\(\int \frac {(d+e x)^3}{(a+c x^2)^{3/2}} \, dx\) [570]

Optimal result

Integrand size = 19, antiderivative size = 106 \[ \int \frac {(d+e x)^3}{\left (a+c x^2\right )^{3/2}} \, dx=-\frac {(a e-c d x) (d+e x)^2}{a c \sqrt {a+c x^2}}-\frac {e \left (2 \left (c d^2-a e^2\right )+c d e x\right ) \sqrt {a+c x^2}}{a c^2}+\frac {3 d e^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 106, 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, 794, 223, 212} \[ \int \frac {(d+e x)^3}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {3 d e^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}}-\frac {e \sqrt {a+c x^2} \left (2 \left (c d^2-a e^2\right )+c d e x\right )}{a c^2}-\frac {(d+e x)^2 (a e-c d x)}{a c \sqrt {a+c x^2}} \]

[In]

Int[(d + e*x)^3/(a + c*x^2)^(3/2),x]

[Out]

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

Rule 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.85 \[ \int \frac {(d+e x)^3}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {2 a^2 e^3+c^2 d^3 x+a c e \left (-3 d^2-3 d e x+e^2 x^2\right )}{a c^2 \sqrt {a+c x^2}}-\frac {3 d e^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{c^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.06

method	result	size
risch	\(\frac {e^{3} \sqrt {c \,x^{2}+a}}{c^{2}}+\frac {\frac {d^{3} c x}{a \sqrt {c \,x^{2}+a}}+3 d \,e^{2} c \left (-\frac {x}{c \sqrt {c \,x^{2}+a}}+\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}\right )-\frac {-a \,e^{3}+3 d^{2} e c}{c \sqrt {c \,x^{2}+a}}}{c}\)	\(112\)
default	\(\frac {d^{3} x}{a \sqrt {c \,x^{2}+a}}+e^{3} \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+a}}+\frac {2 a}{c^{2} \sqrt {c \,x^{2}+a}}\right )+3 d \,e^{2} \left (-\frac {x}{c \sqrt {c \,x^{2}+a}}+\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}\right )-\frac {3 d^{2} e}{c \sqrt {c \,x^{2}+a}}\)	\(115\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.32 \[ \int \frac {(d+e x)^3}{\left (a+c x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (a c d e^{2} x^{2} + a^{2} d e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (a c e^{3} x^{2} - 3 \, a c d^{2} e + 2 \, a^{2} e^{3} + {\left (c^{2} d^{3} - 3 \, a c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a c^{3} x^{2} + a^{2} c^{2}\right )}}, -\frac {3 \, {\left (a c d e^{2} x^{2} + a^{2} d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (a c e^{3} x^{2} - 3 \, a c d^{2} e + 2 \, a^{2} e^{3} + {\left (c^{2} d^{3} - 3 \, a c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{a c^{3} x^{2} + a^{2} c^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x)^3}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {e^{3} x^{2}}{\sqrt {c x^{2} + a} c} + \frac {d^{3} x}{\sqrt {c x^{2} + a} a} - \frac {3 \, d e^{2} x}{\sqrt {c x^{2} + a} c} + \frac {3 \, d e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {3}{2}}} - \frac {3 \, d^{2} e}{\sqrt {c x^{2} + a} c} + \frac {2 \, a e^{3}}{\sqrt {c x^{2} + a} c^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.97 \[ \int \frac {(d+e x)^3}{\left (a+c x^2\right )^{3/2}} \, dx=-\frac {3 \, d e^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{c^{\frac {3}{2}}} + \frac {{\left (\frac {e^{3} x}{c} + \frac {c^{3} d^{3} - 3 \, a c^{2} d e^{2}}{a c^{3}}\right )} x - \frac {3 \, a c^{2} d^{2} e - 2 \, a^{2} c e^{3}}{a c^{3}}}{\sqrt {c x^{2} + a}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((d + e*x)^3/(a + c*x^2)^(3/2),x)

[Out]

int((d + e*x)^3/(a + c*x^2)^(3/2), x)