3.14.70 \(\int \frac {(d+e x)^{7/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=188 \[ \frac {7 e^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{9/2} (b d-a e)^{3/2}}-\frac {7 e^4 \sqrt {d+e x}}{128 b^4 (a+b x) (b d-a e)}-\frac {7 e^3 \sqrt {d+e x}}{64 b^4 (a+b x)^2}-\frac {7 e^2 (d+e x)^{3/2}}{48 b^3 (a+b x)^3}-\frac {7 e (d+e x)^{5/2}}{40 b^2 (a+b x)^4}-\frac {(d+e x)^{7/2}}{5 b (a+b x)^5} \]

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Rubi [A]  time = 0.11, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 47, 51, 63, 208} \begin {gather*} -\frac {7 e^4 \sqrt {d+e x}}{128 b^4 (a+b x) (b d-a e)}-\frac {7 e^3 \sqrt {d+e x}}{64 b^4 (a+b x)^2}-\frac {7 e^2 (d+e x)^{3/2}}{48 b^3 (a+b x)^3}+\frac {7 e^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{9/2} (b d-a e)^{3/2}}-\frac {7 e (d+e x)^{5/2}}{40 b^2 (a+b x)^4}-\frac {(d+e x)^{7/2}}{5 b (a+b x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(7/2)/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(-7*e^3*Sqrt[d + e*x])/(64*b^4*(a + b*x)^2) - (7*e^4*Sqrt[d + e*x])/(128*b^4*(b*d - a*e)*(a + b*x)) - (7*e^2*(
d + e*x)^(3/2))/(48*b^3*(a + b*x)^3) - (7*e*(d + e*x)^(5/2))/(40*b^2*(a + b*x)^4) - (d + e*x)^(7/2)/(5*b*(a +
b*x)^5) + (7*e^5*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(128*b^(9/2)*(b*d - a*e)^(3/2))

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 {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac {(d+e x)^{7/2}}{(a+b x)^6} \, dx\\ &=-\frac {(d+e x)^{7/2}}{5 b (a+b x)^5}+\frac {(7 e) \int \frac {(d+e x)^{5/2}}{(a+b x)^5} \, dx}{10 b}\\ &=-\frac {7 e (d+e x)^{5/2}}{40 b^2 (a+b x)^4}-\frac {(d+e x)^{7/2}}{5 b (a+b x)^5}+\frac {\left (7 e^2\right ) \int \frac {(d+e x)^{3/2}}{(a+b x)^4} \, dx}{16 b^2}\\ &=-\frac {7 e^2 (d+e x)^{3/2}}{48 b^3 (a+b x)^3}-\frac {7 e (d+e x)^{5/2}}{40 b^2 (a+b x)^4}-\frac {(d+e x)^{7/2}}{5 b (a+b x)^5}+\frac {\left (7 e^3\right ) \int \frac {\sqrt {d+e x}}{(a+b x)^3} \, dx}{32 b^3}\\ &=-\frac {7 e^3 \sqrt {d+e x}}{64 b^4 (a+b x)^2}-\frac {7 e^2 (d+e x)^{3/2}}{48 b^3 (a+b x)^3}-\frac {7 e (d+e x)^{5/2}}{40 b^2 (a+b x)^4}-\frac {(d+e x)^{7/2}}{5 b (a+b x)^5}+\frac {\left (7 e^4\right ) \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{128 b^4}\\ &=-\frac {7 e^3 \sqrt {d+e x}}{64 b^4 (a+b x)^2}-\frac {7 e^4 \sqrt {d+e x}}{128 b^4 (b d-a e) (a+b x)}-\frac {7 e^2 (d+e x)^{3/2}}{48 b^3 (a+b x)^3}-\frac {7 e (d+e x)^{5/2}}{40 b^2 (a+b x)^4}-\frac {(d+e x)^{7/2}}{5 b (a+b x)^5}-\frac {\left (7 e^5\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{256 b^4 (b d-a e)}\\ &=-\frac {7 e^3 \sqrt {d+e x}}{64 b^4 (a+b x)^2}-\frac {7 e^4 \sqrt {d+e x}}{128 b^4 (b d-a e) (a+b x)}-\frac {7 e^2 (d+e x)^{3/2}}{48 b^3 (a+b x)^3}-\frac {7 e (d+e x)^{5/2}}{40 b^2 (a+b x)^4}-\frac {(d+e x)^{7/2}}{5 b (a+b x)^5}-\frac {\left (7 e^4\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^4 (b d-a e)}\\ &=-\frac {7 e^3 \sqrt {d+e x}}{64 b^4 (a+b x)^2}-\frac {7 e^4 \sqrt {d+e x}}{128 b^4 (b d-a e) (a+b x)}-\frac {7 e^2 (d+e x)^{3/2}}{48 b^3 (a+b x)^3}-\frac {7 e (d+e x)^{5/2}}{40 b^2 (a+b x)^4}-\frac {(d+e x)^{7/2}}{5 b (a+b x)^5}+\frac {7 e^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{9/2} (b d-a e)^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.28 \begin {gather*} \frac {2 e^5 (d+e x)^{9/2} \, _2F_1\left (\frac {9}{2},6;\frac {11}{2};-\frac {b (d+e x)}{a e-b d}\right )}{9 (a e-b d)^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(7/2)/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(2*e^5*(d + e*x)^(9/2)*Hypergeometric2F1[9/2, 6, 11/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(9*(-(b*d) + a*e)^6)

