (b d-a e)^2}{4 b^5}+\frac {21 e^2 (d+e x)^{3/2} (b d-a e)}{4 b^4}-\frac {9 e (d+e x)^{7/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}+\frac {63 e^2 (d+e x)^{5/2}}{20 b^3} \]

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Rubi [A]  time = 0.11, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 47, 50, 63, 208} \begin {gather*} \frac {21 e^2 (d+e x)^{3/2} (b d-a e)}{4 b^4}+\frac {63 e^2 \sqrt {d+e x} (b d-a e)^2}{4 b^5}-\frac {63 e^2 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{11/2}}-\frac {9 e (d+e x)^{7/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}+\frac {63 e^2 (d+e x)^{5/2}}{20 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^(9/2))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(63*e^2*(b*d - a*e)^2*Sqrt[d + e*x])/(4*b^5) + (21*e^2*(b*d - a*e)*(d + e*x)^(3/2))/(4*b^4) + (63*e^2*(d + e*x
)^(5/2))/(20*b^3) - (9*e*(d + e*x)^(7/2))/(4*b^2*(a + b*x)) - (d + e*x)^(9/2)/(2*b*(a + b*x)^2) - (63*e^2*(b*d
- a*e)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*b^(11/2))

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {(d+e x)^{9/2}}{(a+b x)^3} \, dx\\ &=-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}+\frac {(9 e) \int \frac {(d+e x)^{7/2}}{(a+b x)^2} \, dx}{4 b}\\ &=-\frac {9 e (d+e x)^{7/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}+\frac {\left (63 e^2\right ) \int \frac {(d+e x)^{5/2}}{a+b x} \, dx}{8 b^2}\\ &=\frac {63 e^2 (d+e x)^{5/2}}{20 b^3}-\frac {9 e (d+e x)^{7/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}+\frac {\left (63 e^2 (b d-a e)\right ) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{8 b^3}\\ &=\frac {21 e^2 (b d-a e) (d+e x)^{3/2}}{4 b^4}+\frac {63 e^2 (d+e x)^{5/2}}{20 b^3}-\frac {9 e (d+e x)^{7/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}+\frac {\left (63 e^2 (b d-a e)^2\right ) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{8 b^4}\\ &=\frac {63 e^2 (b d-a e)^2 \sqrt {d+e x}}{4 b^5}+\frac {21 e^2 (b d-a e) (d+e x)^{3/2}}{4 b^4}+\frac {63 e^2 (d+e x)^{5/2}}{20 b^3}-\frac {9 e (d+e x)^{7/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}+\frac {\left (63 e^2 (b d-a e)^3\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 b^5}\\ &=\frac {63 e^2 (b d-a e)^2 \sqrt {d+e x}}{4 b^5}+\frac {21 e^2 (b d-a e) (d+e x)^{3/2}}{4 b^4}+\frac {63 e^2 (d+e x)^{5/2}}{20 b^3}-\frac {9 e (d+e x)^{7/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}+\frac {\left (63 e (b d-a 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 )}{4 b^5}\\ &=\frac {63 e^2 (b d-a e)^2 \sqrt {d+e x}}{4 b^5}+\frac {21 e^2 (b d-a e) (d+e x)^{3/2}}{4 b^4}+\frac {63 e^2 (d+e x)^{5/2}}{20 b^3}-\frac {9 e (d+e x)^{7/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}-\frac {63 e^2 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.30 \begin {gather*} \frac {2 e^2 (d+e x)^{11/2} \, _2F_1\left (3,\frac {11}{2};\frac {13}{2};-\frac {b (d+e x)}{a e-b d}\right )}{11 (a e-b d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^(9/2))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(2*e^2*(d + e*x)^(11/2)*Hypergeometric2F1[3, 11/2, 13/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(11*(-(b*d) + a*e)^
3)

