\(\int \frac {(c d^2-c e^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\) [874]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 29, antiderivative size = 136 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {4 c d \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}+\frac {2 \left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}-\frac {4 \sqrt {2} c^{3/2} d^{3/2} \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{e} \]

[Out]

2/3*(-c*e^2*x^2+c*d^2)^(3/2)/e/(e*x+d)^(3/2)-4*c^(3/2)*d^(3/2)*arctanh(1/2*(-c*e^2*x^2+c*d^2)^(1/2)*2^(1/2)/c^
(1/2)/d^(1/2)/(e*x+d)^(1/2))*2^(1/2)/e+4*c*d*(-c*e^2*x^2+c*d^2)^(1/2)/e/(e*x+d)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {679, 675, 214} \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=-\frac {4 \sqrt {2} c^{3/2} d^{3/2} \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{e}+\frac {4 c d \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}+\frac {2 \left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{3/2}} \]

[In]

Int[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(5/2),x]

[Out]

(4*c*d*Sqrt[c*d^2 - c*e^2*x^2])/(e*Sqrt[d + e*x]) + (2*(c*d^2 - c*e^2*x^2)^(3/2))/(3*e*(d + e*x)^(3/2)) - (4*S
qrt[2]*c^(3/2)*d^(3/2)*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/e

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 675

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(2*c*d + e^2*x^2
), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0]

Rule 679

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 \left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}+(2 c d) \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx \\ & = \frac {4 c d \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}+\frac {2 \left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}+\left (4 c^2 d^2\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx \\ & = \frac {4 c d \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}+\frac {2 \left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}+\left (8 c^2 d^2 e\right ) \text {Subst}\left (\int \frac {1}{-2 c d e^2+e^2 x^2} \, dx,x,\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}}\right ) \\ & = \frac {4 c d \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}+\frac {2 \left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}-\frac {4 \sqrt {2} c^{3/2} d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 c \sqrt {c \left (d^2-e^2 x^2\right )} \left (\frac {7 d-e x}{\sqrt {d+e x}}-\frac {6 \sqrt {2} d^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{\sqrt {d^2-e^2 x^2}}\right )}{3 e} \]

[In]

Integrate[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(5/2),x]

[Out]

(2*c*Sqrt[c*(d^2 - e^2*x^2)]*((7*d - e*x)/Sqrt[d + e*x] - (6*Sqrt[2]*d^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[d]*Sqrt[d +
 e*x])/Sqrt[d^2 - e^2*x^2]])/Sqrt[d^2 - e^2*x^2]))/(3*e)

Maple [A] (verified)

Time = 2.59 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.82

method result size
default \(-\frac {2 \sqrt {c \left (-x^{2} e^{2}+d^{2}\right )}\, c \left (6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,d^{2}+e x \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}-7 \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}\, d \right )}{3 \sqrt {e x +d}\, \sqrt {c \left (-e x +d \right )}\, e \sqrt {c d}}\) \(112\)
risch \(\frac {2 \left (-e x +7 d \right ) \left (-e x +d \right ) \sqrt {-\frac {c \left (x^{2} e^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}\, c^{2}}{3 e \sqrt {-c \left (e x -d \right )}\, \sqrt {-c \left (x^{2} e^{2}-d^{2}\right )}}-\frac {4 d^{2} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-c e x +c d}\, \sqrt {2}}{2 \sqrt {c d}}\right ) \sqrt {-\frac {c \left (x^{2} e^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}\, c^{2}}{e \sqrt {c d}\, \sqrt {-c \left (x^{2} e^{2}-d^{2}\right )}}\) \(177\)

[In]

int((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*(c*(-e^2*x^2+d^2))^(1/2)*c*(6*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*c*d^2+e*x*(c*(-
e*x+d))^(1/2)*(c*d)^(1/2)-7*(c*(-e*x+d))^(1/2)*(c*d)^(1/2)*d)/(e*x+d)^(1/2)/(c*(-e*x+d))^(1/2)/e/(c*d)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.98 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\left [\frac {2 \, {\left (3 \, \sqrt {2} {\left (c d e x + c d^{2}\right )} \sqrt {c d} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (c e x - 7 \, c d\right )} \sqrt {e x + d}\right )}}{3 \, {\left (e^{2} x + d e\right )}}, -\frac {2 \, {\left (6 \, \sqrt {2} {\left (c d e x + c d^{2}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{c e^{2} x^{2} - c d^{2}}\right ) + \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (c e x - 7 \, c d\right )} \sqrt {e x + d}\right )}}{3 \, {\left (e^{2} x + d e\right )}}\right ] \]

[In]

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

[2/3*(3*sqrt(2)*(c*d*e*x + c*d^2)*sqrt(c*d)*log(-(c*e^2*x^2 - 2*c*d*e*x - 3*c*d^2 + 2*sqrt(2)*sqrt(-c*e^2*x^2
+ c*d^2)*sqrt(c*d)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) - sqrt(-c*e^2*x^2 + c*d^2)*(c*e*x - 7*c*d)*sqrt(e
*x + d))/(e^2*x + d*e), -2/3*(6*sqrt(2)*(c*d*e*x + c*d^2)*sqrt(-c*d)*arctan(sqrt(2)*sqrt(-c*e^2*x^2 + c*d^2)*s
qrt(-c*d)*sqrt(e*x + d)/(c*e^2*x^2 - c*d^2)) + sqrt(-c*e^2*x^2 + c*d^2)*(c*e*x - 7*c*d)*sqrt(e*x + d))/(e^2*x
+ d*e)]

Sympy [F]

\[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int \frac {\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((-c*e**2*x**2+c*d**2)**(3/2)/(e*x+d)**(5/2),x)

[Out]

Integral((-c*(-d + e*x)*(d + e*x))**(3/2)/(d + e*x)**(5/2), x)

Maxima [F]

\[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int { \frac {{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {3}{2}}}{{\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

integrate((-c*e^2*x^2 + c*d^2)^(3/2)/(e*x + d)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.07 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {6 \, \sqrt {2} c d^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d}} + \frac {6 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{3} d + {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c^{2}}{c^{3}}\right )} c}{3 \, e} - \frac {4 \, \sqrt {2} {\left (3 \, c^{2} d^{2} \arctan \left (\frac {\sqrt {c d}}{\sqrt {-c d}}\right ) + 4 \, \sqrt {c d} \sqrt {-c d} c d\right )}}{3 \, \sqrt {-c d} e} \]

[In]

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

2/3*(6*sqrt(2)*c*d^2*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c*d))/sqrt(-c*d) + (6*sqrt(-(e*x + d)
*c + 2*c*d)*c^3*d + (-(e*x + d)*c + 2*c*d)^(3/2)*c^2)/c^3)*c/e - 4/3*sqrt(2)*(3*c^2*d^2*arctan(sqrt(c*d)/sqrt(
-c*d)) + 4*sqrt(c*d)*sqrt(-c*d)*c*d)/(sqrt(-c*d)*e)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int \frac {{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^{5/2}} \,d x \]

[In]

int((c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(5/2),x)

[Out]

int((c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(5/2), x)