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IntegrateAlgebraic [A]  time = 1.85, size = 307, normalized size = 1.63 \begin {gather*} -\frac {e^5 \sqrt {d+e x} \left (105 a^4 e^4+490 a^3 b e^3 (d+e x)-420 a^3 b d e^3+630 a^2 b^2 d^2 e^2+896 a^2 b^2 e^2 (d+e x)^2-1470 a^2 b^2 d e^2 (d+e x)-420 a b^3 d^3 e+1470 a b^3 d^2 e (d+e x)+790 a b^3 e (d+e x)^3-1792 a b^3 d e (d+e x)^2+105 b^4 d^4-490 b^4 d^3 (d+e x)+896 b^4 d^2 (d+e x)^2-105 b^4 (d+e x)^4-790 b^4 d (d+e x)^3\right )}{1920 b^4 (b d-a e) (-a e-b (d+e x)+b d)^5}-\frac {7 e^5 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{128 b^{9/2} (a e-b d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(7/2)/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

-1/1920*(e^5*Sqrt[d + e*x]*(105*b^4*d^4 - 420*a*b^3*d^3*e + 630*a^2*b^2*d^2*e^2 - 420*a^3*b*d*e^3 + 105*a^4*e^
4 - 490*b^4*d^3*(d + e*x) + 1470*a*b^3*d^2*e*(d + e*x) - 1470*a^2*b^2*d*e^2*(d + e*x) + 490*a^3*b*e^3*(d + e*x
) + 896*b^4*d^2*(d + e*x)^2 - 1792*a*b^3*d*e*(d + e*x)^2 + 896*a^2*b^2*e^2*(d + e*x)^2 - 790*b^4*d*(d + e*x)^3
+ 790*a*b^3*e*(d + e*x)^3 - 105*b^4*(d + e*x)^4))/(b^4*(b*d - a*e)*(b*d - a*e - b*(d + e*x))^5) - (7*e^5*ArcT
an[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(128*b^(9/2)*(-(b*d) + a*e)^(3/2))

