\(\int \frac {(d+e x)^{5/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [1704]

Optimal result

Integrand size = 30, antiderivative size = 212 \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (b d-a e)^2 (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (b d-a e)^{5/2} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {660, 52, 65, 214} \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 (a+b x) (b d-a e)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{3/2} (b d-a e)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) \sqrt {d+e x} (b d-a e)^2}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

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

[Out]

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

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 (a b+b^2 x\right ) \int \frac {(d+e x)^{5/2}}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (\left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{3/2}}{a b+b^2 x} \, dx}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {2 (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (\left (b^2 d-a b e\right )^2 \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{a b+b^2 x} \, dx}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {2 (b d-a e)^2 (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (\left (b^2 d-a b e\right )^3 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {2 (b d-a e)^2 (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (b^2 d-a b e\right )^3 \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 )}{b^6 e \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {2 (b d-a e)^2 (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (b d-a e)^{5/2} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.60 \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\sqrt {b} \sqrt {d+e x} \left (15 a^2 e^2-5 a b e (7 d+e x)+b^2 \left (23 d^2+11 d e x+3 e^2 x^2\right )\right )-15 (-b d+a e)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{15 b^{7/2} \sqrt {(a+b x)^2}} \]

[In]

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

[Out]

(2*(a + b*x)*(Sqrt[b]*Sqrt[d + e*x]*(15*a^2*e^2 - 5*a*b*e*(7*d + e*x) + b^2*(23*d^2 + 11*d*e*x + 3*e^2*x^2)) -
15*(-(b*d) + a*e)^(5/2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]]))/(15*b^(7/2)*Sqrt[(a + b*x)^2])

Maple [A] (verified)

Time = 2.12 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {2 \left (3 x^{2} b^{2} e^{2}-5 x a b \,e^{2}+11 b^{2} d e x +15 a^{2} e^{2}-35 a b d e +23 b^{2} d^{2}\right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{15 b^{3} \left (b x +a \right )}-\frac {2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{3} \sqrt {\left (a e -b d \right ) b}\, \left (b x +a \right )}\)	\(171\)
default	\(\frac {2 \left (b x +a \right ) \left (3 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{2}-5 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a b e +5 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{2} d -15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} e^{3}+45 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b d \,e^{2}-45 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} d^{2} e +15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{3} d^{3}+15 \sqrt {e x +d}\, a^{2} e^{2} \sqrt {\left (a e -b d \right ) b}-30 \sqrt {e x +d}\, a d e b \sqrt {\left (a e -b d \right ) b}+15 \sqrt {e x +d}\, d^{2} b^{2} \sqrt {\left (a e -b d \right ) b}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, b^{3} \sqrt {\left (a e -b d \right ) b}}\)	\(309\)

[In]

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

[Out]

2/15*(3*b^2*e^2*x^2-5*a*b*e^2*x+11*b^2*d*e*x+15*a^2*e^2-35*a*b*d*e+23*b^2*d^2)*(e*x+d)^(1/2)/b^3*((b*x+a)^2)^(
1/2)/(b*x+a)-2/b^3*(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)/((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 = 0.45 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.37 \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [\frac {15 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (3 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 35 \, a b d e + 15 \, a^{2} e^{2} + {\left (11 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (3 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 35 \, a b d e + 15 \, a^{2} e^{2} + {\left (11 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, b^{3}}\right ] \]

[In]

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

[Out]

[1/15*(15*(b^2*d^2 - 2*a*b*d*e + 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*sqr
t((b*d - a*e)/b))/(b*x + a)) + 2*(3*b^2*e^2*x^2 + 23*b^2*d^2 - 35*a*b*d*e + 15*a^2*e^2 + (11*b^2*d*e - 5*a*b*e
^2)*x)*sqrt(e*x + d))/b^3, -2/15*(15*(b^2*d^2 - 2*a*b*d*e + 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)) - (3*b^2*e^2*x^2 + 23*b^2*d^2 - 35*a*b*d*e + 15*a^2*e^2 + (11*b^2*d*e -
5*a*b*e^2)*x)*sqrt(e*x + d))/b^3]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.09 \[ \int \frac {(d+e x)^{5/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \, {\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{4} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{4} d \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {e x + d} b^{4} d^{2} \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{3} e \mathrm {sgn}\left (b x + a\right ) - 30 \, \sqrt {e x + d} a b^{3} d e \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {e x + d} a^{2} b^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, b^{5}} \]

[In]

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

[Out]

2*(b^3*d^3*sgn(b*x + a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) - a^3*e^3*sgn(b*x + a))*arct
an(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^3) + 2/15*(3*(e*x + d)^(5/2)*b^4*sgn(b*x + a)
+ 5*(e*x + d)^(3/2)*b^4*d*sgn(b*x + a) + 15*sqrt(e*x + d)*b^4*d^2*sgn(b*x + a) - 5*(e*x + d)^(3/2)*a*b^3*e*sg
n(b*x + a) - 30*sqrt(e*x + d)*a*b^3*d*e*sgn(b*x + a) + 15*sqrt(e*x + d)*a^2*b^2*e^2*sgn(b*x + a))/b^5

Mupad [F(-1)]

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

[In]

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

[Out]

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