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

Optimal. Leaf size=146 \[ -\frac {35 e^2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2}}+\frac {35 e^2 \sqrt {d+e x} (b d-a e)}{4 b^4}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {35 e^2 (d+e x)^{3/2}}{12 b^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 146, 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, 50, 63, 208} \begin {gather*} \frac {35 e^2 \sqrt {d+e x} (b d-a e)}{4 b^4}-\frac {35 e^2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2}}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {35 e^2 (d+e x)^{3/2}}{12 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(35*e^2*(b*d - a*e)*Sqrt[d + e*x])/(4*b^4) + (35*e^2*(d + e*x)^(3/2))/(12*b^3) - (7*e*(d + e*x)^(5/2))/(4*b^2*
(a + b*x)) - (d + e*x)^(7/2)/(2*b*(a + b*x)^2) - (35*e^2*(b*d - a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqr
t[b*d - a*e]])/(4*b^(9/2))

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)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {(d+e x)^{7/2}}{(a+b x)^3} \, dx\\ &=-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {(7 e) \int \frac {(d+e x)^{5/2}}{(a+b x)^2} \, dx}{4 b}\\ &=-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {\left (35 e^2\right ) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{8 b^2}\\ &=\frac {35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {\left (35 e^2 (b d-a e)\right ) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{8 b^3}\\ &=\frac {35 e^2 (b d-a e) \sqrt {d+e x}}{4 b^4}+\frac {35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {\left (35 e^2 (b d-a e)^2\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 b^4}\\ &=\frac {35 e^2 (b d-a e) \sqrt {d+e x}}{4 b^4}+\frac {35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {\left (35 e (b d-a e)^2\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^4}\\ &=\frac {35 e^2 (b d-a e) \sqrt {d+e x}}{4 b^4}+\frac {35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}-\frac {35 e^2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.36 \begin {gather*} \frac {2 e^2 (d+e x)^{9/2} \, _2F_1\left (3,\frac {9}{2};\frac {11}{2};-\frac {b (d+e x)}{a e-b d}\right )}{9 (a e-b d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^2*(d + e*x)^(9/2)*Hypergeometric2F1[3, 9/2, 11/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(9*(-(b*d) + a*e)^3)

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IntegrateAlgebraic [A]  time = 0.73, size = 253, normalized size = 1.73 \begin {gather*} \frac {35 \left (-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 ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 b^{9/2} (a e-b d)^{3/2}}+\frac {e^2 \sqrt {d+e x} \left (-105 a^3 e^3-175 a^2 b e^2 (d+e x)+315 a^2 b d e^2-315 a b^2 d^2 e-56 a b^2 e (d+e x)^2+350 a b^2 d e (d+e x)+105 b^3 d^3-175 b^3 d^2 (d+e x)+8 b^3 (d+e x)^3+56 b^3 d (d+e x)^2\right )}{12 b^4 (a e+b (d+e x)-b d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*Sqrt[d + e*x]*(105*b^3*d^3 - 315*a*b^2*d^2*e + 315*a^2*b*d*e^2 - 105*a^3*e^3 - 175*b^3*d^2*(d + e*x) + 35
0*a*b^2*d*e*(d + e*x) - 175*a^2*b*e^2*(d + e*x) + 56*b^3*d*(d + e*x)^2 - 56*a*b^2*e*(d + e*x)^2 + 8*b^3*(d + e
*x)^3))/(12*b^4*(-(b*d) + a*e + b*(d + e*x))^2) + (35*(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[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(4*b^(9/2)*(-(b*d) + a*e)^(3/2))