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fricas [B]  time = 0.43, size = 1158, normalized size = 6.16 \begin {gather*} \left [-\frac {105 \, {\left (b^{5} e^{5} x^{5} + 5 \, a b^{4} e^{5} x^{4} + 10 \, a^{2} b^{3} e^{5} x^{3} + 10 \, a^{3} b^{2} e^{5} x^{2} + 5 \, a^{4} b e^{5} x + a^{5} e^{5}\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 (384 \, b^{6} d^{5} - 432 \, a b^{5} d^{4} e - 8 \, a^{2} b^{4} d^{3} e^{2} - 14 \, a^{3} b^{3} d^{2} e^{3} - 35 \, a^{4} b^{2} d e^{4} + 105 \, a^{5} b e^{5} + 105 \, {\left (b^{6} d e^{4} - a b^{5} e^{5}\right )} x^{4} + 10 \, {\left (121 \, b^{6} d^{2} e^{3} - 200 \, a b^{5} d e^{4} + 79 \, a^{2} b^{4} e^{5}\right )} x^{3} + 2 \, {\left (1052 \, b^{6} d^{3} e^{2} - 1341 \, a b^{5} d^{2} e^{3} - 159 \, a^{2} b^{4} d e^{4} + 448 \, a^{3} b^{3} e^{5}\right )} x^{2} + 2 \, {\left (744 \, b^{6} d^{4} e - 872 \, a b^{5} d^{3} e^{2} - 33 \, a^{2} b^{4} d^{2} e^{3} - 84 \, a^{3} b^{3} d e^{4} + 245 \, a^{4} b^{2} e^{5}\right )} x\right )} \sqrt {e x + d}}{3840 \, {\left (a^{5} b^{7} d^{2} - 2 \, a^{6} b^{6} d e + a^{7} b^{5} e^{2} + {\left (b^{12} d^{2} - 2 \, a b^{11} d e + a^{2} b^{10} e^{2}\right )} x^{5} + 5 \, {\left (a b^{11} d^{2} - 2 \, a^{2} b^{10} d e + a^{3} b^{9} e^{2}\right )} x^{4} + 10 \, {\left (a^{2} b^{10} d^{2} - 2 \, a^{3} b^{9} d e + a^{4} b^{8} e^{2}\right )} x^{3} + 10 \, {\left (a^{3} b^{9} d^{2} - 2 \, a^{4} b^{8} d e + a^{5} b^{7} e^{2}\right )} x^{2} + 5 \, {\left (a^{4} b^{8} d^{2} - 2 \, a^{5} b^{7} d e + a^{6} b^{6} e^{2}\right )} x\right )}}, -\frac {105 \, {\left (b^{5} e^{5} x^{5} + 5 \, a b^{4} e^{5} x^{4} + 10 \, a^{2} b^{3} e^{5} x^{3} + 10 \, a^{3} b^{2} e^{5} x^{2} + 5 \, a^{4} b e^{5} x + a^{5} e^{5}\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 (384 \, b^{6} d^{5} - 432 \, a b^{5} d^{4} e - 8 \, a^{2} b^{4} d^{3} e^{2} - 14 \, a^{3} b^{3} d^{2} e^{3} - 35 \, a^{4} b^{2} d e^{4} + 105 \, a^{5} b e^{5} + 105 \, {\left (b^{6} d e^{4} - a b^{5} e^{5}\right )} x^{4} + 10 \, {\left (121 \, b^{6} d^{2} e^{3} - 200 \, a b^{5} d e^{4} + 79 \, a^{2} b^{4} e^{5}\right )} x^{3} + 2 \, {\left (1052 \, b^{6} d^{3} e^{2} - 1341 \, a b^{5} d^{2} e^{3} - 159 \, a^{2} b^{4} d e^{4} + 448 \, a^{3} b^{3} e^{5}\right )} x^{2} + 2 \, {\left (744 \, b^{6} d^{4} e - 872 \, a b^{5} d^{3} e^{2} - 33 \, a^{2} b^{4} d^{2} e^{3} - 84 \, a^{3} b^{3} d e^{4} + 245 \, a^{4} b^{2} e^{5}\right )} x\right )} \sqrt {e x + d}}{1920 \, {\left (a^{5} b^{7} d^{2} - 2 \, a^{6} b^{6} d e + a^{7} b^{5} e^{2} + {\left (b^{12} d^{2} - 2 \, a b^{11} d e + a^{2} b^{10} e^{2}\right )} x^{5} + 5 \, {\left (a b^{11} d^{2} - 2 \, a^{2} b^{10} d e + a^{3} b^{9} e^{2}\right )} x^{4} + 10 \, {\left (a^{2} b^{10} d^{2} - 2 \, a^{3} b^{9} d e + a^{4} b^{8} e^{2}\right )} x^{3} + 10 \, {\left (a^{3} b^{9} d^{2} - 2 \, a^{4} b^{8} d e + a^{5} b^{7} e^{2}\right )} x^{2} + 5 \, {\left (a^{4} b^{8} d^{2} - 2 \, a^{5} b^{7} d e + a^{6} b^{6} e^{2}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

