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

Optimal result

Integrand size = 30, antiderivative size = 158 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {3 e \sqrt {d+e x}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{3/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e^2 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{5/2} \sqrt {b d-a e} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {660, 43, 65, 214} \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {3 e^2 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {b d-a e}}-\frac {3 e \sqrt {d+e x}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{3/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.70 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-\sqrt {b} \sqrt {d+e x} (2 b d+3 a e+5 b e x)+\frac {3 e^2 (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {-b d+a e}}}{4 b^{5/2} (a+b x) \sqrt {(a+b x)^2}} \]

[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])

Maple [A] (verified)

Time = 2.27 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.23

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

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.42 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (2 \, b^{3} d^{2} + a b^{2} d e - 3 \, a^{2} b e^{2} + 5 \, {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{4} d - a^{3} b^{3} e + {\left (b^{6} d - a b^{5} e\right )} x^{2} + 2 \, {\left (a b^{5} d - a^{2} b^{4} e\right )} x\right )}}, \frac {3 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (2 \, b^{3} d^{2} + a b^{2} d e - 3 \, a^{2} b e^{2} + 5 \, {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{4} d - a^{3} b^{3} e + {\left (b^{6} d - a b^{5} e\right )} x^{2} + 2 \, {\left (a b^{5} d - a^{2} b^{4} e\right )} x\right )}}\right ] \]

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

Sympy [F]

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

[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)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.78 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {3 \, e^{2} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{2} \mathrm {sgn}\left (b x + a\right )} - \frac {5 \, {\left (e x + d\right )}^{\frac {3}{2}} b e^{2} - 3 \, \sqrt {e x + d} b d e^{2} + 3 \, \sqrt {e x + d} a e^{3}}{4 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2} b^{2} \mathrm {sgn}\left (b x + a\right )} \]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^{3/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

[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)