\(\int \frac {(c d^2-c e^2 x^2)^{3/2}}{\sqrt {d+e x}} \, dx\) [872]

Optimal result

Integrand size = 29, antiderivative size = 78 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=-\frac {8 d \left (c d^2-c e^2 x^2\right )^{5/2}}{35 c e (d+e x)^{5/2}}-\frac {2 \left (c d^2-c e^2 x^2\right )^{5/2}}{7 c e (d+e x)^{3/2}} \]

[Out]

-8/35*d*(-c*e^2*x^2+c*d^2)^(5/2)/c/e/(e*x+d)^(5/2)-2/7*(-c*e^2*x^2+c*d^2)^(5/2)/c/e/(e*x+d)^(3/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {671, 663} \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=-\frac {2 \left (c d^2-c e^2 x^2\right )^{5/2}}{7 c e (d+e x)^{3/2}}-\frac {8 d \left (c d^2-c e^2 x^2\right )^{5/2}}{35 c e (d+e x)^{5/2}} \]

[In]

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

[Out]

(-8*d*(c*d^2 - c*e^2*x^2)^(5/2))/(35*c*e*(d + e*x)^(5/2)) - (2*(c*d^2 - c*e^2*x^2)^(5/2))/(7*c*e*(d + e*x)^(3/
2))

Rule 663

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

Rule 671

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (c d^2-c e^2 x^2\right )^{5/2}}{7 c e (d+e x)^{3/2}}+\frac {1}{7} (4 d) \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx \\ & = -\frac {8 d \left (c d^2-c e^2 x^2\right )^{5/2}}{35 c e (d+e x)^{5/2}}-\frac {2 \left (c d^2-c e^2 x^2\right )^{5/2}}{7 c e (d+e x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=-\frac {2 c (d-e x)^2 (9 d+5 e x) \sqrt {c \left (d^2-e^2 x^2\right )}}{35 e \sqrt {d+e x}} \]

[In]

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

[Out]

(-2*c*(d - e*x)^2*(9*d + 5*e*x)*Sqrt[c*(d^2 - e^2*x^2)])/(35*e*Sqrt[d + e*x])

Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.56

method	result	size
gosper	\(-\frac {2 \left (-e x +d \right ) \left (5 e x +9 d \right ) \left (-c \,x^{2} e^{2}+c \,d^{2}\right )^{\frac {3}{2}}}{35 e \left (e x +d \right )^{\frac {3}{2}}}\)	\(44\)
default	\(-\frac {2 \sqrt {c \left (-x^{2} e^{2}+d^{2}\right )}\, c \left (-e x +d \right )^{2} \left (5 e x +9 d \right )}{35 \sqrt {e x +d}\, e}\)	\(46\)
risch	\(-\frac {2 \sqrt {-\frac {c \left (x^{2} e^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}\, c^{2} \left (5 e^{3} x^{3}-d \,e^{2} x^{2}-13 d^{2} e x +9 d^{3}\right ) \left (-e x +d \right )}{35 \sqrt {-c \left (x^{2} e^{2}-d^{2}\right )}\, e \sqrt {-c \left (e x -d \right )}}\)	\(107\)

[In]

int((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/35*(-e*x+d)*(5*e*x+9*d)*(-c*e^2*x^2+c*d^2)^(3/2)/e/(e*x+d)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=-\frac {2 \, {\left (5 \, c e^{3} x^{3} - c d e^{2} x^{2} - 13 \, c d^{2} e x + 9 \, c d^{3}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{35 \, {\left (e^{2} x + d e\right )}} \]

[In]

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

-2/35*(5*c*e^3*x^3 - c*d*e^2*x^2 - 13*c*d^2*e*x + 9*c*d^3)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)/(e^2*x + d*e
)

Sympy [F]

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

[In]

integrate((-c*e**2*x**2+c*d**2)**(3/2)/(e*x+d)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.71 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=-\frac {2 \, {\left (5 \, c^{\frac {3}{2}} e^{3} x^{3} - c^{\frac {3}{2}} d e^{2} x^{2} - 13 \, c^{\frac {3}{2}} d^{2} e x + 9 \, c^{\frac {3}{2}} d^{3}\right )} \sqrt {-e x + d}}{35 \, e} \]

[In]

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

-2/35*(5*c^(3/2)*e^3*x^3 - c^(3/2)*d*e^2*x^2 - 13*c^(3/2)*d^2*e*x + 9*c^(3/2)*d^3)*sqrt(-e*x + d)/e

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (66) = 132\).

Time = 0.28 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.09 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (35 \, {\left (2 \, \sqrt {2} \sqrt {c d} d - \frac {{\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}}}{c}\right )} c d^{2} - {\left (\frac {22 \, \sqrt {2} \sqrt {c d} d^{3}}{e^{2}} - \frac {35 \, {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c^{2} d^{2} - 42 \, {\left ({\left (e x + d\right )} c - 2 \, c d\right )}^{2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c d - 15 \, {\left ({\left (e x + d\right )} c - 2 \, c d\right )}^{3} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{c^{3} e^{2}}\right )} c e^{2}\right )}}{105 \, e} \]

[In]

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

2/105*(35*(2*sqrt(2)*sqrt(c*d)*d - (-(e*x + d)*c + 2*c*d)^(3/2)/c)*c*d^2 - (22*sqrt(2)*sqrt(c*d)*d^3/e^2 - (35
*(-(e*x + d)*c + 2*c*d)^(3/2)*c^2*d^2 - 42*((e*x + d)*c - 2*c*d)^2*sqrt(-(e*x + d)*c + 2*c*d)*c*d - 15*((e*x +
d)*c - 2*c*d)^3*sqrt(-(e*x + d)*c + 2*c*d))/(c^3*e^2))*c*e^2)/e

Mupad [B] (verification not implemented)

Time = 9.95 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.77 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=-\frac {\sqrt {c\,d^2-c\,e^2\,x^2}\,\left (\frac {18\,c\,d^3}{35\,e}+\frac {2\,c\,e^2\,x^3}{7}-\frac {26\,c\,d^2\,x}{35}-\frac {2\,c\,d\,e\,x^2}{35}\right )}{\sqrt {d+e\,x}} \]

[In]

int((c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(1/2),x)

[Out]

-((c*d^2 - c*e^2*x^2)^(1/2)*((18*c*d^3)/(35*e) + (2*c*e^2*x^3)/7 - (26*c*d^2*x)/35 - (2*c*d*e*x^2)/35))/(d + e
*x)^(1/2)