\(\int \frac {1}{(d+e x)^{3/2} (a^2+2 a b x+b^2 x^2)} \, dx\) [1651]

Optimal result

Integrand size = 28, antiderivative size = 99 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {3 e}{(b d-a e)^2 \sqrt {d+e x}}-\frac {1}{(b d-a e) (a+b x) \sqrt {d+e x}}+\frac {3 \sqrt {b} e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 44, 53, 65, 214} \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {3 \sqrt {b} e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{5/2}}-\frac {3 e}{\sqrt {d+e x} (b d-a e)^2}-\frac {1}{(a+b x) \sqrt {d+e x} (b d-a e)} \]

[In]

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

[Out]

(-3*e)/((b*d - a*e)^2*Sqrt[d + e*x]) - 1/((b*d - a*e)*(a + b*x)*Sqrt[d + e*x]) + (3*Sqrt[b]*e*ArcTanh[(Sqrt[b]
*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(5/2)

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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(a+b x)^2 (d+e x)^{3/2}} \, dx \\ & = -\frac {1}{(b d-a e) (a+b x) \sqrt {d+e x}}-\frac {(3 e) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{2 (b d-a e)} \\ & = -\frac {3 e}{(b d-a e)^2 \sqrt {d+e x}}-\frac {1}{(b d-a e) (a+b x) \sqrt {d+e x}}-\frac {(3 b e) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{2 (b d-a e)^2} \\ & = -\frac {3 e}{(b d-a e)^2 \sqrt {d+e x}}-\frac {1}{(b d-a e) (a+b x) \sqrt {d+e x}}-\frac {(3 b) \text {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)^2} \\ & = -\frac {3 e}{(b d-a e)^2 \sqrt {d+e x}}-\frac {1}{(b d-a e) (a+b x) \sqrt {d+e x}}+\frac {3 \sqrt {b} e \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {2 a e+b (d+3 e x)}{(b d-a e)^2 (a+b x) \sqrt {d+e x}}-\frac {3 \sqrt {b} e \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{5/2}} \]

[In]

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

[Out]

-((2*a*e + b*(d + 3*e*x))/((b*d - a*e)^2*(a + b*x)*Sqrt[d + e*x])) - (3*Sqrt[b]*e*ArcTan[(Sqrt[b]*Sqrt[d + e*x
])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(5/2)

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {-\frac {b \sqrt {e x +d}}{b x +a}-\frac {3 e b \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}-\frac {2 e}{\sqrt {e x +d}}}{\left (a e -b d \right )^{2}}\)	\(78\)
derivativedivides	\(2 e \left (-\frac {1}{\left (a e -b d \right )^{2} \sqrt {e x +d}}-\frac {b \left (\frac {\sqrt {e x +d}}{2 b \left (e x +d \right )+2 a e -2 b d}+\frac {3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{2}}\right )\)	\(100\)
default	\(2 e \left (-\frac {1}{\left (a e -b d \right )^{2} \sqrt {e x +d}}-\frac {b \left (\frac {\sqrt {e x +d}}{2 b \left (e x +d \right )+2 a e -2 b d}+\frac {3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{2}}\right )\)	\(100\)

[In]

int(1/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (85) = 170\).

Time = 0.39 (sec) , antiderivative size = 423, normalized size of antiderivative = 4.27 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\left [\frac {3 \, {\left (b e^{2} x^{2} + a d e + {\left (b d e + a e^{2}\right )} x\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (3 \, b e x + b d + 2 \, a e\right )} \sqrt {e x + d}}{2 \, {\left (a b^{2} d^{3} - 2 \, a^{2} b d^{2} e + a^{3} d e^{2} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} + {\left (b^{3} d^{3} - a b^{2} d^{2} e - a^{2} b d e^{2} + a^{3} e^{3}\right )} x\right )}}, \frac {3 \, {\left (b e^{2} x^{2} + a d e + {\left (b d e + a e^{2}\right )} x\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (3 \, b e x + b d + 2 \, a e\right )} \sqrt {e x + d}}{a b^{2} d^{3} - 2 \, a^{2} b d^{2} e + a^{3} d e^{2} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} + {\left (b^{3} d^{3} - a b^{2} d^{2} e - a^{2} b d e^{2} + a^{3} e^{3}\right )} x}\right ] \]

[In]

integrate(1/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")

[Out]

[1/2*(3*(b*e^2*x^2 + a*d*e + (b*d*e + a*e^2)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*s
qrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) - 2*(3*b*e*x + b*d + 2*a*e)*sqrt(e*x + d))/(a*b^2*d^3 - 2*a^2*b*d
^2*e + a^3*d*e^2 + (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x^2 + (b^3*d^3 - a*b^2*d^2*e - a^2*b*d*e^2 + a^3*e^
3)*x), (3*(b*e^2*x^2 + a*d*e + (b*d*e + a*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)) - (3*b*e*x + b*d + 2*a*e)*sqrt(e*x + d))/(a*b^2*d^3 - 2*a^2*b*d^2*e + a^3*d*e^2
+ (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x^2 + (b^3*d^3 - a*b^2*d^2*e - a^2*b*d*e^2 + a^3*e^3)*x)]

Sympy [F]

\[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\int \frac {1}{\left (a + b x\right )^{2} \left (d + e x\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(e*x+d)**(3/2)/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

Integral(1/((a + b*x)**2*(d + e*x)**(3/2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(e*x+d)^(3/2)/(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 detail

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.44 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {3 \, b e \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e}} - \frac {3 \, {\left (e x + d\right )} b e - 2 \, b d e + 2 \, a e^{2}}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} {\left ({\left (e x + d\right )}^{\frac {3}{2}} b - \sqrt {e x + d} b d + \sqrt {e x + d} a e\right )}} \]

[In]

integrate(1/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {\frac {2\,e}{a\,e-b\,d}+\frac {3\,b\,e\,\left (d+e\,x\right )}{{\left (a\,e-b\,d\right )}^2}}{b\,{\left (d+e\,x\right )}^{3/2}+\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}}-\frac {3\,\sqrt {b}\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{{\left (a\,e-b\,d\right )}^{5/2}}\right )}{{\left (a\,e-b\,d\right )}^{5/2}} \]

[In]

int(1/((d + e*x)^(3/2)*(a^2 + b^2*x^2 + 2*a*b*x)),x)

[Out]

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