[-1/3840*(105*(b^5*e^5*x^5 + 5*a*b^4*e^5*x^4 + 10*a^2*b^3*e^5*x^3 + 10*a^3*b^2*e^5*x^2 + 5*a^4*b*e^5*x + a^5*e
^5)*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*(384*b^
6*d^5 - 432*a*b^5*d^4*e - 8*a^2*b^4*d^3*e^2 - 14*a^3*b^3*d^2*e^3 - 35*a^4*b^2*d*e^4 + 105*a^5*b*e^5 + 105*(b^6
*d*e^4 - a*b^5*e^5)*x^4 + 10*(121*b^6*d^2*e^3 - 200*a*b^5*d*e^4 + 79*a^2*b^4*e^5)*x^3 + 2*(1052*b^6*d^3*e^2 -
1341*a*b^5*d^2*e^3 - 159*a^2*b^4*d*e^4 + 448*a^3*b^3*e^5)*x^2 + 2*(744*b^6*d^4*e - 872*a*b^5*d^3*e^2 - 33*a^2*
b^4*d^2*e^3 - 84*a^3*b^3*d*e^4 + 245*a^4*b^2*e^5)*x)*sqrt(e*x + d))/(a^5*b^7*d^2 - 2*a^6*b^6*d*e + a^7*b^5*e^2
+ (b^12*d^2 - 2*a*b^11*d*e + a^2*b^10*e^2)*x^5 + 5*(a*b^11*d^2 - 2*a^2*b^10*d*e + a^3*b^9*e^2)*x^4 + 10*(a^2*
b^10*d^2 - 2*a^3*b^9*d*e + a^4*b^8*e^2)*x^3 + 10*(a^3*b^9*d^2 - 2*a^4*b^8*d*e + a^5*b^7*e^2)*x^2 + 5*(a^4*b^8*
d^2 - 2*a^5*b^7*d*e + a^6*b^6*e^2)*x), -1/1920*(105*(b^5*e^5*x^5 + 5*a*b^4*e^5*x^4 + 10*a^2*b^3*e^5*x^3 + 10*a
^3*b^2*e^5*x^2 + 5*a^4*b*e^5*x + a^5*e^5)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*
x + b*d)) + (384*b^6*d^5 - 432*a*b^5*d^4*e - 8*a^2*b^4*d^3*e^2 - 14*a^3*b^3*d^2*e^3 - 35*a^4*b^2*d*e^4 + 105*a
^5*b*e^5 + 105*(b^6*d*e^4 - a*b^5*e^5)*x^4 + 10*(121*b^6*d^2*e^3 - 200*a*b^5*d*e^4 + 79*a^2*b^4*e^5)*x^3 + 2*(
1052*b^6*d^3*e^2 - 1341*a*b^5*d^2*e^3 - 159*a^2*b^4*d*e^4 + 448*a^3*b^3*e^5)*x^2 + 2*(744*b^6*d^4*e - 872*a*b^
5*d^3*e^2 - 33*a^2*b^4*d^2*e^3 - 84*a^3*b^3*d*e^4 + 245*a^4*b^2*e^5)*x)*sqrt(e*x + d))/(a^5*b^7*d^2 - 2*a^6*b^
6*d*e + a^7*b^5*e^2 + (b^12*d^2 - 2*a*b^11*d*e + a^2*b^10*e^2)*x^5 + 5*(a*b^11*d^2 - 2*a^2*b^10*d*e + a^3*b^9*
e^2)*x^4 + 10*(a^2*b^10*d^2 - 2*a^3*b^9*d*e + a^4*b^8*e^2)*x^3 + 10*(a^3*b^9*d^2 - 2*a^4*b^8*d*e + a^5*b^7*e^2
)*x^2 + 5*(a^4*b^8*d^2 - 2*a^5*b^7*d*e + a^6*b^6*e^2)*x)]

