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

Optimal. Leaf size=202 \[ \frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 e^2 \sqrt {b d-a e} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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)

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Rubi [A]
time = 0.07, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {660, 43, 52, 65, 214} \begin {gather*} -\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 {b d-a e} \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}}+\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}} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.40, size = 138, normalized size = 0.68 \begin {gather*} \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 \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[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])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(412\) vs. \(2(134)=268\).
time = 0.86, size = 413, normalized size = 2.04

method	result	size
risch	\(\frac {2 e^{2} \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{b^{3} \left (b x +a \right )}+\frac {\left (\frac {9 e^{3} \left (e x +d \right )^{\frac {3}{2}} a}{4 b^{2} \left (b e x +a e \right )^{2}}-\frac {9 e^{2} \left (e x +d \right )^{\frac {3}{2}} d}{4 b \left (b e x +a e \right )^{2}}+\frac {7 e^{4} \sqrt {e x +d}\, a^{2}}{4 b^{3} \left (b e x +a e \right )^{2}}-\frac {7 e^{3} \sqrt {e x +d}\, a d}{2 b^{2} \left (b e x +a e \right )^{2}}+\frac {7 e^{2} \sqrt {e x +d}\, d^{2}}{4 b \left (b e x +a e \right )^{2}}-\frac {15 e^{3} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a}{4 b^{3} \sqrt {b \left (a e -b d \right )}}+\frac {15 e^{2} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) d}{4 b^{2} \sqrt {b \left (a e -b d \right )}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\)	\(272\)
default	\(\frac {\left (-15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a \,b^{2} e^{3} x^{2}+15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) b^{3} d \,e^{2} x^{2}+8 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, b^{2} e^{2} x^{2}-30 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2} b \,e^{3} x +30 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a \,b^{2} d \,e^{2} x +9 \left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, a b e -9 \left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, b^{2} d +16 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a b \,e^{2} x -15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{3} e^{3}+15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2} b d \,e^{2}+15 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a^{2} e^{2}-14 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a b d e +7 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, b^{2} d^{2}\right ) \left (b x +a \right )}{4 \sqrt {b \left (a e -b d \right )}\, b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(413\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-15*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a*b^2*e^3*x^2+15*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/
2))*b^3*d*e^2*x^2+8*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*b^2*e^2*x^2-30*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2
))*a^2*b*e^3*x+30*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a*b^2*d*e^2*x+9*(e*x+d)^(3/2)*(b*(a*e-b*d))^(1/2
)*a*b*e-9*(e*x+d)^(3/2)*(b*(a*e-b*d))^(1/2)*b^2*d+16*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*a*b*e^2*x-15*arctan(b*(
e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a^3*e^3+15*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a^2*b*d*e^2+15*(e*x+d
)^(1/2)*(b*(a*e-b*d))^(1/2)*a^2*e^2-14*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*a*b*d*e+7*(e*x+d)^(1/2)*(b*(a*e-b*d))
^(1/2)*b^2*d^2)*(b*x+a)/(b*(a*e-b*d))^(1/2)/b^3/((b*x+a)^2)^(3/2)

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

[Out]

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

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Fricas [A]
time = 2.32, size = 330, normalized size = 1.63 \begin {gather*} \left [\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {\frac {b d - a e}{b}} e^{2} \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 (2 \, b^{2} d^{2} - {\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} e^{2} + {\left (9 \, b^{2} d x + 5 \, a b d\right )} e\right )} \sqrt {x e + d}}{8 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{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 ) e^{2} + {\left (2 \, b^{2} d^{2} - {\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} e^{2} + {\left (9 \, b^{2} d x + 5 \, a b d\right )} e\right )} \sqrt {x e + d}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*x^2 + 2*a*b*x + a^2)*sqrt((b*d - a*e)/b)*e^2*log((2*b*d - 2*sqrt(x*e + d)*b*sqrt((b*d - a*e)/b)
+ (b*x - a)*e)/(b*x + a)) - 2*(2*b^2*d^2 - (8*b^2*x^2 + 25*a*b*x + 15*a^2)*e^2 + (9*b^2*d*x + 5*a*b*d)*e)*sqrt
(x*e + d))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-(b*d - a*e)/b)*arctan(-sq
rt(x*e + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e))*e^2 + (2*b^2*d^2 - (8*b^2*x^2 + 25*a*b*x + 15*a^2)*e^2 + (9*b^
2*d*x + 5*a*b*d)*e)*sqrt(x*e + d))/(b^5*x^2 + 2*a*b^4*x + a^2*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}}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]
time = 1.40, size = 198, normalized size = 0.98 \begin {gather*} \frac {15 \, {\left (b d e^{2} - a e^{3}\right )} \arctan \left (\frac {\sqrt {x e + 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 {2 \, \sqrt {x e + d} e^{2}}{b^{3} \mathrm {sgn}\left (b x + a\right )} - \frac {9 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{2} d e^{2} - 7 \, \sqrt {x e + d} b^{2} d^{2} e^{2} - 9 \, {\left (x e + d\right )}^{\frac {3}{2}} a b e^{3} + 14 \, \sqrt {x e + d} a b d e^{3} - 7 \, \sqrt {x e + d} a^{2} e^{4}}{4 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{3} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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