3.18.14 \(\int \frac {(d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [1714]

Optimal. Leaf size=158 \[ -\frac {3 e \sqrt {d+e x}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{3/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{5/2} \sqrt {b d-a e} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 158, 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 = {660, 43, 65, 214} \begin {gather*} -\frac {3 e \sqrt {d+e x}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{3/2}}{2 b (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 b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {b d-a e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 660

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

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Mathematica [A]
time = 0.34, size = 110, normalized size = 0.70 \begin {gather*} \frac {-\sqrt {b} \sqrt {d+e x} (2 b d+3 a e+5 b e x)+\frac {3 e^2 (a+b x)^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {-b d+a e}}}{4 b^{5/2} (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.68, size = 194, normalized size = 1.23

method	result	size
default	\(-\frac {\left (-3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) b^{2} e^{2} x^{2}-6 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a b \,e^{2} x +5 \left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, b -3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2} e^{2}+3 \sqrt {b \left (a e -b d \right )}\, \sqrt {e x +d}\, a e -3 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, b d \right ) \left (b x +a \right )}{4 \sqrt {b \left (a e -b d \right )}\, b^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(194\)

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)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.21, size = 378, normalized size = 2.39 \begin {gather*} \left [\frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b^{2} d - a b e} e^{2} \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 (2 \, b^{3} d^{2} - {\left (5 \, a b^{2} x + 3 \, a^{2} b\right )} e^{2} + {\left (5 \, b^{3} d x + a b^{2} d\right )} e\right )} \sqrt {x e + d}}{8 \, {\left (b^{6} d x^{2} + 2 \, a b^{5} d x + a^{2} b^{4} d - {\left (a b^{5} x^{2} + 2 \, a^{2} b^{4} x + a^{3} b^{3}\right )} e\right )}}, \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\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^{2} - {\left (2 \, b^{3} d^{2} - {\left (5 \, a b^{2} x + 3 \, a^{2} b\right )} e^{2} + {\left (5 \, b^{3} d x + a b^{2} d\right )} e\right )} \sqrt {x e + d}}{4 \, {\left (b^{6} d x^{2} + 2 \, a b^{5} d x + a^{2} b^{4} d - {\left (a b^{5} x^{2} + 2 \, a^{2} b^{4} x + a^{3} b^{3}\right )} e\right )}}\right ] \end {gather*}

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)^(3/2),x, algorithm="fricas")

[Out]

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

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

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)**(3/2),x)

[Out]

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

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Giac [A]
time = 1.01, size = 128, normalized size = 0.81 \begin {gather*} \frac {3 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{4 \, \sqrt {-b^{2} d + a b e} b^{2} \mathrm {sgn}\left (b x + a\right )} - \frac {5 \, {\left (x e + d\right )}^{\frac {3}{2}} b e^{2} - 3 \, \sqrt {x e + d} b d e^{2} + 3 \, \sqrt {x e + d} a e^{3}}{4 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{2} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

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)^(3/2),x, algorithm="giac")

[Out]

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

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

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)^(3/2),x)

[Out]

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

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