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IntegrateAlgebraic [A]  time = 0.80, size = 306, normalized size = 1.75 \begin {gather*} \frac {e^2 \sqrt {d+e x} \left (315 a^4 e^4+525 a^3 b e^3 (d+e x)-1260 a^3 b d e^3+1890 a^2 b^2 d^2 e^2+168 a^2 b^2 e^2 (d+e x)^2-1575 a^2 b^2 d e^2 (d+e x)-1260 a b^3 d^3 e+1575 a b^3 d^2 e (d+e x)-24 a b^3 e (d+e x)^3-336 a b^3 d e (d+e x)^2+315 b^4 d^4-525 b^4 d^3 (d+e x)+168 b^4 d^2 (d+e x)^2+8 b^4 (d+e x)^4+24 b^4 d (d+e x)^3\right )}{20 b^5 (a e+b (d+e x)-b d)^2}-\frac {63 e^2 (b d-a e)^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 b^{11/2} \sqrt {a e-b d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)*(d + e*x)^(9/2))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(e^2*Sqrt[d + e*x]*(315*b^4*d^4 - 1260*a*b^3*d^3*e + 1890*a^2*b^2*d^2*e^2 - 1260*a^3*b*d*e^3 + 315*a^4*e^4 - 5
25*b^4*d^3*(d + e*x) + 1575*a*b^3*d^2*e*(d + e*x) - 1575*a^2*b^2*d*e^2*(d + e*x) + 525*a^3*b*e^3*(d + e*x) + 1
68*b^4*d^2*(d + e*x)^2 - 336*a*b^3*d*e*(d + e*x)^2 + 168*a^2*b^2*e^2*(d + e*x)^2 + 24*b^4*d*(d + e*x)^3 - 24*a
*b^3*e*(d + e*x)^3 + 8*b^4*(d + e*x)^4))/(20*b^5*(-(b*d) + a*e + b*(d + e*x))^2) - (63*e^2*(b*d - a*e)^3*ArcTa
n[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(4*b^(11/2)*Sqrt[-(b*d) + a*e])

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fricas [B]  time = 0.44, size = 730, normalized size = 4.17 \begin {gather*} \left [\frac {315 \, {\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} b d e^{3} + a^{4} e^{4} + {\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \, {\left (a b^{3} d^{2} e^{2} - 2 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (8 \, b^{4} e^{4} x^{4} - 10 \, b^{4} d^{4} - 45 \, a b^{3} d^{3} e + 483 \, a^{2} b^{2} d^{2} e^{2} - 735 \, a^{3} b d e^{3} + 315 \, a^{4} e^{4} + 8 \, {\left (7 \, b^{4} d e^{3} - 3 \, a b^{3} e^{4}\right )} x^{3} + 24 \, {\left (12 \, b^{4} d^{2} e^{2} - 17 \, a b^{3} d e^{3} + 7 \, a^{2} b^{2} e^{4}\right )} x^{2} - {\left (85 \, b^{4} d^{3} e - 831 \, a b^{3} d^{2} e^{2} + 1239 \, a^{2} b^{2} d e^{3} - 525 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{40 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac {315 \, {\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} b d e^{3} + a^{4} e^{4} + {\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \, {\left (a b^{3} d^{2} e^{2} - 2 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (8 \, b^{4} e^{4} x^{4} - 10 \, b^{4} d^{4} - 45 \, a b^{3} d^{3} e + 483 \, a^{2} b^{2} d^{2} e^{2} - 735 \, a^{3} b d e^{3} + 315 \, a^{4} e^{4} + 8 \, {\left (7 \, b^{4} d e^{3} - 3 \, a b^{3} e^{4}\right )} x^{3} + 24 \, {\left (12 \, b^{4} d^{2} e^{2} - 17 \, a b^{3} d e^{3} + 7 \, a^{2} b^{2} e^{4}\right )} x^{2} - {\left (85 \, b^{4} d^{3} e - 831 \, a b^{3} d^{2} e^{2} + 1239 \, a^{2} b^{2} d e^{3} - 525 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{20 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[1/40*(315*(a^2*b^2*d^2*e^2 - 2*a^3*b*d*e^3 + a^4*e^4 + (b^4*d^2*e^2 - 2*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 2*(a
*b^3*d^2*e^2 - 2*a^2*b^2*d*e^3 + a^3*b*e^4)*x)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*
b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(8*b^4*e^4*x^4 - 10*b^4*d^4 - 45*a*b^3*d^3*e + 483*a^2*b^2*d^2*e^2 - 735
*a^3*b*d*e^3 + 315*a^4*e^4 + 8*(7*b^4*d*e^3 - 3*a*b^3*e^4)*x^3 + 24*(12*b^4*d^2*e^2 - 17*a*b^3*d*e^3 + 7*a^2*b
^2*e^4)*x^2 - (85*b^4*d^3*e - 831*a*b^3*d^2*e^2 + 1239*a^2*b^2*d*e^3 - 525*a^3*b*e^4)*x)*sqrt(e*x + d))/(b^7*x
^2 + 2*a*b^6*x + a^2*b^5), -1/20*(315*(a^2*b^2*d^2*e^2 - 2*a^3*b*d*e^3 + a^4*e^4 + (b^4*d^2*e^2 - 2*a*b^3*d*e^
3 + a^2*b^2*e^4)*x^2 + 2*(a*b^3*d^2*e^2 - 2*a^2*b^2*d*e^3 + a^3*b*e^4)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*
x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (8*b^4*e^4*x^4 - 10*b^4*d^4 - 45*a*b^3*d^3*e + 483*a^2*b^2*d^2*e^
2 - 735*a^3*b*d*e^3 + 315*a^4*e^4 + 8*(7*b^4*d*e^3 - 3*a*b^3*e^4)*x^3 + 24*(12*b^4*d^2*e^2 - 17*a*b^3*d*e^3 +
7*a^2*b^2*e^4)*x^2 - (85*b^4*d^3*e - 831*a*b^3*d^2*e^2 + 1239*a^2*b^2*d*e^3 - 525*a^3*b*e^4)*x)*sqrt(e*x + d))
/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)]

