\(\int \frac {(d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx\) [1647]

Optimal result

Integrand size = 28, antiderivative size = 110 \[ \int \frac {(d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {5 e (b d-a e) \sqrt {d+e x}}{b^3}+\frac {5 e (d+e x)^{3/2}}{3 b^2}-\frac {(d+e x)^{5/2}}{b (a+b x)}-\frac {5 e (b d-a e)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 43, 52, 65, 214} \[ \int \frac {(d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {5 e (b d-a e)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}+\frac {5 e \sqrt {d+e x} (b d-a e)}{b^3}-\frac {(d+e x)^{5/2}}{b (a+b x)}+\frac {5 e (d+e x)^{3/2}}{3 b^2} \]

[In]

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

[Out]

(5*e*(b*d - a*e)*Sqrt[d + e*x])/b^3 + (5*e*(d + e*x)^(3/2))/(3*b^2) - (d + e*x)^(5/2)/(b*(a + b*x)) - (5*e*(b*
d - a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {\sqrt {d+e x} \left (15 a^2 e^2+10 a b e (-2 d+e x)+b^2 \left (3 d^2-14 d e x-2 e^2 x^2\right )\right )}{3 b^3 (a+b x)}+\frac {5 e (-b d+a e)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{b^{7/2}} \]

[In]

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

[Out]

-1/3*(Sqrt[d + e*x]*(15*a^2*e^2 + 10*a*b*e*(-2*d + e*x) + b^2*(3*d^2 - 14*d*e*x - 2*e^2*x^2)))/(b^3*(a + b*x))
+ (5*e*(-(b*d) + a*e)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/b^(7/2)

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.09

method	result	size
risch	\(-\frac {2 e \left (-b e x +6 a e -7 b d \right ) \sqrt {e x +d}}{3 b^{3}}+\frac {\left (2 a^{2} e^{2}-4 a b d e +2 b^{2} d^{2}\right ) e \left (-\frac {\sqrt {e x +d}}{2 \left (b \left (e x +d \right )+a e -b d \right )}+\frac {5 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{b^{3}}\)	\(120\)
pseudoelliptic	\(-\frac {5 \left (-e \left (a e -b d \right )^{2} \left (b x +a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )+\left (\frac {\left (-\frac {2}{3} x^{2} e^{2}-\frac {14}{3} d e x +d^{2}\right ) b^{2}}{5}-\frac {4 e \left (-\frac {e x}{2}+d \right ) a b}{3}+a^{2} e^{2}\right ) \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3} \left (b x +a \right )}\)	\(127\)
derivativedivides	\(2 e \left (-\frac {-\frac {\left (e x +d \right )^{\frac {3}{2}} b}{3}+2 a e \sqrt {e x +d}-2 d b \sqrt {e x +d}}{b^{3}}+\frac {\frac {\left (-\frac {1}{2} a^{2} e^{2}+a b d e -\frac {1}{2} b^{2} d^{2}\right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {5 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}}{b^{3}}\right )\)	\(152\)
default	\(2 e \left (-\frac {-\frac {\left (e x +d \right )^{\frac {3}{2}} b}{3}+2 a e \sqrt {e x +d}-2 d b \sqrt {e x +d}}{b^{3}}+\frac {\frac {\left (-\frac {1}{2} a^{2} e^{2}+a b d e -\frac {1}{2} b^{2} d^{2}\right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {5 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}}{b^{3}}\right )\)	\(152\)

[In]

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

[Out]

-2/3*e*(-b*e*x+6*a*e-7*b*d)*(e*x+d)^(1/2)/b^3+1/b^3*(2*a^2*e^2-4*a*b*d*e+2*b^2*d^2)*e*(-1/2*(e*x+d)^(1/2)/(b*(
e*x+d)+a*e-b*d)+5/2/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 330, normalized size of antiderivative = 3.00 \[ \int \frac {(d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\left [-\frac {15 \, {\left (a b d e - a^{2} e^{2} + {\left (b^{2} d e - a b e^{2}\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (2 \, b^{2} e^{2} x^{2} - 3 \, b^{2} d^{2} + 20 \, a b d e - 15 \, a^{2} e^{2} + 2 \, {\left (7 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, -\frac {15 \, {\left (a b d e - a^{2} e^{2} + {\left (b^{2} d e - a b e^{2}\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (2 \, b^{2} e^{2} x^{2} - 3 \, b^{2} d^{2} + 20 \, a b d e - 15 \, a^{2} e^{2} + 2 \, {\left (7 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \]