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fricas [B]  time = 0.45, size = 520, normalized size = 3.56 \begin {gather*} \left [-\frac {105 \, {\left (a^{2} b d e^{2} - a^{3} e^{3} + {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (a b^{2} d e^{2} - a^{2} b e^{3}\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^{3} e^{3} x^{3} - 6 \, b^{3} d^{3} - 21 \, a b^{2} d^{2} e + 140 \, a^{2} b d e^{2} - 105 \, a^{3} e^{3} + 8 \, {\left (10 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} - {\left (39 \, b^{3} d^{2} e - 238 \, a b^{2} d e^{2} + 175 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, -\frac {105 \, {\left (a^{2} b d e^{2} - a^{3} e^{3} + {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (a b^{2} d e^{2} - a^{2} b e^{3}\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^{3} e^{3} x^{3} - 6 \, b^{3} d^{3} - 21 \, a b^{2} d^{2} e + 140 \, a^{2} b d e^{2} - 105 \, a^{3} e^{3} + 8 \, {\left (10 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} - {\left (39 \, b^{3} d^{2} e - 238 \, a b^{2} d e^{2} + 175 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{12 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(105*(a^2*b*d*e^2 - a^3*e^3 + (b^3*d*e^2 - a*b^2*e^3)*x^2 + 2*(a*b^2*d*e^2 - a^2*b*e^3)*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^3*e^3*x^3 - 6*b^
3*d^3 - 21*a*b^2*d^2*e + 140*a^2*b*d*e^2 - 105*a^3*e^3 + 8*(10*b^3*d*e^2 - 7*a*b^2*e^3)*x^2 - (39*b^3*d^2*e -
238*a*b^2*d*e^2 + 175*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), -1/12*(105*(a^2*b*d*e^2 -
a^3*e^3 + (b^3*d*e^2 - a*b^2*e^3)*x^2 + 2*(a*b^2*d*e^2 - a^2*b*e^3)*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^3*e^3*x^3 - 6*b^3*d^3 - 21*a*b^2*d^2*e + 140*a^2*b*d*e^2 - 105*
a^3*e^3 + 8*(10*b^3*d*e^2 - 7*a*b^2*e^3)*x^2 - (39*b^3*d^2*e - 238*a*b^2*d*e^2 + 175*a^2*b*e^3)*x)*sqrt(e*x +
d))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]

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giac [B]  time = 0.20, size = 265, normalized size = 1.82 \begin {gather*} \frac {35 \, {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\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^{4}} - \frac {13 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{2} - 11 \, \sqrt {x e + d} b^{3} d^{3} e^{2} - 26 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{2} d e^{3} + 33 \, \sqrt {x e + d} a b^{2} d^{2} e^{3} + 13 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b e^{4} - 33 \, \sqrt {x e + d} a^{2} b d e^{4} + 11 \, \sqrt {x e + d} a^{3} e^{5}}{4 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{4}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{6} e^{2} + 9 \, \sqrt {x e + d} b^{6} d e^{2} - 9 \, \sqrt {x e + d} a b^{5} e^{3}\right )}}{3 \, b^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/4*(b^2*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*
b^4) - 1/4*(13*(x*e + d)^(3/2)*b^3*d^2*e^2 - 11*sqrt(x*e + d)*b^3*d^3*e^2 - 26*(x*e + d)^(3/2)*a*b^2*d*e^3 + 3
3*sqrt(x*e + d)*a*b^2*d^2*e^3 + 13*(x*e + d)^(3/2)*a^2*b*e^4 - 33*sqrt(x*e + d)*a^2*b*d*e^4 + 11*sqrt(x*e + d)
*a^3*e^5)/(((x*e + d)*b - b*d + a*e)^2*b^4) + 2/3*((x*e + d)^(3/2)*b^6*e^2 + 9*sqrt(x*e + d)*b^6*d*e^2 - 9*sqr
t(x*e + d)*a*b^5*e^3)/b^9

