3.17.57 \(\int \frac {(d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [1657]

Optimal. Leaf size=126 \[ -\frac {5 e^2 \sqrt {d+e x}}{8 b^3 (a+b x)}-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}-\frac {5 e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{7/2} \sqrt {b d-a e}} \]

[Out]

-5/12*e*(e*x+d)^(3/2)/b^2/(b*x+a)^2-1/3*(e*x+d)^(5/2)/b/(b*x+a)^3-5/8*e^3*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+
b*d)^(1/2))/b^(7/2)/(-a*e+b*d)^(1/2)-5/8*e^2*(e*x+d)^(1/2)/b^3/(b*x+a)

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Rubi [A]
time = 0.04, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 43, 65, 214} \begin {gather*} -\frac {5 e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{7/2} \sqrt {b d-a e}}-\frac {5 e^2 \sqrt {d+e x}}{8 b^3 (a+b x)}-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*e^2*Sqrt[d + e*x])/(8*b^3*(a + b*x)) - (5*e*(d + e*x)^(3/2))/(12*b^2*(a + b*x)^2) - (d + e*x)^(5/2)/(3*b*(
a + b*x)^3) - (5*e^3*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(7/2)*Sqrt[b*d - a*e])

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 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*} \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {(d+e x)^{5/2}}{(a+b x)^4} \, dx\\ &=-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}+\frac {(5 e) \int \frac {(d+e x)^{3/2}}{(a+b x)^3} \, dx}{6 b}\\ &=-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}+\frac {\left (5 e^2\right ) \int \frac {\sqrt {d+e x}}{(a+b x)^2} \, dx}{8 b^2}\\ &=-\frac {5 e^2 \sqrt {d+e x}}{8 b^3 (a+b x)}-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}+\frac {\left (5 e^3\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 b^3}\\ &=-\frac {5 e^2 \sqrt {d+e x}}{8 b^3 (a+b x)}-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}+\frac {\left (5 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 )}{8 b^3}\\ &=-\frac {5 e^2 \sqrt {d+e x}}{8 b^3 (a+b x)}-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}-\frac {5 e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{7/2} \sqrt {b d-a e}}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 119, normalized size = 0.94 \begin {gather*} -\frac {\sqrt {d+e x} \left (15 a^2 e^2+10 a b e (d+4 e x)+b^2 \left (8 d^2+26 d e x+33 e^2 x^2\right )\right )}{24 b^3 (a+b x)^3}+\frac {5 e^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{8 b^{7/2} \sqrt {-b d+a e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(Sqrt[d + e*x]*(15*a^2*e^2 + 10*a*b*e*(d + 4*e*x) + b^2*(8*d^2 + 26*d*e*x + 33*e^2*x^2)))/(b^3*(a + b*x)
^3) + (5*e^3*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(8*b^(7/2)*Sqrt[-(b*d) + a*e])

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Maple [A]
time = 0.66, size = 130, normalized size = 1.03

method	result	size
derivativedivides	\(2 e^{3} \left (\frac {-\frac {11 \left (e x +d \right )^{\frac {5}{2}}}{16 b}-\frac {5 \left (a e -b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{6 b^{2}}-\frac {5 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {e x +d}}{16 b^{3}}}{\left (\left (e x +d \right ) b +a e -b d \right )^{3}}+\frac {5 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{16 b^{3} \sqrt {b \left (a e -b d \right )}}\right )\)	\(130\)
default	\(2 e^{3} \left (\frac {-\frac {11 \left (e x +d \right )^{\frac {5}{2}}}{16 b}-\frac {5 \left (a e -b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{6 b^{2}}-\frac {5 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {e x +d}}{16 b^{3}}}{\left (\left (e x +d \right ) b +a e -b d \right )^{3}}+\frac {5 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{16 b^{3} \sqrt {b \left (a e -b d \right )}}\right )\)	\(130\)

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)^2,x,method=_RETURNVERBOSE)

[Out]

2*e^3*((-11/16/b*(e*x+d)^(5/2)-5/6*(a*e-b*d)/b^2*(e*x+d)^(3/2)-5/16/b^3*(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(1
/2))/((e*x+d)*b+a*e-b*d)^3+5/16/b^3/(b*(a*e-b*d))^(1/2)*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \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)^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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (107) = 214\).
time = 3.37, size = 538, normalized size = 4.27 \begin {gather*} \left [\frac {15 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \sqrt {b^{2} d - a b e} e^{3} \log \left (\frac {2 \, b d + {\left (b x - a\right )} e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {x e + d}}{b x + a}\right ) - 2 \, {\left (8 \, b^{4} d^{3} - {\left (33 \, a b^{3} x^{2} + 40 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} e^{3} + {\left (33 \, b^{4} d x^{2} + 14 \, a b^{3} d x + 5 \, a^{2} b^{2} d\right )} e^{2} + 2 \, {\left (13 \, b^{4} d^{2} x + a b^{3} d^{2}\right )} e\right )} \sqrt {x e + d}}{48 \, {\left (b^{8} d x^{3} + 3 \, a b^{7} d x^{2} + 3 \, a^{2} b^{6} d x + a^{3} b^{5} d - {\left (a b^{7} x^{3} + 3 \, a^{2} b^{6} x^{2} + 3 \, a^{3} b^{5} x + a^{4} b^{4}\right )} e\right )}}, \frac {15 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {x e + d}}{b x e + b d}\right ) e^{3} - {\left (8 \, b^{4} d^{3} - {\left (33 \, a b^{3} x^{2} + 40 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} e^{3} + {\left (33 \, b^{4} d x^{2} + 14 \, a b^{3} d x + 5 \, a^{2} b^{2} d\right )} e^{2} + 2 \, {\left (13 \, b^{4} d^{2} x + a b^{3} d^{2}\right )} e\right )} \sqrt {x e + d}}{24 \, {\left (b^{8} d x^{3} + 3 \, a b^{7} d x^{2} + 3 \, a^{2} b^{6} d x + a^{3} b^{5} d - {\left (a b^{7} x^{3} + 3 \, a^{2} b^{6} x^{2} + 3 \, a^{3} b^{5} x + a^{4} b^{4}\right )} e\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)^2,x, algorithm="fricas")