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giac [B]  time = 0.22, size = 374, normalized size = 2.14 \begin {gather*} \frac {63 \, {\left (b^{3} d^{3} e^{2} - 3 \, a b^{2} d^{2} e^{3} + 3 \, a^{2} b d e^{4} - a^{3} e^{5}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{5}} - \frac {17 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} d^{3} e^{2} - 15 \, \sqrt {x e + d} b^{4} d^{4} e^{2} - 51 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{3} d^{2} e^{3} + 60 \, \sqrt {x e + d} a b^{3} d^{3} e^{3} + 51 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b^{2} d e^{4} - 90 \, \sqrt {x e + d} a^{2} b^{2} d^{2} e^{4} - 17 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{3} b e^{5} + 60 \, \sqrt {x e + d} a^{3} b d e^{5} - 15 \, \sqrt {x e + d} a^{4} e^{6}}{4 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{5}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {5}{2}} b^{12} e^{2} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{12} d e^{2} + 30 \, \sqrt {x e + d} b^{12} d^{2} e^{2} - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{11} e^{3} - 60 \, \sqrt {x e + d} a b^{11} d e^{3} + 30 \, \sqrt {x e + d} a^{2} b^{10} e^{4}\right )}}{5 \, b^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

63/4*(b^3*d^3*e^2 - 3*a*b^2*d^2*e^3 + 3*a^2*b*d*e^4 - a^3*e^5)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(s
qrt(-b^2*d + a*b*e)*b^5) - 1/4*(17*(x*e + d)^(3/2)*b^4*d^3*e^2 - 15*sqrt(x*e + d)*b^4*d^4*e^2 - 51*(x*e + d)^(
3/2)*a*b^3*d^2*e^3 + 60*sqrt(x*e + d)*a*b^3*d^3*e^3 + 51*(x*e + d)^(3/2)*a^2*b^2*d*e^4 - 90*sqrt(x*e + d)*a^2*
b^2*d^2*e^4 - 17*(x*e + d)^(3/2)*a^3*b*e^5 + 60*sqrt(x*e + d)*a^3*b*d*e^5 - 15*sqrt(x*e + d)*a^4*e^6)/(((x*e +
d)*b - b*d + a*e)^2*b^5) + 2/5*((x*e + d)^(5/2)*b^12*e^2 + 5*(x*e + d)^(3/2)*b^12*d*e^2 + 30*sqrt(x*e + d)*b^
12*d^2*e^2 - 5*(x*e + d)^(3/2)*a*b^11*e^3 - 60*sqrt(x*e + d)*a*b^11*d*e^3 + 30*sqrt(x*e + d)*a^2*b^10*e^4)/b^1
5

