\(\int x \sqrt {-c+d x} \sqrt {c+d x} (a+b x^2) \, dx\) [339]

Optimal result

Integrand size = 29, antiderivative size = 67 \[ \int x \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {\left (b c^2+a d^2\right ) (-c+d x)^{3/2} (c+d x)^{3/2}}{3 d^4}+\frac {b (-c+d x)^{5/2} (c+d x)^{5/2}}{5 d^4} \]

[Out]

1/3*(a*d^2+b*c^2)*(d*x-c)^(3/2)*(d*x+c)^(3/2)/d^4+1/5*b*(d*x-c)^(5/2)*(d*x+c)^(5/2)/d^4

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {471, 75} \[ \int x \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {(d x-c)^{3/2} (c+d x)^{3/2} \left (5 a d^2+2 b c^2\right )}{15 d^4}+\frac {b x^2 (d x-c)^{3/2} (c+d x)^{3/2}}{5 d^2} \]

[In]

Int[x*Sqrt[-c + d*x]*Sqrt[c + d*x]*(a + b*x^2),x]

[Out]

((2*b*c^2 + 5*a*d^2)*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(15*d^4) + (b*x^2*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(5*
d^2)

Rule 75

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 471

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)
*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*((a2 + b2*x^(n/2))^(p + 1)/(b1*b2*e*(
m + n*(p + 1) + 1))), x] - Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(b1*b2*(m + n*(p + 1) + 1)), I
nt[(e*x)^m*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n, p}, x] &&
EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && NeQ[m + n*(p + 1) + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b x^2 (-c+d x)^{3/2} (c+d x)^{3/2}}{5 d^2}-\frac {1}{5} \left (-5 a-\frac {2 b c^2}{d^2}\right ) \int x \sqrt {-c+d x} \sqrt {c+d x} \, dx \\ & = \frac {\left (2 b c^2+5 a d^2\right ) (-c+d x)^{3/2} (c+d x)^{3/2}}{15 d^4}+\frac {b x^2 (-c+d x)^{3/2} (c+d x)^{3/2}}{5 d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.73 \[ \int x \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {(-c+d x)^{3/2} (c+d x)^{3/2} \left (2 b c^2+5 a d^2+3 b d^2 x^2\right )}{15 d^4} \]

[In]

Integrate[x*Sqrt[-c + d*x]*Sqrt[c + d*x]*(a + b*x^2),x]

[Out]

((-c + d*x)^(3/2)*(c + d*x)^(3/2)*(2*b*c^2 + 5*a*d^2 + 3*b*d^2*x^2))/(15*d^4)

Maple [A] (verified)

Time = 4.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.66

method	result	size
gosper	\(\frac {\left (d x -c \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}} \left (3 b \,d^{2} x^{2}+5 a \,d^{2}+2 b \,c^{2}\right )}{15 d^{4}}\)	\(44\)
default	\(-\frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right ) \left (3 b \,d^{2} x^{2}+5 a \,d^{2}+2 b \,c^{2}\right )}{15 d^{4}}\)	\(56\)
risch	\(\frac {\sqrt {d x +c}\, \left (-3 b \,d^{4} x^{4}-5 a \,d^{4} x^{2}+b \,c^{2} d^{2} x^{2}+5 a \,c^{2} d^{2}+2 b \,c^{4}\right ) \left (-d x +c \right )}{15 \sqrt {d x -c}\, d^{4}}\)	\(73\)

[In]

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

[Out]

1/15/d^4*(d*x-c)^(3/2)*(d*x+c)^(3/2)*(3*b*d^2*x^2+5*a*d^2+2*b*c^2)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99 \[ \int x \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {{\left (3 \, b d^{4} x^{4} - 2 \, b c^{4} - 5 \, a c^{2} d^{2} - {\left (b c^{2} d^{2} - 5 \, a d^{4}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{15 \, d^{4}} \]

[In]

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

[Out]

1/15*(3*b*d^4*x^4 - 2*b*c^4 - 5*a*c^2*d^2 - (b*c^2*d^2 - 5*a*d^4)*x^2)*sqrt(d*x + c)*sqrt(d*x - c)/d^4

Sympy [F]

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

[In]

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

[Out]

Integral(x*(a + b*x**2)*sqrt(-c + d*x)*sqrt(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.04 \[ \int x \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {{\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} b x^{2}}{5 \, d^{2}} + \frac {2 \, {\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} b c^{2}}{15 \, d^{4}} + \frac {{\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} a}{3 \, d^{2}} \]

[In]

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

[Out]

1/5*(d^2*x^2 - c^2)^(3/2)*b*x^2/d^2 + 2/15*(d^2*x^2 - c^2)^(3/2)*b*c^2/d^4 + 1/3*(d^2*x^2 - c^2)^(3/2)*a/d^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (55) = 110\).

