3.1724 \(\int \frac {(d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=260 \[ -\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e^4 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac {5 e^3 \sqrt {d+e x}}{64 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

-5/24*e*(e*x+d)^(3/2)/b^2/(b*x+a)^2/((b*x+a)^2)^(1/2)-1/4*(e*x+d)^(5/2)/b/(b*x+a)^3/((b*x+a)^2)^(1/2)+5/64*e^4
*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(7/2)/(-a*e+b*d)^(3/2)/((b*x+a)^2)^(1/2)-5/64*e^3*(
e*x+d)^(1/2)/b^3/(-a*e+b*d)/((b*x+a)^2)^(1/2)-5/32*e^2*(e*x+d)^(1/2)/b^3/(b*x+a)/((b*x+a)^2)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {646, 47, 51, 63, 208} \[ -\frac {5 e^3 \sqrt {d+e x}}{64 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e^4 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(5/2)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-5*e^3*Sqrt[d + e*x])/(64*b^3*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e^2*Sqrt[d + e*x])/(32*b^3*(a +
b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e*(d + e*x)^(3/2))/(24*b^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
) - (d + e*x)^(5/2)/(4*b*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*e^4*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d
+ e*x])/Sqrt[b*d - a*e]])/(64*b^(7/2)*(b*d - a*e)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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]

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{5/2}}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 b^2 e \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{3/2}}{\left (a b+b^2 x\right )^4} \, dx}{8 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 e^2 \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{\left (a b+b^2 x\right )^3} \, dx}{16 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 e^3 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^2 \sqrt {d+e x}} \, dx}{64 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {5 e^3 \sqrt {d+e x}}{64 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (5 e^4 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{128 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {5 e^3 \sqrt {d+e x}}{64 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (5 e^3 \left (a b+b^2 x\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a b-\frac {b^2 d}{e}+\frac {b^2 x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{64 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {5 e^3 \sqrt {d+e x}}{64 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e^4 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{7/2} (b d-a e)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 67, normalized size = 0.26 \[ -\frac {2 e^4 (a+b x) (d+e x)^{7/2} \, _2F_1\left (\frac {7}{2},5;\frac {9}{2};\frac {b (d+e x)}{b d-a e}\right )}{7 \sqrt {(a+b x)^2} (b d-a e)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(5/2)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-2*e^4*(a + b*x)*(d + e*x)^(7/2)*Hypergeometric2F1[7/2, 5, 9/2, (b*(d + e*x))/(b*d - a*e)])/(7*(b*d - a*e)^5*
Sqrt[(a + b*x)^2])

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fricas [B]  time = 0.82, size = 894, normalized size = 3.44 \[ \left [-\frac {15 \, {\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 (48 \, b^{5} d^{4} - 56 \, a b^{4} d^{3} e - 2 \, a^{2} b^{3} d^{2} e^{2} - 5 \, a^{3} b^{2} d e^{3} + 15 \, a^{4} b e^{4} + 15 \, {\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} + {\left (118 \, b^{5} d^{2} e^{2} - 191 \, a b^{4} d e^{3} + 73 \, a^{2} b^{3} e^{4}\right )} x^{2} + {\left (136 \, b^{5} d^{3} e - 172 \, a b^{4} d^{2} e^{2} - 19 \, a^{2} b^{3} d e^{3} + 55 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{384 \, {\left (a^{4} b^{6} d^{2} - 2 \, a^{5} b^{5} d e + a^{6} b^{4} e^{2} + {\left (b^{10} d^{2} - 2 \, a b^{9} d e + a^{2} b^{8} e^{2}\right )} x^{4} + 4 \, {\left (a b^{9} d^{2} - 2 \, a^{2} b^{8} d e + a^{3} b^{7} e^{2}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} d^{2} - 2 \, a^{3} b^{7} d e + a^{4} b^{6} e^{2}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} d^{2} - 2 \, a^{4} b^{6} d e + a^{5} b^{5} e^{2}\right )} x\right )}}, -\frac {15 \, {\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 (48 \, b^{5} d^{4} - 56 \, a b^{4} d^{3} e - 2 \, a^{2} b^{3} d^{2} e^{2} - 5 \, a^{3} b^{2} d e^{3} + 15 \, a^{4} b e^{4} + 15 \, {\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} + {\left (118 \, b^{5} d^{2} e^{2} - 191 \, a b^{4} d e^{3} + 73 \, a^{2} b^{3} e^{4}\right )} x^{2} + {\left (136 \, b^{5} d^{3} e - 172 \, a b^{4} d^{2} e^{2} - 19 \, a^{2} b^{3} d e^{3} + 55 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{192 \, {\left (a^{4} b^{6} d^{2} - 2 \, a^{5} b^{5} d e + a^{6} b^{4} e^{2} + {\left (b^{10} d^{2} - 2 \, a b^{9} d e + a^{2} b^{8} e^{2}\right )} x^{4} + 4 \, {\left (a b^{9} d^{2} - 2 \, a^{2} b^{8} d e + a^{3} b^{7} e^{2}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} d^{2} - 2 \, a^{3} b^{7} d e + a^{4} b^{6} e^{2}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} d^{2} - 2 \, a^{4} b^{6} d e + a^{5} b^{5} e^{2}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(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)*
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*(48*b^5*d^4 - 56*a*b^4*d^3*e -
2*a^2*b^3*d^2*e^2 - 5*a^3*b^2*d*e^3 + 15*a^4*b*e^4 + 15*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (118*b^5*d^2*e^2 - 191*a
*b^4*d*e^3 + 73*a^2*b^3*e^4)*x^2 + (136*b^5*d^3*e - 172*a*b^4*d^2*e^2 - 19*a^2*b^3*d*e^3 + 55*a^3*b^2*e^4)*x)*
sqrt(e*x + d))/(a^4*b^6*d^2 - 2*a^5*b^5*d*e + a^6*b^4*e^2 + (b^10*d^2 - 2*a*b^9*d*e + a^2*b^8*e^2)*x^4 + 4*(a*
b^9*d^2 - 2*a^2*b^8*d*e + a^3*b^7*e^2)*x^3 + 6*(a^2*b^8*d^2 - 2*a^3*b^7*d*e + a^4*b^6*e^2)*x^2 + 4*(a^3*b^7*d^
2 - 2*a^4*b^6*d*e + a^5*b^5*e^2)*x), -1/192*(15*(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)) + (48*b^5*d^4 -
56*a*b^4*d^3*e - 2*a^2*b^3*d^2*e^2 - 5*a^3*b^2*d*e^3 + 15*a^4*b*e^4 + 15*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (118*b^
5*d^2*e^2 - 191*a*b^4*d*e^3 + 73*a^2*b^3*e^4)*x^2 + (136*b^5*d^3*e - 172*a*b^4*d^2*e^2 - 19*a^2*b^3*d*e^3 + 55
*a^3*b^2*e^4)*x)*sqrt(e*x + d))/(a^4*b^6*d^2 - 2*a^5*b^5*d*e + a^6*b^4*e^2 + (b^10*d^2 - 2*a*b^9*d*e + a^2*b^8
*e^2)*x^4 + 4*(a*b^9*d^2 - 2*a^2*b^8*d*e + a^3*b^7*e^2)*x^3 + 6*(a^2*b^8*d^2 - 2*a^3*b^7*d*e + a^4*b^6*e^2)*x^
2 + 4*(a^3*b^7*d^2 - 2*a^4*b^6*d*e + a^5*b^5*e^2)*x)]