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maple [B]  time = 0.10, size = 543, normalized size = 3.10 \begin {gather*} \frac {15 \sqrt {e x +d}\, a^{4} e^{6}}{4 \left (b e x +a e \right )^{2} b^{5}}-\frac {15 \sqrt {e x +d}\, a^{3} d \,e^{5}}{\left (b e x +a e \right )^{2} b^{4}}+\frac {45 \sqrt {e x +d}\, a^{2} d^{2} e^{4}}{2 \left (b e x +a e \right )^{2} b^{3}}-\frac {15 \sqrt {e x +d}\, a \,d^{3} e^{3}}{\left (b e x +a e \right )^{2} b^{2}}+\frac {15 \sqrt {e x +d}\, d^{4} e^{2}}{4 \left (b e x +a e \right )^{2} b}+\frac {17 \left (e x +d \right )^{\frac {3}{2}} a^{3} e^{5}}{4 \left (b e x +a e \right )^{2} b^{4}}-\frac {63 a^{3} e^{5} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{5}}-\frac {51 \left (e x +d \right )^{\frac {3}{2}} a^{2} d \,e^{4}}{4 \left (b e x +a e \right )^{2} b^{3}}+\frac {189 a^{2} d \,e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{4}}+\frac {51 \left (e x +d \right )^{\frac {3}{2}} a \,d^{2} e^{3}}{4 \left (b e x +a e \right )^{2} b^{2}}-\frac {189 a \,d^{2} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{3}}-\frac {17 \left (e x +d \right )^{\frac {3}{2}} d^{3} e^{2}}{4 \left (b e x +a e \right )^{2} b}+\frac {63 d^{3} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{2}}+\frac {12 \sqrt {e x +d}\, a^{2} e^{4}}{b^{5}}-\frac {24 \sqrt {e x +d}\, a d \,e^{3}}{b^{4}}+\frac {12 \sqrt {e x +d}\, d^{2} e^{2}}{b^{3}}-\frac {2 \left (e x +d \right )^{\frac {3}{2}} a \,e^{3}}{b^{4}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} d \,e^{2}}{b^{3}}+\frac {2 \left (e x +d \right )^{\frac {5}{2}} e^{2}}{5 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

2/5*e^2*(e*x+d)^(5/2)/b^3-2*e^3/b^4*(e*x+d)^(3/2)*a+2*e^2/b^3*(e*x+d)^(3/2)*d+12*e^4/b^5*a^2*(e*x+d)^(1/2)-24*
e^3/b^4*a*d*(e*x+d)^(1/2)+12*e^2/b^3*d^2*(e*x+d)^(1/2)+17/4*e^5/b^4/(b*e*x+a*e)^2*(e*x+d)^(3/2)*a^3-51/4*e^4/b
^3/(b*e*x+a*e)^2*(e*x+d)^(3/2)*a^2*d+51/4*e^3/b^2/(b*e*x+a*e)^2*(e*x+d)^(3/2)*a*d^2-17/4*e^2/b/(b*e*x+a*e)^2*(
e*x+d)^(3/2)*d^3+15/4*e^6/b^5/(b*e*x+a*e)^2*(e*x+d)^(1/2)*a^4-15*e^5/b^4/(b*e*x+a*e)^2*(e*x+d)^(1/2)*a^3*d+45/
2*e^4/b^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*a^2*d^2-15*e^3/b^2/(b*e*x+a*e)^2*(e*x+d)^(1/2)*a*d^3+15/4*e^2/b/(b*e*x+a
*e)^2*(e*x+d)^(1/2)*d^4-63/4*e^5/b^5/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^3+189/4
*e^4/b^4/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^2*d-189/4*e^3/b^3/((a*e-b*d)*b)^(1/
2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*d^2+63/4*e^2/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a
*e-b*d)*b)^(1/2)*b)*d^3