Time = 0.35 (sec) , antiderivative size = 361, normalized size of antiderivative = 5.39 \[ \int x \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {5 \, {\left ({\left ({\left (d x + c\right )} {\left (2 \, {\left (d x + c\right )} {\left (\frac {3 \, {\left (d x + c\right )}}{d^{3}} - \frac {13 \, c}{d^{3}}\right )} + \frac {43 \, c^{2}}{d^{3}}\right )} - \frac {39 \, c^{3}}{d^{3}}\right )} \sqrt {d x + c} \sqrt {d x - c} - \frac {18 \, c^{4} \log \left ({\left | -\sqrt {d x + c} + \sqrt {d x - c} \right |}\right )}{d^{3}}\right )} b c + 20 \, {\left (\sqrt {d x + c} \sqrt {d x - c} {\left ({\left (d x + c\right )} {\left (\frac {2 \, {\left (d x + c\right )}}{d^{2}} - \frac {7 \, c}{d^{2}}\right )} + \frac {9 \, c^{2}}{d^{2}}\right )} + \frac {6 \, c^{3} \log \left ({\left | -\sqrt {d x + c} + \sqrt {d x - c} \right |}\right )}{d^{2}}\right )} a d + {\left ({\left ({\left (2 \, {\left (d x + c\right )} {\left (3 \, {\left (d x + c\right )} {\left (\frac {4 \, {\left (d x + c\right )}}{d^{4}} - \frac {21 \, c}{d^{4}}\right )} + \frac {133 \, c^{2}}{d^{4}}\right )} - \frac {295 \, c^{3}}{d^{4}}\right )} {\left (d x + c\right )} + \frac {195 \, c^{4}}{d^{4}}\right )} \sqrt {d x + c} \sqrt {d x - c} + \frac {90 \, c^{5} \log \left ({\left | -\sqrt {d x + c} + \sqrt {d x - c} \right |}\right )}{d^{4}}\right )} b d - \frac {60 \, {\left (2 \, c^{2} \log \left ({\left | -\sqrt {d x + c} + \sqrt {d x - c} \right |}\right ) - \sqrt {d x + c} \sqrt {d x - c} {\left (d x - 2 \, c\right )}\right )} a c}{d}}{120 \, d} \]

[In]

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

[Out]

1/120*(5*(((d*x + c)*(2*(d*x + c)*(3*(d*x + c)/d^3 - 13*c/d^3) + 43*c^2/d^3) - 39*c^3/d^3)*sqrt(d*x + c)*sqrt(
d*x - c) - 18*c^4*log(abs(-sqrt(d*x + c) + sqrt(d*x - c)))/d^3)*b*c + 20*(sqrt(d*x + c)*sqrt(d*x - c)*((d*x +
c)*(2*(d*x + c)/d^2 - 7*c/d^2) + 9*c^2/d^2) + 6*c^3*log(abs(-sqrt(d*x + c) + sqrt(d*x - c)))/d^2)*a*d + (((2*(
d*x + c)*(3*(d*x + c)*(4*(d*x + c)/d^4 - 21*c/d^4) + 133*c^2/d^4) - 295*c^3/d^4)*(d*x + c) + 195*c^4/d^4)*sqrt
(d*x + c)*sqrt(d*x - c) + 90*c^5*log(abs(-sqrt(d*x + c) + sqrt(d*x - c)))/d^4)*b*d - 60*(2*c^2*log(abs(-sqrt(d
*x + c) + sqrt(d*x - c))) - sqrt(d*x + c)*sqrt(d*x - c)*(d*x - 2*c))*a*c/d)/d

Mupad [B] (verification not implemented)

Time = 5.75 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.24 \[ \int x \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\sqrt {d\,x-c}\,\left (\frac {b\,x^4\,\sqrt {c+d\,x}}{5}-\frac {\left (2\,b\,c^4+5\,a\,c^2\,d^2\right )\,\sqrt {c+d\,x}}{15\,d^4}+\frac {x^2\,\left (5\,a\,d^4-b\,c^2\,d^2\right )\,\sqrt {c+d\,x}}{15\,d^4}\right ) \]

[In]

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

[Out]

(d*x - c)^(1/2)*((b*x^4*(c + d*x)^(1/2))/5 - ((2*b*c^4 + 5*a*c^2*d^2)*(c + d*x)^(1/2))/(15*d^4) + (x^2*(5*a*d^
4 - b*c^2*d^2)*(c + d*x)^(1/2))/(15*d^4))