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giac [B]  time = 0.24, size = 360, normalized size = 1.91 \begin {gather*} -\frac {7 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{5}}{128 \, {\left (b^{5} d - a b^{4} e\right )} \sqrt {-b^{2} d + a b e}} - \frac {105 \, {\left (x e + d\right )}^{\frac {9}{2}} b^{4} e^{5} + 790 \, {\left (x e + d\right )}^{\frac {7}{2}} b^{4} d e^{5} - 896 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{4} d^{2} e^{5} + 490 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} d^{3} e^{5} - 105 \, \sqrt {x e + d} b^{4} d^{4} e^{5} - 790 \, {\left (x e + d\right )}^{\frac {7}{2}} a b^{3} e^{6} + 1792 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{3} d e^{6} - 1470 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{3} d^{2} e^{6} + 420 \, \sqrt {x e + d} a b^{3} d^{3} e^{6} - 896 \, {\left (x e + d\right )}^{\frac {5}{2}} a^{2} b^{2} e^{7} + 1470 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b^{2} d e^{7} - 630 \, \sqrt {x e + d} a^{2} b^{2} d^{2} e^{7} - 490 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{3} b e^{8} + 420 \, \sqrt {x e + d} a^{3} b d e^{8} - 105 \, \sqrt {x e + d} a^{4} e^{9}}{1920 \, {\left (b^{5} d - a b^{4} e\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((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")

[Out]

-7/128*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^5/((b^5*d - a*b^4*e)*sqrt(-b^2*d + a*b*e)) - 1/1920*(105
*(x*e + d)^(9/2)*b^4*e^5 + 790*(x*e + d)^(7/2)*b^4*d*e^5 - 896*(x*e + d)^(5/2)*b^4*d^2*e^5 + 490*(x*e + d)^(3/
2)*b^4*d^3*e^5 - 105*sqrt(x*e + d)*b^4*d^4*e^5 - 790*(x*e + d)^(7/2)*a*b^3*e^6 + 1792*(x*e + d)^(5/2)*a*b^3*d*
e^6 - 1470*(x*e + d)^(3/2)*a*b^3*d^2*e^6 + 420*sqrt(x*e + d)*a*b^3*d^3*e^6 - 896*(x*e + d)^(5/2)*a^2*b^2*e^7 +
1470*(x*e + d)^(3/2)*a^2*b^2*d*e^7 - 630*sqrt(x*e + d)*a^2*b^2*d^2*e^7 - 490*(x*e + d)^(3/2)*a^3*b*e^8 + 420*
sqrt(x*e + d)*a^3*b*d*e^8 - 105*sqrt(x*e + d)*a^4*e^9)/((b^5*d - a*b^4*e)*((x*e + d)*b - b*d + a*e)^5)

