3.1652 \(\int \frac {1}{(d+e x)^{5/2} (a^2+2 a b x+b^2 x^2)} \, dx\)

Optimal. Leaf size=124 \[ \frac {5 b^{3/2} e \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}}-\frac {5 b e}{\sqrt {d+e x} (b d-a e)^3}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}-\frac {5 e}{3 (d+e x)^{3/2} (b d-a e)^2} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 51, 63, 208} \[ \frac {5 b^{3/2} e \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}}-\frac {5 b e}{\sqrt {d+e x} (b d-a e)^3}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}-\frac {5 e}{3 (d+e x)^{3/2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(5/2)*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

(-5*e)/(3*(b*d - a*e)^2*(d + e*x)^(3/2)) - 1/((b*d - a*e)*(a + b*x)*(d + e*x)^(3/2)) - (5*b*e)/((b*d - a*e)^3*
Sqrt[d + e*x]) + (5*b^(3/2)*e*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(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 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 {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx &=\int \frac {1}{(a+b x)^2 (d+e x)^{5/2}} \, dx\\ &=-\frac {1}{(b d-a e) (a+b x) (d+e x)^{3/2}}-\frac {(5 e) \int \frac {1}{(a+b x) (d+e x)^{5/2}} \, dx}{2 (b d-a e)}\\ &=-\frac {5 e}{3 (b d-a e)^2 (d+e x)^{3/2}}-\frac {1}{(b d-a e) (a+b x) (d+e x)^{3/2}}-\frac {(5 b e) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{2 (b d-a e)^2}\\ &=-\frac {5 e}{3 (b d-a e)^2 (d+e x)^{3/2}}-\frac {1}{(b d-a e) (a+b x) (d+e x)^{3/2}}-\frac {5 b e}{(b d-a e)^3 \sqrt {d+e x}}-\frac {\left (5 b^2 e\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{2 (b d-a e)^3}\\ &=-\frac {5 e}{3 (b d-a e)^2 (d+e x)^{3/2}}-\frac {1}{(b d-a e) (a+b x) (d+e x)^{3/2}}-\frac {5 b e}{(b d-a e)^3 \sqrt {d+e x}}-\frac {\left (5 b^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 )}{(b d-a e)^3}\\ &=-\frac {5 e}{3 (b d-a e)^2 (d+e x)^{3/2}}-\frac {1}{(b d-a e) (a+b x) (d+e x)^{3/2}}-\frac {5 b e}{(b d-a e)^3 \sqrt {d+e x}}+\frac {5 b^{3/2} e \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.40 \[ -\frac {2 e \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};-\frac {b (d+e x)}{a e-b d}\right )}{3 (d+e x)^{3/2} (a e-b d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(5/2)*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

(-2*e*Hypergeometric2F1[-3/2, 2, -1/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(3*(-(b*d) + a*e)^2*(d + e*x)^(3/2))

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fricas [B]  time = 0.97, size = 782, normalized size = 6.31 \[ \left [-\frac {15 \, {\left (b^{2} e^{3} x^{3} + a b d^{2} e + {\left (2 \, b^{2} d e^{2} + a b e^{3}\right )} x^{2} + {\left (b^{2} d^{2} e + 2 \, a b d e^{2}\right )} x\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) + 2 \, {\left (15 \, b^{2} e^{2} x^{2} + 3 \, b^{2} d^{2} + 14 \, a b d e - 2 \, a^{2} e^{2} + 10 \, {\left (2 \, b^{2} d e + a b e^{2}\right )} x\right )} \sqrt {e x + d}}{6 \, {\left (a b^{3} d^{5} - 3 \, a^{2} b^{2} d^{4} e + 3 \, a^{3} b d^{3} e^{2} - a^{4} d^{2} e^{3} + {\left (b^{4} d^{3} e^{2} - 3 \, a b^{3} d^{2} e^{3} + 3 \, a^{2} b^{2} d e^{4} - a^{3} b e^{5}\right )} x^{3} + {\left (2 \, b^{4} d^{4} e - 5 \, a b^{3} d^{3} e^{2} + 3 \, a^{2} b^{2} d^{2} e^{3} + a^{3} b d e^{4} - a^{4} e^{5}\right )} x^{2} + {\left (b^{4} d^{5} - a b^{3} d^{4} e - 3 \, a^{2} b^{2} d^{3} e^{2} + 5 \, a^{3} b d^{2} e^{3} - 2 \, a^{4} d e^{4}\right )} x\right )}}, \frac {15 \, {\left (b^{2} e^{3} x^{3} + a b d^{2} e + {\left (2 \, b^{2} d e^{2} + a b e^{3}\right )} x^{2} + {\left (b^{2} d^{2} e + 2 \, a b d e^{2}\right )} x\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (15 \, b^{2} e^{2} x^{2} + 3 \, b^{2} d^{2} + 14 \, a b d e - 2 \, a^{2} e^{2} + 10 \, {\left (2 \, b^{2} d e + a b e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (a b^{3} d^{5} - 3 \, a^{2} b^{2} d^{4} e + 3 \, a^{3} b d^{3} e^{2} - a^{4} d^{2} e^{3} + {\left (b^{4} d^{3} e^{2} - 3 \, a b^{3} d^{2} e^{3} + 3 \, a^{2} b^{2} d e^{4} - a^{3} b e^{5}\right )} x^{3} + {\left (2 \, b^{4} d^{4} e - 5 \, a b^{3} d^{3} e^{2} + 3 \, a^{2} b^{2} d^{2} e^{3} + a^{3} b d e^{4} - a^{4} e^{5}\right )} x^{2} + {\left (b^{4} d^{5} - a b^{3} d^{4} e - 3 \, a^{2} b^{2} d^{3} e^{2} + 5 \, a^{3} b d^{2} e^{3} - 2 \, a^{4} d e^{4}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")