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giac [A]  time = 0.39, size = 353, normalized size = 1.36 \[ -\frac {5 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{4}}{64 \, {\left (b^{4} d \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b^{3} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {15 \, {\left (x e + d\right )}^{\frac {7}{2}} b^{3} e^{4} + 73 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{3} d e^{4} - 55 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{4} + 15 \, \sqrt {x e + d} b^{3} d^{3} e^{4} - 73 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{2} e^{5} + 110 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{2} d e^{5} - 45 \, \sqrt {x e + d} a b^{2} d^{2} e^{5} - 55 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b e^{6} + 45 \, \sqrt {x e + d} a^{2} b d e^{6} - 15 \, \sqrt {x e + d} a^{3} e^{7}}{192 \, {\left (b^{4} d \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b^{3} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

-5/64*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^4/((b^4*d*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a*b^3*e*sg
n((x*e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) - 1/192*(15*(x*e + d)^(7/2)*b^3*e^4 + 73*(x*e + d)^(5/
2)*b^3*d*e^4 - 55*(x*e + d)^(3/2)*b^3*d^2*e^4 + 15*sqrt(x*e + d)*b^3*d^3*e^4 - 73*(x*e + d)^(5/2)*a*b^2*e^5 +
110*(x*e + d)^(3/2)*a*b^2*d*e^5 - 45*sqrt(x*e + d)*a*b^2*d^2*e^5 - 55*(x*e + d)^(3/2)*a^2*b*e^6 + 45*sqrt(x*e
+ d)*a^2*b*d*e^6 - 15*sqrt(x*e + d)*a^3*e^7)/((b^4*d*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a*b^3*e*sgn((x*e + d
)*b*e - b*d*e + a*e^2))*((x*e + d)*b - b*d + a*e)^4)

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maple [B]  time = 0.07, size = 477, normalized size = 1.83 \[ \frac {\left (15 b^{4} e^{4} x^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+60 a \,b^{3} e^{4} x^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+90 a^{2} b^{2} e^{4} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+60 a^{3} b \,e^{4} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 a^{4} e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-15 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{3} e^{3}+45 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{2} b d \,e^{2}-45 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d^{2} e +15 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{3}-55 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a^{2} b \,e^{2}+110 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d e -55 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{2}-73 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e +73 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} d +15 \left (e x +d \right )^{\frac {7}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3}\right ) \left (b x +a \right )}{192 \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/192*(15*b^4*e^4*x^4*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+60*a*b^3*e^4*x^3*arctan((e*x+d)^(1/2)/((a*e-
b*d)*b)^(1/2)*b)+15*(e*x+d)^(7/2)*((a*e-b*d)*b)^(1/2)*b^3+90*a^2*b^2*e^4*x^2*arctan((e*x+d)^(1/2)/((a*e-b*d)*b
)^(1/2)*b)-73*(e*x+d)^(5/2)*((a*e-b*d)*b)^(1/2)*a*b^2*e+73*(e*x+d)^(5/2)*((a*e-b*d)*b)^(1/2)*b^3*d+60*a^3*b*e^
4*x*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)-55*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*a^2*b*e^2+110*(e*x+d)^(3/
2)*((a*e-b*d)*b)^(1/2)*a*b^2*d*e-55*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*b^3*d^2+15*a^4*e^4*arctan((e*x+d)^(1/2)/
((a*e-b*d)*b)^(1/2)*b)-15*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a^3*e^3+45*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a^2*b
*d*e^2-45*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a*b^2*d^2*e+15*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*b^3*d^3)*(b*x+a)/
((a*e-b*d)*b)^(1/2)/b^3/(a*e-b*d)/((b*x+a)^2)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{\frac {5}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(5/2)/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d+e\,x\right )}^{5/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(5/2)/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

int((d + e*x)^(5/2)/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)

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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((e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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