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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)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^2,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 = 0.14, size = 361, normalized size = 2.06 \begin {gather*} \left (\frac {2\,e^2\,{\left (3\,b^3\,d-3\,a\,b^2\,e\right )}^2}{b^9}-\frac {6\,e^2\,{\left (a\,e-b\,d\right )}^2}{b^5}\right )\,\sqrt {d+e\,x}+\frac {{\left (d+e\,x\right )}^{3/2}\,\left (\frac {17\,a^3\,b\,e^5}{4}-\frac {51\,a^2\,b^2\,d\,e^4}{4}+\frac {51\,a\,b^3\,d^2\,e^3}{4}-\frac {17\,b^4\,d^3\,e^2}{4}\right )+\sqrt {d+e\,x}\,\left (\frac {15\,a^4\,e^6}{4}-15\,a^3\,b\,d\,e^5+\frac {45\,a^2\,b^2\,d^2\,e^4}{2}-15\,a\,b^3\,d^3\,e^3+\frac {15\,b^4\,d^4\,e^2}{4}\right )}{b^7\,{\left (d+e\,x\right )}^2-\left (2\,b^7\,d-2\,a\,b^6\,e\right )\,\left (d+e\,x\right )+b^7\,d^2+a^2\,b^5\,e^2-2\,a\,b^6\,d\,e}+\frac {2\,e^2\,{\left (d+e\,x\right )}^{5/2}}{5\,b^3}+\frac {2\,e^2\,\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,b^6}-\frac {63\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^2\,{\left (a\,e-b\,d\right )}^{5/2}\,\sqrt {d+e\,x}}{a^3\,e^5-3\,a^2\,b\,d\,e^4+3\,a\,b^2\,d^2\,e^3-b^3\,d^3\,e^2}\right )\,{\left (a\,e-b\,d\right )}^{5/2}}{4\,b^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)*(d + e*x)^(9/2))/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)

[Out]

((2*e^2*(3*b^3*d - 3*a*b^2*e)^2)/b^9 - (6*e^2*(a*e - b*d)^2)/b^5)*(d + e*x)^(1/2) + ((d + e*x)^(3/2)*((17*a^3*
b*e^5)/4 - (17*b^4*d^3*e^2)/4 + (51*a*b^3*d^2*e^3)/4 - (51*a^2*b^2*d*e^4)/4) + (d + e*x)^(1/2)*((15*a^4*e^6)/4
+ (15*b^4*d^4*e^2)/4 - 15*a*b^3*d^3*e^3 + (45*a^2*b^2*d^2*e^4)/2 - 15*a^3*b*d*e^5))/(b^7*(d + e*x)^2 - (2*b^7
*d - 2*a*b^6*e)*(d + e*x) + b^7*d^2 + a^2*b^5*e^2 - 2*a*b^6*d*e) + (2*e^2*(d + e*x)^(5/2))/(5*b^3) + (2*e^2*(3
*b^3*d - 3*a*b^2*e)*(d + e*x)^(3/2))/(3*b^6) - (63*e^2*atan((b^(1/2)*e^2*(a*e - b*d)^(5/2)*(d + e*x)^(1/2))/(a
^3*e^5 - b^3*d^3*e^2 + 3*a*b^2*d^2*e^3 - 3*a^2*b*d*e^4))*(a*e - b*d)^(5/2))/(4*b^(11/2))

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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)**(9/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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