[Out]

[-1/6*(15*(b^2*e^3*x^3 + a*b*d^2*e + (2*b^2*d*e^2 + a*b*e^3)*x^2 + (b^2*d^2*e + 2*a*b*d*e^2)*x)*sqrt(b/(b*d -
a*e))*log((b*e*x + 2*b*d - a*e - 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) + 2*(15*b^2*e^2*x
^2 + 3*b^2*d^2 + 14*a*b*d*e - 2*a^2*e^2 + 10*(2*b^2*d*e + a*b*e^2)*x)*sqrt(e*x + d))/(a*b^3*d^5 - 3*a^2*b^2*d^
4*e + 3*a^3*b*d^3*e^2 - a^4*d^2*e^3 + (b^4*d^3*e^2 - 3*a*b^3*d^2*e^3 + 3*a^2*b^2*d*e^4 - a^3*b*e^5)*x^3 + (2*b
^4*d^4*e - 5*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*e^3 + a^3*b*d*e^4 - a^4*e^5)*x^2 + (b^4*d^5 - a*b^3*d^4*e - 3*a^2*b
^2*d^3*e^2 + 5*a^3*b*d^2*e^3 - 2*a^4*d*e^4)*x), 1/3*(15*(b^2*e^3*x^3 + a*b*d^2*e + (2*b^2*d*e^2 + a*b*e^3)*x^2
+ (b^2*d^2*e + 2*a*b*d*e^2)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b
*e*x + b*d)) - (15*b^2*e^2*x^2 + 3*b^2*d^2 + 14*a*b*d*e - 2*a^2*e^2 + 10*(2*b^2*d*e + a*b*e^2)*x)*sqrt(e*x + d
))/(a*b^3*d^5 - 3*a^2*b^2*d^4*e + 3*a^3*b*d^3*e^2 - a^4*d^2*e^3 + (b^4*d^3*e^2 - 3*a*b^3*d^2*e^3 + 3*a^2*b^2*d
*e^4 - a^3*b*e^5)*x^3 + (2*b^4*d^4*e - 5*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*e^3 + a^3*b*d*e^4 - a^4*e^5)*x^2 + (b^4
*d^5 - a*b^3*d^4*e - 3*a^2*b^2*d^3*e^2 + 5*a^3*b*d^2*e^3 - 2*a^4*d*e^4)*x)]

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giac [B]  time = 0.19, size = 224, normalized size = 1.81 \[ -\frac {5 \, b^{2} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} - \frac {\sqrt {x e + d} b^{2} e}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}} - \frac {2 \, {\left (6 \, {\left (x e + d\right )} b e + b d e - a e^{2}\right )}}{3 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} {\left (x e + d\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

-5*b^2*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqr
t(-b^2*d + a*b*e)) - sqrt(x*e + d)*b^2*e/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*((x*e + d)*b - b
*d + a*e)) - 2/3*(6*(x*e + d)*b*e + b*d*e - a*e^2)/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*(x*e +
d)^(3/2))

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maple [A]  time = 0.06, size = 125, normalized size = 1.01 \[ \frac {5 b^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}}+\frac {\sqrt {e x +d}\, b^{2} e}{\left (a e -b d \right )^{3} \left (b e x +a e \right )}+\frac {4 b e}{\left (a e -b d \right )^{3} \sqrt {e x +d}}-\frac {2 e}{3 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2),x)

[Out]

e*b^2/(a*e-b*d)^3*(e*x+d)^(1/2)/(b*e*x+a*e)+5*e*b^2/(a*e-b*d)^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e
-b*d)*b)^(1/2)*b)-2/3*e/(a*e-b*d)^2/(e*x+d)^(3/2)+4*e/(a*e-b*d)^3*b/(e*x+d)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^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 = 0.67, size = 161, normalized size = 1.30 \[ \frac {\frac {10\,b\,e\,\left (d+e\,x\right )}{3\,{\left (a\,e-b\,d\right )}^2}-\frac {2\,e}{3\,\left (a\,e-b\,d\right )}+\frac {5\,b^2\,e\,{\left (d+e\,x\right )}^2}{{\left (a\,e-b\,d\right )}^3}}{b\,{\left (d+e\,x\right )}^{5/2}+\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}}+\frac {5\,b^{3/2}\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{{\left (a\,e-b\,d\right )}^{7/2}}\right )}{{\left (a\,e-b\,d\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)^(5/2)*(a^2 + b^2*x^2 + 2*a*b*x)),x)

[Out]

((10*b*e*(d + e*x))/(3*(a*e - b*d)^2) - (2*e)/(3*(a*e - b*d)) + (5*b^2*e*(d + e*x)^2)/(a*e - b*d)^3)/(b*(d + e
*x)^(5/2) + (a*e - b*d)*(d + e*x)^(3/2)) + (5*b^(3/2)*e*atan((b^(1/2)*(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))/(a*e - b*d)^(7/2)))/(a*e - b*d)^(7/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x\right )^{2} \left (d + e x\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

Integral(1/((a + b*x)**2*(d + e*x)**(5/2)), x)

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