3.1716 \(\int \frac {1}{\sqrt {d+e x} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=172 \[ -\frac {3 e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}}+\frac {3 e \sqrt {d+e x}}{4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {\sqrt {d+e x}}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {646, 51, 63, 208} \[ -\frac {3 e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}}+\frac {3 e \sqrt {d+e x}}{4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {\sqrt {d+e x}}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d + e*x]*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(3*e*Sqrt[d + e*x])/(4*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - Sqrt[d + e*x]/(2*(b*d - a*e)*(a + b*x)*S
qrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*e^2*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*Sqrt[b]*
(b*d - a*e)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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]

Rule 646

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

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Mathematica [C]  time = 0.02, size = 65, normalized size = 0.38 \[ -\frac {2 e^2 (a+b x) \sqrt {d+e x} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {b (d+e x)}{b d-a e}\right )}{\sqrt {(a+b x)^2} (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[d + e*x]*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(-2*e^2*(a + b*x)*Sqrt[d + e*x]*Hypergeometric2F1[1/2, 3, 3/2, (b*(d + e*x))/(b*d - a*e)])/((b*d - a*e)^3*Sqrt
[(a + b*x)^2])

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fricas [B]  time = 1.06, size = 549, normalized size = 3.19 \[ \left [\frac {3 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (2 \, b^{3} d^{2} - 7 \, a b^{2} d e + 5 \, a^{2} b e^{2} - 3 \, {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{4} d^{3} - 3 \, a^{3} b^{3} d^{2} e + 3 \, a^{4} b^{2} d e^{2} - a^{5} b e^{3} + {\left (b^{6} d^{3} - 3 \, a b^{5} d^{2} e + 3 \, a^{2} b^{4} d e^{2} - a^{3} b^{3} e^{3}\right )} x^{2} + 2 \, {\left (a b^{5} d^{3} - 3 \, a^{2} b^{4} d^{2} e + 3 \, a^{3} b^{3} d e^{2} - a^{4} b^{2} e^{3}\right )} x\right )}}, \frac {3 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (2 \, b^{3} d^{2} - 7 \, a b^{2} d e + 5 \, a^{2} b e^{2} - 3 \, {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{4} d^{3} - 3 \, a^{3} b^{3} d^{2} e + 3 \, a^{4} b^{2} d e^{2} - a^{5} b e^{3} + {\left (b^{6} d^{3} - 3 \, a b^{5} d^{2} e + 3 \, a^{2} b^{4} d e^{2} - a^{3} b^{3} e^{3}\right )} x^{2} + 2 \, {\left (a b^{5} d^{3} - 3 \, a^{2} b^{4} d^{2} e + 3 \, a^{3} b^{3} d e^{2} - a^{4} b^{2} e^{3}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(3*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*
b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(2*b^3*d^2 - 7*a*b^2*d*e + 5*a^2*b*e^2 - 3*(b^3*d*e - a*b^2*e^2)*x)*sqrt(e*
x + d))/(a^2*b^4*d^3 - 3*a^3*b^3*d^2*e + 3*a^4*b^2*d*e^2 - a^5*b*e^3 + (b^6*d^3 - 3*a*b^5*d^2*e + 3*a^2*b^4*d*
e^2 - a^3*b^3*e^3)*x^2 + 2*(a*b^5*d^3 - 3*a^2*b^4*d^2*e + 3*a^3*b^3*d*e^2 - a^4*b^2*e^3)*x), 1/4*(3*(b^2*e^2*x
^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (2
*b^3*d^2 - 7*a*b^2*d*e + 5*a^2*b*e^2 - 3*(b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x + d))/(a^2*b^4*d^3 - 3*a^3*b^3*d^2*
e + 3*a^4*b^2*d*e^2 - a^5*b*e^3 + (b^6*d^3 - 3*a*b^5*d^2*e + 3*a^2*b^4*d*e^2 - a^3*b^3*e^3)*x^2 + 2*(a*b^5*d^3
- 3*a^2*b^4*d^2*e + 3*a^3*b^3*d*e^2 - a^4*b^2*e^3)*x)]

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giac [B]  time = 0.26, size = 284, normalized size = 1.65 \[ \frac {3 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{4 \, {\left (b^{2} d^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 2 \, a b d e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{2} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt {-b^{2} d + a b e}} + \frac {3 \, {\left (x e + d\right )}^{\frac {3}{2}} b e^{2} - 5 \, \sqrt {x e + d} b d e^{2} + 5 \, \sqrt {x e + d} a e^{3}}{4 \, {\left (b^{2} d^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 2 \, a b d e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{2} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

3/4*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^2*d^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 2*a*b*d*e*
sgn((x*e + d)*b*e - b*d*e + a*e^2) + a^2*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) + 1/4*(
3*(x*e + d)^(3/2)*b*e^2 - 5*sqrt(x*e + d)*b*d*e^2 + 5*sqrt(x*e + d)*a*e^3)/((b^2*d^2*sgn((x*e + d)*b*e - b*d*e
+ a*e^2) - 2*a*b*d*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) + a^2*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2))*((x*e +
d)*b - b*d + a*e)^2)

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maple [A]  time = 0.09, size = 203, normalized size = 1.18 \[ \frac {\left (3 b^{2} e^{2} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+6 a b \,e^{2} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+3 a^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+3 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b e x +5 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a e -2 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b d \right ) \left (b x +a \right )}{4 \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right )^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(1/2),x)

[Out]

1/4*(3*b^2*e^2*x^2*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+6*a*b*e^2*x*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^
(1/2)*b)+3*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*x*b*e+3*a^2*e^2*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+5*(e*
x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a*e-2*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*b*d)*(b*x+a)/((a*e-b*d)*b)^(1/2)/(a*e-b
*d)^2/((b*x+a)^2)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} \sqrt {e x + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*sqrt(e*x + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {d+e\,x}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)),x)

[Out]

int(1/((d + e*x)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {d + e x} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**(1/2),x)

[Out]

Integral(1/(sqrt(d + e*x)*((a + b*x)**2)**(3/2)), x)

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