\(\int \frac {(d+e x)^2}{(a+c x^2)^{3/2}} \, dx\) [571]

Optimal result

Integrand size = 19, antiderivative size = 83 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^{3/2}} \, dx=-\frac {(a e-c d x) (d+e x)}{a c \sqrt {a+c x^2}}-\frac {d e \sqrt {a+c x^2}}{a c}+\frac {e^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {753, 655, 223, 212} \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {e^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}}-\frac {d e \sqrt {a+c x^2}}{a c}-\frac {(d+e x) (a e-c d x)}{a c \sqrt {a+c x^2}} \]

[In]

Int[(d + e*x)^2/(a + c*x^2)^(3/2),x]

[Out]

-(((a*e - c*d*x)*(d + e*x))/(a*c*Sqrt[a + c*x^2])) - (d*e*Sqrt[a + c*x^2])/(a*c) + (e^2*ArcTanh[(Sqrt[c]*x)/Sq
rt[a + c*x^2]])/c^(3/2)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 753

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
+ c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {(a e-c d x) (d+e x)}{a c \sqrt {a+c x^2}}+\frac {\int \frac {a e^2-c d e x}{\sqrt {a+c x^2}} \, dx}{a c} \\ & = -\frac {(a e-c d x) (d+e x)}{a c \sqrt {a+c x^2}}-\frac {d e \sqrt {a+c x^2}}{a c}+\frac {e^2 \int \frac {1}{\sqrt {a+c x^2}} \, dx}{c} \\ & = -\frac {(a e-c d x) (d+e x)}{a c \sqrt {a+c x^2}}-\frac {d e \sqrt {a+c x^2}}{a c}+\frac {e^2 \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{c} \\ & = -\frac {(a e-c d x) (d+e x)}{a c \sqrt {a+c x^2}}-\frac {d e \sqrt {a+c x^2}}{a c}+\frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.83 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {-2 a d e+c d^2 x-a e^2 x}{a c \sqrt {a+c x^2}}-\frac {e^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{c^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.94 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {d^{2} x}{a \sqrt {c \,x^{2}+a}}+e^{2} \left (-\frac {x}{c \sqrt {c \,x^{2}+a}}+\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}\right )-\frac {2 d e}{c \sqrt {c \,x^{2}+a}}\)	\(75\)

[In]

int((e*x+d)^2/(c*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

d^2*x/a/(c*x^2+a)^(1/2)+e^2*(-x/c/(c*x^2+a)^(1/2)+1/c^(3/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2)))-2*d*e/c/(c*x^2+a)^(
1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.42 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^{3/2}} \, dx=\left [\frac {{\left (a c e^{2} x^{2} + a^{2} e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - 2 \, {\left (2 \, a c d e - {\left (c^{2} d^{2} - a c e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a c^{3} x^{2} + a^{2} c^{2}\right )}}, -\frac {{\left (a c e^{2} x^{2} + a^{2} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (2 \, a c d e - {\left (c^{2} d^{2} - a c e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{a c^{3} x^{2} + a^{2} c^{2}}\right ] \]

[In]

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

[Out]

[1/2*((a*c*e^2*x^2 + a^2*e^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(2*a*c*d*e - (c^2*d^
2 - a*c*e^2)*x)*sqrt(c*x^2 + a))/(a*c^3*x^2 + a^2*c^2), -((a*c*e^2*x^2 + a^2*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/s
qrt(c*x^2 + a)) + (2*a*c*d*e - (c^2*d^2 - a*c*e^2)*x)*sqrt(c*x^2 + a))/(a*c^3*x^2 + a^2*c^2)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.82 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {d^{2} x}{\sqrt {c x^{2} + a} a} - \frac {e^{2} x}{\sqrt {c x^{2} + a} c} + \frac {e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {3}{2}}} - \frac {2 \, d e}{\sqrt {c x^{2} + a} c} \]

[In]

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

[Out]

d^2*x/(sqrt(c*x^2 + a)*a) - e^2*x/(sqrt(c*x^2 + a)*c) + e^2*arcsinh(c*x/sqrt(a*c))/c^(3/2) - 2*d*e/(sqrt(c*x^2
+ a)*c)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^{3/2}} \, dx=-\frac {e^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{c^{\frac {3}{2}}} - \frac {\frac {2 \, d e}{c} - \frac {{\left (c^{2} d^{2} - a c e^{2}\right )} x}{a c^{2}}}{\sqrt {c x^{2} + a}} \]

[In]

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

[Out]

-e^2*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2) - (2*d*e/c - (c^2*d^2 - a*c*e^2)*x/(a*c^2))/sqrt(c*x^2 + a
)

Mupad [B] (verification not implemented)

Time = 9.60 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {e^2\,\ln \left (\sqrt {c}\,x+\sqrt {c\,x^2+a}\right )}{c^{3/2}}+\frac {d^2\,x}{a\,\sqrt {c\,x^2+a}}-\frac {e^2\,x}{c\,\sqrt {c\,x^2+a}}-\frac {2\,d\,e}{c\,\sqrt {c\,x^2+a}} \]

[In]

int((d + e*x)^2/(a + c*x^2)^(3/2),x)

[Out]

(e^2*log(c^(1/2)*x + (a + c*x^2)^(1/2)))/c^(3/2) + (d^2*x)/(a*(a + c*x^2)^(1/2)) - (e^2*x)/(c*(a + c*x^2)^(1/2
)) - (2*d*e)/(c*(a + c*x^2)^(1/2))