3.1648 \(\int \frac {(d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=85 \[ -\frac {3 e \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2}}-\frac {(d+e x)^{3/2}}{b (a+b x)}+\frac {3 e \sqrt {d+e x}}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ -\frac {3 e \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2}}-\frac {(d+e x)^{3/2}}{b (a+b x)}+\frac {3 e \sqrt {d+e x}}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(3/2)/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.59 \[ \frac {2 e (d+e x)^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\frac {b (d+e x)}{a e-b d}\right )}{5 (a e-b d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(3/2)/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

(2*e*(d + e*x)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(5*(-(b*d) + a*e)^2)

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fricas [A]  time = 0.90, size = 210, normalized size = 2.47 \[ \left [\frac {3 \, {\left (b e x + a e\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (2 \, b e x - b d + 3 \, a e\right )} \sqrt {e x + d}}{2 \, {\left (b^{3} x + a b^{2}\right )}}, -\frac {3 \, {\left (b e x + a e\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (2 \, b e x - b d + 3 \, a e\right )} \sqrt {e x + d}}{b^{3} x + a b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*(b*e*x + a*e)*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*e*x - b*d + 3*a*e)*sqrt(e*x + d))/(b^3*x + a*b^2), -(3*(b*e*x + a*e)*sqrt(-(b*d - a*e)/b)*ar
ctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (2*b*e*x - b*d + 3*a*e)*sqrt(e*x + d))/(b^3*x + a*b^
2)]

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giac [A]  time = 0.21, size = 122, normalized size = 1.44 \[ \frac {3 \, {\left (b d e - a e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{2}} + \frac {2 \, \sqrt {x e + d} e}{b^{2}} - \frac {\sqrt {x e + d} b d e - \sqrt {x e + d} a e^{2}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 148, normalized size = 1.74 \[ -\frac {3 a \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}+\frac {3 d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {\sqrt {e x +d}\, a \,e^{2}}{\left (b e x +a e \right ) b^{2}}-\frac {\sqrt {e x +d}\, d e}{\left (b e x +a e \right ) b}+\frac {2 \sqrt {e x +d}\, e}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(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 = 0.11, size = 109, normalized size = 1.28 \[ \frac {\left (a\,e^2-b\,d\,e\right )\,\sqrt {d+e\,x}}{b^3\,\left (d+e\,x\right )-b^3\,d+a\,b^2\,e}+\frac {2\,e\,\sqrt {d+e\,x}}{b^2}-\frac {3\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,e\,\sqrt {a\,e-b\,d}\,\sqrt {d+e\,x}}{a\,e^2-b\,d\,e}\right )\,\sqrt {a\,e-b\,d}}{b^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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