[Out]

[1/48*(15*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(b^2*d - a*b*e)*e^3*log((2*b*d + (b*x - a)*e - 2*sqrt(
b^2*d - a*b*e)*sqrt(x*e + d))/(b*x + a)) - 2*(8*b^4*d^3 - (33*a*b^3*x^2 + 40*a^2*b^2*x + 15*a^3*b)*e^3 + (33*b
^4*d*x^2 + 14*a*b^3*d*x + 5*a^2*b^2*d)*e^2 + 2*(13*b^4*d^2*x + a*b^3*d^2)*e)*sqrt(x*e + d))/(b^8*d*x^3 + 3*a*b
^7*d*x^2 + 3*a^2*b^6*d*x + a^3*b^5*d - (a*b^7*x^3 + 3*a^2*b^6*x^2 + 3*a^3*b^5*x + a^4*b^4)*e), 1/24*(15*(b^3*x
^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(x*e + d)/(b*x*e + b*
d))*e^3 - (8*b^4*d^3 - (33*a*b^3*x^2 + 40*a^2*b^2*x + 15*a^3*b)*e^3 + (33*b^4*d*x^2 + 14*a*b^3*d*x + 5*a^2*b^2
*d)*e^2 + 2*(13*b^4*d^2*x + a*b^3*d^2)*e)*sqrt(x*e + d))/(b^8*d*x^3 + 3*a*b^7*d*x^2 + 3*a^2*b^6*d*x + a^3*b^5*
d - (a*b^7*x^3 + 3*a^2*b^6*x^2 + 3*a^3*b^5*x + a^4*b^4)*e)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**2,x)

[Out]

Timed out

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Giac [A]
time = 0.93, size = 165, normalized size = 1.31 \begin {gather*} \frac {5 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{3}}{8 \, \sqrt {-b^{2} d + a b e} b^{3}} - \frac {33 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{2} e^{3} - 40 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{2} d e^{3} + 15 \, \sqrt {x e + d} b^{2} d^{2} e^{3} + 40 \, {\left (x e + d\right )}^{\frac {3}{2}} a b e^{4} - 30 \, \sqrt {x e + d} a b d e^{4} + 15 \, \sqrt {x e + d} a^{2} e^{5}}{24 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{3}} \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)^2,x, algorithm="giac")

[Out]

5/8*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^3/(sqrt(-b^2*d + a*b*e)*b^3) - 1/24*(33*(x*e + d)^(5/2)*b^2
*e^3 - 40*(x*e + d)^(3/2)*b^2*d*e^3 + 15*sqrt(x*e + d)*b^2*d^2*e^3 + 40*(x*e + d)^(3/2)*a*b*e^4 - 30*sqrt(x*e
+ d)*a*b*d*e^4 + 15*sqrt(x*e + d)*a^2*e^5)/(((x*e + d)*b - b*d + a*e)^3*b^3)

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Mupad [B]
time = 0.17, size = 222, normalized size = 1.76 \begin {gather*} \frac {5\,e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{8\,b^{7/2}\,\sqrt {a\,e-b\,d}}-\frac {\frac {11\,e^3\,{\left (d+e\,x\right )}^{5/2}}{8\,b}+\frac {5\,e^3\,\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{8\,b^3}+\frac {5\,e^3\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,b^2}}{\left (d+e\,x\right )\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )+b^3\,{\left (d+e\,x\right )}^3-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^2+a^3\,e^3-b^3\,d^3+3\,a\,b^2\,d^2\,e-3\,a^2\,b\,d\,e^2} \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)^2,x)

[Out]

(5*e^3*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(8*b^(7/2)*(a*e - b*d)^(1/2)) - ((11*e^3*(d + e*x)^(
5/2))/(8*b) + (5*e^3*(d + e*x)^(1/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*d*e))/(8*b^3) + (5*e^3*(a*e - b*d)*(d + e*x)^(
3/2))/(3*b^2))/((d + e*x)*(3*b^3*d^2 + 3*a^2*b*e^2 - 6*a*b^2*d*e) + b^3*(d + e*x)^3 - (3*b^3*d - 3*a*b^2*e)*(d
+ e*x)^2 + a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2)

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