3.14.49 \(\int \frac {(d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=110 \[ -\frac {5 e (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}+\frac {5 e \sqrt {d+e x} (b d-a e)}{b^3}-\frac {(d+e x)^{5/2}}{b (a+b x)}+\frac {5 e (d+e x)^{3/2}}{3 b^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 47, 50, 63, 208} \begin {gather*} \frac {5 e \sqrt {d+e x} (b d-a e)}{b^3}-\frac {5 e (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}-\frac {(d+e x)^{5/2}}{b (a+b x)}+\frac {5 e (d+e x)^{3/2}}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

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 63

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 50, normalized size = 0.45 \begin {gather*} \frac {2 e (d+e x)^{7/2} \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};-\frac {b (d+e x)}{a e-b d}\right )}{7 (a e-b d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(d + e*x)^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(7*(-(b*d) + a*e)^2)

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IntegrateAlgebraic [A]  time = 0.39, size = 187, normalized size = 1.70 \begin {gather*} \frac {e \sqrt {d+e x} \left (-15 a^2 e^2-10 a b e (d+e x)+30 a b d e-15 b^2 d^2+2 b^2 (d+e x)^2+10 b^2 d (d+e x)\right )}{3 b^3 (a e+b (d+e x)-b d)}+\frac {5 \left (-a^3 e^4+3 a^2 b d e^3-3 a b^2 d^2 e^2+b^3 d^3 e\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{b^{7/2} (a e-b d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[d + e*x]*(-15*b^2*d^2 + 30*a*b*d*e - 15*a^2*e^2 + 10*b^2*d*(d + e*x) - 10*a*b*e*(d + e*x) + 2*b^2*(d +
e*x)^2))/(3*b^3*(-(b*d) + a*e + b*(d + e*x))) + (5*(b^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4)*Ar
cTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(b^(7/2)*(-(b*d) + a*e)^(3/2))

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fricas [A]  time = 0.42, size = 330, normalized size = 3.00 \begin {gather*} \left [-\frac {15 \, {\left (a b d e - a^{2} e^{2} + {\left (b^{2} d e - a b e^{2}\right )} x\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 (2 \, b^{2} e^{2} x^{2} - 3 \, b^{2} d^{2} + 20 \, a b d e - 15 \, a^{2} e^{2} + 2 \, {\left (7 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, -\frac {15 \, {\left (a b d e - a^{2} e^{2} + {\left (b^{2} d e - a b e^{2}\right )} x\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 (2 \, b^{2} e^{2} x^{2} - 3 \, b^{2} d^{2} + 20 \, a b d e - 15 \, a^{2} e^{2} + 2 \, {\left (7 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (b^{4} x + a 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),x, algorithm="fricas")

[Out]

[-1/6*(15*(a*b*d*e - a^2*e^2 + (b^2*d*e - a*b*e^2)*x)*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*(2*b^2*e^2*x^2 - 3*b^2*d^2 + 20*a*b*d*e - 15*a^2*e^2 + 2*(7*b^2*d
*e - 5*a*b*e^2)*x)*sqrt(e*x + d))/(b^4*x + a*b^3), -1/3*(15*(a*b*d*e - a^2*e^2 + (b^2*d*e - a*b*e^2)*x)*sqrt(-
(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (2*b^2*e^2*x^2 - 3*b^2*d^2 + 20*a*b
*d*e - 15*a^2*e^2 + 2*(7*b^2*d*e - 5*a*b*e^2)*x)*sqrt(e*x + d))/(b^4*x + a*b^3)]

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giac [B]  time = 0.18, size = 191, normalized size = 1.74 \begin {gather*} \frac {5 \, {\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\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 {\sqrt {x e + d} b^{2} d^{2} e - 2 \, \sqrt {x e + d} a b d e^{2} + \sqrt {x e + d} a^{2} e^{3}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{3}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{4} e + 6 \, \sqrt {x e + d} b^{4} d e - 6 \, \sqrt {x e + d} a b^{3} e^{2}\right )}}{3 \, b^{6}} \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),x, algorithm="giac")

[Out]

5*(b^2*d^2*e - 2*a*b*d*e^2 + a^2*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^3)
- (sqrt(x*e + d)*b^2*d^2*e - 2*sqrt(x*e + d)*a*b*d*e^2 + sqrt(x*e + d)*a^2*e^3)/(((x*e + d)*b - b*d + a*e)*b^3
) + 2/3*((x*e + d)^(3/2)*b^4*e + 6*sqrt(x*e + d)*b^4*d*e - 6*sqrt(x*e + d)*a*b^3*e^2)/b^6

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maple [B]  time = 0.06, size = 258, normalized size = 2.35 \begin {gather*} \frac {5 a^{2} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}-\frac {10 a d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}+\frac {5 d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}-\frac {\sqrt {e x +d}\, a^{2} e^{3}}{\left (b e x +a e \right ) b^{3}}+\frac {2 \sqrt {e x +d}\, a d \,e^{2}}{\left (b e x +a e \right ) b^{2}}-\frac {\sqrt {e x +d}\, d^{2} e}{\left (b e x +a e \right ) b}-\frac {4 \sqrt {e x +d}\, a \,e^{2}}{b^{3}}+\frac {4 \sqrt {e x +d}\, d e}{b^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} e}{3 b^{2}} \end {gather*}

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),x)

[Out]

2/3*e*(e*x+d)^(3/2)/b^2-4/b^3*a*e^2*(e*x+d)^(1/2)+4*e/b^2*(e*x+d)^(1/2)*d-1/b^3*(e*x+d)^(1/2)/(b*e*x+a*e)*a^2*
e^3+2/b^2*(e*x+d)^(1/2)/(b*e*x+a*e)*a*d*e^2-e/b*(e*x+d)^(1/2)/(b*e*x+a*e)*d^2+5/b^3/((a*e-b*d)*b)^(1/2)*arctan
((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^2*e^3-10/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1
/2)*b)*a*d*e^2+5*e/b/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*d^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),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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 0.59, size = 161, normalized size = 1.46 \begin {gather*} \frac {2\,e\,{\left (d+e\,x\right )}^{3/2}}{3\,b^2}-\frac {\sqrt {d+e\,x}\,\left (a^2\,e^3-2\,a\,b\,d\,e^2+b^2\,d^2\,e\right )}{b^4\,\left (d+e\,x\right )-b^4\,d+a\,b^3\,e}+\frac {5\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,e\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^3-2\,a\,b\,d\,e^2+b^2\,d^2\,e}\right )\,{\left (a\,e-b\,d\right )}^{3/2}}{b^{7/2}}+\frac {2\,e\,\left (2\,b^2\,d-2\,a\,b\,e\right )\,\sqrt {d+e\,x}}{b^4} \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),x)

[Out]

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

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sympy [F(-1)]  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),x)

[Out]

Timed out

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