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maple [B]  time = 0.06, size = 360, normalized size = 1.91 \begin {gather*} -\frac {7 \sqrt {e x +d}\, a^{3} e^{8}}{128 \left (b e x +a e \right )^{5} b^{4}}+\frac {21 \sqrt {e x +d}\, a^{2} d \,e^{7}}{128 \left (b e x +a e \right )^{5} b^{3}}-\frac {21 \sqrt {e x +d}\, a \,d^{2} e^{6}}{128 \left (b e x +a e \right )^{5} b^{2}}+\frac {7 \sqrt {e x +d}\, d^{3} e^{5}}{128 \left (b e x +a e \right )^{5} b}-\frac {49 \left (e x +d \right )^{\frac {3}{2}} a^{2} e^{7}}{192 \left (b e x +a e \right )^{5} b^{3}}+\frac {49 \left (e x +d \right )^{\frac {3}{2}} a d \,e^{6}}{96 \left (b e x +a e \right )^{5} b^{2}}-\frac {49 \left (e x +d \right )^{\frac {3}{2}} d^{2} e^{5}}{192 \left (b e x +a e \right )^{5} b}-\frac {7 \left (e x +d \right )^{\frac {5}{2}} a \,e^{6}}{15 \left (b e x +a e \right )^{5} b^{2}}+\frac {7 \left (e x +d \right )^{\frac {5}{2}} d \,e^{5}}{15 \left (b e x +a e \right )^{5} b}+\frac {7 \left (e x +d \right )^{\frac {9}{2}} e^{5}}{128 \left (b e x +a e \right )^{5} \left (a e -b d \right )}-\frac {79 \left (e x +d \right )^{\frac {7}{2}} e^{5}}{192 \left (b e x +a e \right )^{5} b}+\frac {7 e^{5} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{128 \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^3,x)

[Out]

7/128*e^5/(b*e*x+a*e)^5/(a*e-b*d)*(e*x+d)^(9/2)-79/192*e^5/(b*e*x+a*e)^5/b*(e*x+d)^(7/2)-7/15*e^6/(b*e*x+a*e)^
5/b^2*(e*x+d)^(5/2)*a+7/15*e^5/(b*e*x+a*e)^5/b*(e*x+d)^(5/2)*d-49/192*e^7/(b*e*x+a*e)^5/b^3*(e*x+d)^(3/2)*a^2+
49/96*e^6/(b*e*x+a*e)^5/b^2*(e*x+d)^(3/2)*a*d-49/192*e^5/(b*e*x+a*e)^5/b*(e*x+d)^(3/2)*d^2-7/128*e^8/(b*e*x+a*
e)^5/b^4*(e*x+d)^(1/2)*a^3+21/128*e^7/(b*e*x+a*e)^5/b^3*(e*x+d)^(1/2)*a^2*d-21/128*e^6/(b*e*x+a*e)^5/b^2*(e*x+
d)^(1/2)*a*d^2+7/128*e^5/(b*e*x+a*e)^5/b*(e*x+d)^(1/2)*d^3+7/128*e^5/(a*e-b*d)/b^4/((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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(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 = 0.66, size = 439, normalized size = 2.34 \begin {gather*} \frac {7\,e^5\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{128\,b^{9/2}\,{\left (a\,e-b\,d\right )}^{3/2}}-\frac {\frac {79\,e^5\,{\left (d+e\,x\right )}^{7/2}}{192\,b}-\frac {7\,e^5\,{\left (d+e\,x\right )}^{9/2}}{128\,\left (a\,e-b\,d\right )}+\frac {49\,e^5\,{\left (d+e\,x\right )}^{3/2}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{192\,b^3}+\frac {7\,e^5\,\sqrt {d+e\,x}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{128\,b^4}+\frac {7\,e^5\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{5/2}}{15\,b^2}}{\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(7/2)/(a^2 + b^2*x^2 + 2*a*b*x)^3,x)

[Out]

(7*e^5*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(128*b^(9/2)*(a*e - b*d)^(3/2)) - ((79*e^5*(d + e*x)
^(7/2))/(192*b) - (7*e^5*(d + e*x)^(9/2))/(128*(a*e - b*d)) + (49*e^5*(d + e*x)^(3/2)*(a^2*e^2 + b^2*d^2 - 2*a
*b*d*e))/(192*b^3) + (7*e^5*(d + e*x)^(1/2)*(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2))/(128*b^4) + (
7*e^5*(a*e - b*d)*(d + e*x)^(5/2))/(15*b^2))/((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)

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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((e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**3,x)

[Out]

Timed out

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