[In]

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

[Out]

[-1/6*(15*(a*b*d*e - a^2*e^2 + (b^2*d*e - a*b*e^2)*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*(2*b^2*e^2*x^2 - 3*b^2*d^2 + 20*a*b*d*e - 15*a^2*e^2 + 2*(7*b^2*d
*e - 5*a*b*e^2)*x)*sqrt(e*x + d))/(b^4*x + a*b^3), -1/3*(15*(a*b*d*e - a^2*e^2 + (b^2*d*e - a*b*e^2)*x)*sqrt(-
(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (2*b^2*e^2*x^2 - 3*b^2*d^2 + 20*a*b
*d*e - 15*a^2*e^2 + 2*(7*b^2*d*e - 5*a*b*e^2)*x)*sqrt(e*x + d))/(b^4*x + a*b^3)]

Sympy [F]

\[ \int \frac {(d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\int \frac {\left (d + e x\right )^{\frac {5}{2}}}{\left (a + b x\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((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 detail

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.65 \[ \int \frac {(d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {5 \, {\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} - \frac {\sqrt {e x + d} b^{2} d^{2} e - 2 \, \sqrt {e x + d} a b d e^{2} + \sqrt {e x + d} a^{2} e^{3}}{{\left ({\left (e x + d\right )} b - b d + a e\right )} b^{3}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} b^{4} e + 6 \, \sqrt {e x + d} b^{4} d e - 6 \, \sqrt {e x + d} a b^{3} e^{2}\right )}}{3 \, b^{6}} \]

[In]

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

[Out]

5*(b^2*d^2*e - 2*a*b*d*e^2 + a^2*e^3)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^3)
- (sqrt(e*x + d)*b^2*d^2*e - 2*sqrt(e*x + d)*a*b*d*e^2 + sqrt(e*x + d)*a^2*e^3)/(((e*x + d)*b - b*d + a*e)*b^3
) + 2/3*((e*x + d)^(3/2)*b^4*e + 6*sqrt(e*x + d)*b^4*d*e - 6*sqrt(e*x + d)*a*b^3*e^2)/b^6

Mupad [B] (verification not implemented)

Time = 9.41 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.46 \[ \int \frac {(d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {2\,e\,{\left (d+e\,x\right )}^{3/2}}{3\,b^2}-\frac {\sqrt {d+e\,x}\,\left (a^2\,e^3-2\,a\,b\,d\,e^2+b^2\,d^2\,e\right )}{b^4\,\left (d+e\,x\right )-b^4\,d+a\,b^3\,e}+\frac {5\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,e\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^3-2\,a\,b\,d\,e^2+b^2\,d^2\,e}\right )\,{\left (a\,e-b\,d\right )}^{3/2}}{b^{7/2}}+\frac {2\,e\,\left (2\,b^2\,d-2\,a\,b\,e\right )\,\sqrt {d+e\,x}}{b^4} \]

[In]

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

[Out]

(2*e*(d + e*x)^(3/2))/(3*b^2) - ((d + e*x)^(1/2)*(a^2*e^3 + b^2*d^2*e - 2*a*b*d*e^2))/(b^4*(d + e*x) - b^4*d +
a*b^3*e) + (5*e*atan((b^(1/2)*e*(a*e - b*d)^(3/2)*(d + e*x)^(1/2))/(a^2*e^3 + b^2*d^2*e - 2*a*b*d*e^2))*(a*e
- b*d)^(3/2))/b^(7/2) + (2*e*(2*b^2*d - 2*a*b*e)*(d + e*x)^(1/2))/b^4