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maple [B]  time = 0.07, size = 380, normalized size = 2.60 \begin {gather*} -\frac {11 \sqrt {e x +d}\, a^{3} e^{5}}{4 \left (b e x +a e \right )^{2} b^{4}}+\frac {33 \sqrt {e x +d}\, a^{2} d \,e^{4}}{4 \left (b e x +a e \right )^{2} b^{3}}-\frac {33 \sqrt {e x +d}\, a \,d^{2} e^{3}}{4 \left (b e x +a e \right )^{2} b^{2}}+\frac {11 \sqrt {e x +d}\, d^{3} e^{2}}{4 \left (b e x +a e \right )^{2} b}-\frac {13 \left (e x +d \right )^{\frac {3}{2}} a^{2} e^{4}}{4 \left (b e x +a e \right )^{2} b^{3}}+\frac {35 a^{2} 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 {13 \left (e x +d \right )^{\frac {3}{2}} a d \,e^{3}}{2 \left (b e x +a e \right )^{2} b^{2}}-\frac {35 a d \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}\, b^{3}}-\frac {13 \left (e x +d \right )^{\frac {3}{2}} d^{2} e^{2}}{4 \left (b e x +a e \right )^{2} b}+\frac {35 d^{2} 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 {6 \sqrt {e x +d}\, a \,e^{3}}{b^{4}}+\frac {6 \sqrt {e x +d}\, d \,e^{2}}{b^{3}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} e^{2}}{3 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*e^2*(e*x+d)^(3/2)/b^3-6*e^3/b^4*a*(e*x+d)^(1/2)+6*e^2/b^3*(e*x+d)^(1/2)*d-13/4*e^4/b^3/(b*e*x+a*e)^2*(e*x+
d)^(3/2)*a^2+13/2*e^3/b^2/(b*e*x+a*e)^2*(e*x+d)^(3/2)*a*d-13/4*e^2/b/(b*e*x+a*e)^2*(e*x+d)^(3/2)*d^2-11/4*e^5/
b^4/(b*e*x+a*e)^2*(e*x+d)^(1/2)*a^3+33/4*e^4/b^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*a^2*d-33/4*e^3/b^2/(b*e*x+a*e)^2*
(e*x+d)^(1/2)*a*d^2+11/4*e^2/b/(b*e*x+a*e)^2*(e*x+d)^(1/2)*d^3+35/4*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^2-35/2*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+35/4*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^2

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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)^(7/2)/(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 = 268, normalized size = 1.84 \begin {gather*} \frac {2\,e^2\,{\left (d+e\,x\right )}^{3/2}}{3\,b^3}-\frac {\sqrt {d+e\,x}\,\left (\frac {11\,a^3\,e^5}{4}-\frac {33\,a^2\,b\,d\,e^4}{4}+\frac {33\,a\,b^2\,d^2\,e^3}{4}-\frac {11\,b^3\,d^3\,e^2}{4}\right )+{\left (d+e\,x\right )}^{3/2}\,\left (\frac {13\,a^2\,b\,e^4}{4}-\frac {13\,a\,b^2\,d\,e^3}{2}+\frac {13\,b^3\,d^2\,e^2}{4}\right )}{b^6\,{\left (d+e\,x\right )}^2-\left (2\,b^6\,d-2\,a\,b^5\,e\right )\,\left (d+e\,x\right )+b^6\,d^2+a^2\,b^4\,e^2-2\,a\,b^5\,d\,e}+\frac {2\,e^2\,\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,\sqrt {d+e\,x}}{b^6}+\frac {35\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^2\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^4-2\,a\,b\,d\,e^3+b^2\,d^2\,e^2}\right )\,{\left (a\,e-b\,d\right )}^{3/2}}{4\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*e^2*(d + e*x)^(3/2))/(3*b^3) - ((d + e*x)^(1/2)*((11*a^3*e^5)/4 - (11*b^3*d^3*e^2)/4 + (33*a*b^2*d^2*e^3)/4
- (33*a^2*b*d*e^4)/4) + (d + e*x)^(3/2)*((13*a^2*b*e^4)/4 + (13*b^3*d^2*e^2)/4 - (13*a*b^2*d*e^3)/2))/(b^6*(d
+ e*x)^2 - (2*b^6*d - 2*a*b^5*e)*(d + e*x) + b^6*d^2 + a^2*b^4*e^2 - 2*a*b^5*d*e) + (2*e^2*(3*b^3*d - 3*a*b^2
*e)*(d + e*x)^(1/2))/b^6 + (35*e^2*atan((b^(1/2)*e^2*(a*e - b*d)^(3/2)*(d + e*x)^(1/2))/(a^2*e^4 + b^2*d^2*e^2
- 2*a*b*d*e^3))*(a*e - b*d)^(3/2))/(4*b^(9/2))

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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)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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