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

Optimal. Leaf size=119 \[ \frac {2 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {646, 51, 63, 208} \begin {gather*} \frac {2 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.02, size = 62, normalized size = 0.52 \begin {gather*} -\frac {2 (a+b x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )}{\sqrt {(a+b x)^2} \sqrt {d+e x} (a e-b d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 14.84, size = 111, normalized size = 0.93 \begin {gather*} \frac {(-a e-b e x) \left (-\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{(a e-b d)^{3/2}}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 214, normalized size = 1.80 \begin {gather*} \left [-\frac {{\left (e x + d\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, \sqrt {e x + d}}{b d^{2} - a d e + {\left (b d e - a e^{2}\right )} x}, -\frac {2 \, {\left ({\left (e x + d\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - \sqrt {e x + d}\right )}}{b d^{2} - a d e + {\left (b d e - a e^{2}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-((e*x + d)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(
b*x + a)) - 2*sqrt(e*x + d))/(b*d^2 - a*d*e + (b*d*e - a*e^2)*x), -2*((e*x + d)*sqrt(-b/(b*d - a*e))*arctan(-(
b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) - sqrt(e*x + d))/(b*d^2 - a*d*e + (b*d*e - a*e^2)
*x)]

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giac [A]  time = 0.24, size = 81, normalized size = 0.68 \begin {gather*} 2 \, {\left (\frac {b \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} {\left (b d - a e\right )}} + \frac {1}{{\left (b d - a e\right )} \sqrt {x e + d}}\right )} \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 90, normalized size = 0.76 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (\sqrt {e x +d}\, b \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}\right )}{\sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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