\(\int \frac {(d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [1713]

Optimal result

Integrand size = 30, antiderivative size = 202 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 e^2 \sqrt {b d-a e} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

-5/4*e*(e*x+d)^(3/2)/b^2/((b*x+a)^2)^(1/2)-1/2*(e*x+d)^(5/2)/b/(b*x+a)/((b*x+a)^2)^(1/2)-15/4*e^2*(b*x+a)*arct
anh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))*(-a*e+b*d)^(1/2)/b^(7/2)/((b*x+a)^2)^(1/2)+15/4*e^2*(b*x+a)*(e*x+d
)^(1/2)/b^3/((b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {660, 43, 52, 65, 214} \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {15 e^2 (a+b x) \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

Int[(d + e*x)^(5/2)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(15*e^2*(a + b*x)*Sqrt[d + e*x])/(4*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e*(d + e*x)^(3/2))/(4*b^2*Sqrt[a^2
+ 2*a*b*x + b^2*x^2]) - (d + e*x)^(5/2)/(2*b*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (15*e^2*Sqrt[b*d - a*
e]*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*b^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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]

Rule 660

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*} \text {integral}& = \frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{5/2}}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 e \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{3/2}}{\left (a b+b^2 x\right )^2} \, dx}{4 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (15 e^2 \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{a b+b^2 x} \, dx}{8 b^2 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (15 e^2 \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (15 e \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \text {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 )}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 e^2 \sqrt {b d-a e} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.68 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {b} \sqrt {d+e x} \left (15 a^2 e^2-5 a b e (d-5 e x)+b^2 \left (-2 d^2-9 d e x+8 e^2 x^2\right )\right )-15 e^2 \sqrt {-b d+a e} (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{4 b^{7/2} (a+b x) \sqrt {(a+b x)^2}} \]

[In]

Integrate[(d + e*x)^(5/2)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(Sqrt[b]*Sqrt[d + e*x]*(15*a^2*e^2 - 5*a*b*e*(d - 5*e*x) + b^2*(-2*d^2 - 9*d*e*x + 8*e^2*x^2)) - 15*e^2*Sqrt[-
(b*d) + a*e]*(a + b*x)^2*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(4*b^(7/2)*(a + b*x)*Sqrt[(a + b*
x)^2])

Maple [A] (verified)

Time = 2.31 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {2 e^{2} \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{b^{3} \left (b x +a \right )}-\frac {\left (2 a e -2 b d \right ) e^{2} \left (\frac {-\frac {9 \left (e x +d \right )^{\frac {3}{2}} b}{8}+\left (-\frac {7 a e}{8}+\frac {7 b d}{8}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{3} \left (b x +a \right )}\)	\(149\)
default	\(\frac {\left (-15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} e^{3} x^{2}+15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{3} d \,e^{2} x^{2}+8 b^{2} e^{2} x^{2} \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}-30 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b \,e^{3} x +30 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} d \,e^{2} x +9 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a b e -9 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{2} d +16 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a b \,e^{2} x -15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} e^{3}+15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b d \,e^{2}+15 \sqrt {e x +d}\, a^{2} e^{2} \sqrt {\left (a e -b d \right ) b}-14 \sqrt {e x +d}\, a d e b \sqrt {\left (a e -b d \right ) b}+7 \sqrt {e x +d}\, d^{2} b^{2} \sqrt {\left (a e -b d \right ) b}\right ) \left (b x +a \right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(413\)

[In]

int((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/b^3*e^2*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)-1/b^3*(2*a*e-2*b*d)*e^2*((-9/8*(e*x+d)^(3/2)*b+(-7/8*a*e+7/8
*b*d)*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b*d)^2+15/8/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)
))*((b*x+a)^2)^(1/2)/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.70 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (8 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} - 5 \, a b d e + 15 \, a^{2} e^{2} - {\left (9 \, b^{2} d e - 25 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (8 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} - 5 \, a b d e + 15 \, a^{2} e^{2} - {\left (9 \, b^{2} d e - 25 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \]

[In]

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(15*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*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^2*e^2*x^2 - 2*b^2*d^2 - 5*a*b*d*e + 15*a^2*e^2 - (9*b^2*d*e - 25*a*
b*e^2)*x)*sqrt(e*x + d))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(
-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (8*b^2*e^2*x^2 - 2*b^2*d^2 - 5*a*b
*d*e + 15*a^2*e^2 - (9*b^2*d*e - 25*a*b*e^2)*x)*sqrt(e*x + d))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)]

Sympy [F]

\[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {5}{2}}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral((d + e*x)**(5/2)/((a + b*x)**2)**(3/2), x)

Maxima [F]

\[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {5}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(5/2)/(b^2*x^2 + 2*a*b*x + a^2)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.97 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {2 \, \sqrt {e x + d} e^{2}}{b^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {15 \, {\left (b d e^{2} - a e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{3} \mathrm {sgn}\left (b x + a\right )} - \frac {9 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{2} d e^{2} - 7 \, \sqrt {e x + d} b^{2} d^{2} e^{2} - 9 \, {\left (e x + d\right )}^{\frac {3}{2}} a b e^{3} + 14 \, \sqrt {e x + d} a b d e^{3} - 7 \, \sqrt {e x + d} a^{2} e^{4}}{4 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2} b^{3} \mathrm {sgn}\left (b x + a\right )} \]

[In]

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

2*sqrt(e*x + d)*e^2/(b^3*sgn(b*x + a)) + 15/4*(b*d*e^2 - a*e^3)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(
sqrt(-b^2*d + a*b*e)*b^3*sgn(b*x + a)) - 1/4*(9*(e*x + d)^(3/2)*b^2*d*e^2 - 7*sqrt(e*x + d)*b^2*d^2*e^2 - 9*(e
*x + d)^(3/2)*a*b*e^3 + 14*sqrt(e*x + d)*a*b*d*e^3 - 7*sqrt(e*x + d)*a^2*e^4)/(((e*x + d)*b - b*d + a*e)^2*b^3
*sgn(b*x + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^{5/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

[In]

int((d + e*x)^(5/2)/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int((d + e*x)^(5/2)/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)