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

Optimal. Leaf size=212 \[ \frac {2 (b d-a e)^2 (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{5/2}}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (b d-a e)^{5/2} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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)

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Rubi [A]
time = 0.08, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {660, 52, 65, 214} \begin {gather*} \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) (b d-a e)^{5/2} \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}}+\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}} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.16, size = 127, normalized size = 0.60 \begin {gather*} \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} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{15 b^{7/2} \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[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])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(308\) vs. \(2(148)=296\).
time = 0.58, size = 309, normalized size = 1.46

method	result	size
risch	\(\frac {2 \left (3 b^{2} x^{2} e^{2}-5 a b \,e^{2} x +11 b^{2} d e x +15 a^{2} e^{2}-35 a b d e +23 b^{2} d^{2}\right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{15 b^{3} \left (b x +a \right )}+\frac {\left (-\frac {2 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) e^{3} a^{3}}{b^{3} \sqrt {b \left (a e -b d \right )}}+\frac {6 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2} d \,e^{2}}{b^{2} \sqrt {b \left (a e -b d \right )}}-\frac {6 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a \,d^{2} e}{b \sqrt {b \left (a e -b d \right )}}+\frac {2 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) d^{3}}{\sqrt {b \left (a e -b d \right )}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\)	\(272\)
default	\(\frac {2 \left (b x +a \right ) \left (3 \left (e x +d \right )^{\frac {5}{2}} \sqrt {b \left (a e -b d \right )}\, b^{2}-5 \left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, a b e +5 \left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, b^{2} d -15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{3} e^{3}+45 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2} b d \,e^{2}-45 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a \,b^{2} d^{2} e +15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) b^{3} d^{3}+15 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a^{2} e^{2}-30 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a b d e +15 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, b^{2} d^{2}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, b^{3} \sqrt {b \left (a e -b d \right )}}\)	\(309\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(b*x+a)*(3*(e*x+d)^(5/2)*(b*(a*e-b*d))^(1/2)*b^2-5*(e*x+d)^(3/2)*(b*(a*e-b*d))^(1/2)*a*b*e+5*(e*x+d)^(3/2
)*(b*(a*e-b*d))^(1/2)*b^2*d-15*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a^3*e^3+45*arctan(b*(e*x+d)^(1/2)/(
b*(a*e-b*d))^(1/2))*a^2*b*d*e^2-45*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a*b^2*d^2*e+15*arctan(b*(e*x+d)
^(1/2)/(b*(a*e-b*d))^(1/2))*b^3*d^3+15*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*a^2*e^2-30*(e*x+d)^(1/2)*(b*(a*e-b*d)
)^(1/2)*a*b*d*e+15*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*b^2*d^2)/((b*x+a)^2)^(1/2)/b^3/(b*(a*e-b*d))^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.36, size = 293, normalized size = 1.38 \begin {gather*} \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 {2 \, b d - 2 \, \sqrt {x e + d} b \sqrt {\frac {b d - a e}{b}} + {\left (b x - a\right )} e}{b x + a}\right ) + 2 \, {\left (23 \, b^{2} d^{2} + {\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} e^{2} + {\left (11 \, b^{2} d x - 35 \, a b d\right )} e\right )} \sqrt {x e + 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 {x e + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (23 \, b^{2} d^{2} + {\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} e^{2} + {\left (11 \, b^{2} d x - 35 \, a b d\right )} e\right )} \sqrt {x e + d}\right )}}{15 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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((2*b*d - 2*sqrt(x*e + d)*b*sqrt((b*d - a*e)/
b) + (b*x - a)*e)/(b*x + a)) + 2*(23*b^2*d^2 + (3*b^2*x^2 - 5*a*b*x + 15*a^2)*e^2 + (11*b^2*d*x - 35*a*b*d)*e)
*sqrt(x*e + 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(x*e + d)*b*sq
rt(-(b*d - a*e)/b)/(b*d - a*e)) - (23*b^2*d^2 + (3*b^2*x^2 - 5*a*b*x + 15*a^2)*e^2 + (11*b^2*d*x - 35*a*b*d)*e
)*sqrt(x*e + d))/b^3]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{\frac {5}{2}}}{\sqrt {\left (a + b x\right )^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]
time = 1.59, size = 240, normalized size = 1.13 \begin {gather*} \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 {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{4} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} d \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {x e + d} b^{4} d^{2} \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{3} e \mathrm {sgn}\left (b x + a\right ) - 30 \, \sqrt {x e + d} a b^{3} d e \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {x e + d} a^{2} b^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^{5/2}}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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