\(\int \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2} \, dx\) [864]

Optimal result

Integrand size = 29, antiderivative size = 78 \[ \int \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2} \, dx=-\frac {8 d \left (c d^2-c e^2 x^2\right )^{3/2}}{15 c e (d+e x)^{3/2}}-\frac {2 \left (c d^2-c e^2 x^2\right )^{3/2}}{5 c e \sqrt {d+e x}} \]

[Out]

-8/15*d*(-c*e^2*x^2+c*d^2)^(3/2)/c/e/(e*x+d)^(3/2)-2/5*(-c*e^2*x^2+c*d^2)^(3/2)/c/e/(e*x+d)^(1/2)

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} \[ \int \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2} \, dx=-\frac {2 \left (c d^2-c e^2 x^2\right )^{3/2}}{5 c e \sqrt {d+e x}}-\frac {8 d \left (c d^2-c e^2 x^2\right )^{3/2}}{15 c e (d+e x)^{3/2}} \]

[In]

Int[Sqrt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2],x]

[Out]

(-8*d*(c*d^2 - c*e^2*x^2)^(3/2))/(15*c*e*(d + e*x)^(3/2)) - (2*(c*d^2 - c*e^2*x^2)^(3/2))/(5*c*e*Sqrt[d + e*x]
)

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 )^{3/2}}{5 c e \sqrt {d+e x}}+\frac {1}{5} (4 d) \int \frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}} \, dx \\ & = -\frac {8 d \left (c d^2-c e^2 x^2\right )^{3/2}}{15 c e (d+e x)^{3/2}}-\frac {2 \left (c d^2-c e^2 x^2\right )^{3/2}}{5 c e \sqrt {d+e x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.68 \[ \int \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2} \, dx=-\frac {2 \left (7 d^2-4 d e x-3 e^2 x^2\right ) \sqrt {c \left (d^2-e^2 x^2\right )}}{15 e \sqrt {d+e x}} \]

[In]

Integrate[Sqrt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2],x]

[Out]

(-2*(7*d^2 - 4*d*e*x - 3*e^2*x^2)*Sqrt[c*(d^2 - e^2*x^2)])/(15*e*Sqrt[d + e*x])

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.55

method	result	size
default	\(-\frac {2 \sqrt {c \left (-x^{2} e^{2}+d^{2}\right )}\, \left (-e x +d \right ) \left (3 e x +7 d \right )}{15 \sqrt {e x +d}\, e}\)	\(43\)
gosper	\(-\frac {2 \left (-e x +d \right ) \left (3 e x +7 d \right ) \sqrt {-c \,x^{2} e^{2}+c \,d^{2}}}{15 e \sqrt {e x +d}}\)	\(44\)
risch	\(-\frac {2 \sqrt {-\frac {c \left (x^{2} e^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}\, c \left (-3 x^{2} e^{2}-4 d e x +7 d^{2}\right ) \left (-e x +d \right )}{15 \sqrt {-c \left (x^{2} e^{2}-d^{2}\right )}\, e \sqrt {-c \left (e x -d \right )}}\)	\(94\)

[In]

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

[Out]

-2/15/(e*x+d)^(1/2)*(c*(-e^2*x^2+d^2))^(1/2)*(-e*x+d)*(3*e*x+7*d)/e

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.72 \[ \int \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2} \, dx=\frac {2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (3 \, e^{2} x^{2} + 4 \, d e x - 7 \, d^{2}\right )} \sqrt {e x + d}}{15 \, {\left (e^{2} x + d e\right )}} \]

[In]

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

[Out]

2/15*sqrt(-c*e^2*x^2 + c*d^2)*(3*e^2*x^2 + 4*d*e*x - 7*d^2)*sqrt(e*x + d)/(e^2*x + d*e)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.69 \[ \int \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2} \, dx=\frac {2 \, {\left (3 \, \sqrt {c} e^{2} x^{2} + 4 \, \sqrt {c} d e x - 7 \, \sqrt {c} d^{2}\right )} {\left (e x + d\right )} \sqrt {-e x + d}}{15 \, {\left (e^{2} x + d e\right )}} \]

[In]

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

[Out]

2/15*(3*sqrt(c)*e^2*x^2 + 4*sqrt(c)*d*e*x - 7*sqrt(c)*d^2)*(e*x + d)*sqrt(-e*x + d)/(e^2*x + d*e)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.40 \[ \int \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2} \, dx=-\frac {2 \, {\left (2 \, \sqrt {2} \sqrt {c d} d^{2} - 5 \, {\left (2 \, \sqrt {2} \sqrt {c d} d - \frac {{\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}}}{c}\right )} d + \frac {5 \, {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c d - 3 \, {\left ({\left (e x + d\right )} c - 2 \, c d\right )}^{2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{c^{2}}\right )}}{15 \, e} \]

[In]

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

[Out]

-2/15*(2*sqrt(2)*sqrt(c*d)*d^2 - 5*(2*sqrt(2)*sqrt(c*d)*d - (-(e*x + d)*c + 2*c*d)^(3/2)/c)*d + (5*(-(e*x + d)
*c + 2*c*d)^(3/2)*c*d - 3*((e*x + d)*c - 2*c*d)^2*sqrt(-(e*x + d)*c + 2*c*d))/c^2)/e

Mupad [B] (verification not implemented)

Time = 10.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2} \, dx=\frac {\sqrt {c\,d^2-c\,e^2\,x^2}\,\left (\frac {2\,x^2\,\sqrt {d+e\,x}}{5}-\frac {14\,d^2\,\sqrt {d+e\,x}}{15\,e^2}+\frac {8\,d\,x\,\sqrt {d+e\,x}}{15\,e}\right )}{x+\frac {d}{e}} \]

[In]

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

[Out]

((c*d^2 - c*e^2*x^2)^(1/2)*((2*x^2*(d + e*x)^(1/2))/5 - (14*d^2*(d + e*x)^(1/2))/(15*e^2) + (8*d*x*(d + e*x)^(
1/2))/(15*e)))/(x + d/e)