\(\int x (a+b x^2)^2 \sqrt {c+d x^2} \, dx\) [600]

Optimal result

Integrand size = 22, antiderivative size = 77 \[ \int x \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {(b c-a d)^2 \left (c+d x^2\right )^{3/2}}{3 d^3}-\frac {2 b (b c-a d) \left (c+d x^2\right )^{5/2}}{5 d^3}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d^3} \]

[Out]

1/3*(-a*d+b*c)^2*(d*x^2+c)^(3/2)/d^3-2/5*b*(-a*d+b*c)*(d*x^2+c)^(5/2)/d^3+1/7*b^2*(d*x^2+c)^(7/2)/d^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {455, 45} \[ \int x \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=-\frac {2 b \left (c+d x^2\right )^{5/2} (b c-a d)}{5 d^3}+\frac {\left (c+d x^2\right )^{3/2} (b c-a d)^2}{3 d^3}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d^3} \]

[In]

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

[Out]

((b*c - a*d)^2*(c + d*x^2)^(3/2))/(3*d^3) - (2*b*(b*c - a*d)*(c + d*x^2)^(5/2))/(5*d^3) + (b^2*(c + d*x^2)^(7/
2))/(7*d^3)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int x \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {\left (c+d x^2\right )^{3/2} \left (35 a^2 d^2+14 a b d \left (-2 c+3 d x^2\right )+b^2 \left (8 c^2-12 c d x^2+15 d^2 x^4\right )\right )}{105 d^3} \]

[In]

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

[Out]

((c + d*x^2)^(3/2)*(35*a^2*d^2 + 14*a*b*d*(-2*c + 3*d*x^2) + b^2*(8*c^2 - 12*c*d*x^2 + 15*d^2*x^4)))/(105*d^3)

Maple [A] (verified)

Time = 2.87 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(\frac {\left (\left (\frac {3}{7} b^{2} x^{4}+\frac {6}{5} a b \,x^{2}+a^{2}\right ) d^{2}-\frac {4 \left (\frac {3 b \,x^{2}}{7}+a \right ) b c d}{5}+\frac {8 b^{2} c^{2}}{35}\right ) \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 d^{3}}\)	\(60\)
gosper	\(\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} \left (15 b^{2} d^{2} x^{4}+42 x^{2} a b \,d^{2}-12 x^{2} b^{2} c d +35 a^{2} d^{2}-28 a b c d +8 b^{2} c^{2}\right )}{105 d^{3}}\)	\(69\)
trager	\(\frac {\left (15 b^{2} d^{3} x^{6}+42 a b \,d^{3} x^{4}+3 b^{2} c \,d^{2} x^{4}+35 a^{2} d^{3} x^{2}+14 a b c \,d^{2} x^{2}-4 b^{2} c^{2} d \,x^{2}+35 c \,a^{2} d^{2}-28 a b \,c^{2} d +8 b^{2} c^{3}\right ) \sqrt {d \,x^{2}+c}}{105 d^{3}}\)	\(108\)
risch	\(\frac {\left (15 b^{2} d^{3} x^{6}+42 a b \,d^{3} x^{4}+3 b^{2} c \,d^{2} x^{4}+35 a^{2} d^{3} x^{2}+14 a b c \,d^{2} x^{2}-4 b^{2} c^{2} d \,x^{2}+35 c \,a^{2} d^{2}-28 a b \,c^{2} d +8 b^{2} c^{3}\right ) \sqrt {d \,x^{2}+c}}{105 d^{3}}\)	\(108\)
default	\(b^{2} \left (\frac {x^{4} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{7 d}-\frac {4 c \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{5 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 d^{2}}\right )}{7 d}\right )+\frac {a^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 d}+2 a b \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{5 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 d^{2}}\right )\)	\(117\)

[In]

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

[Out]

1/3*((3/7*b^2*x^4+6/5*a*b*x^2+a^2)*d^2-4/5*(3/7*b*x^2+a)*b*c*d+8/35*b^2*c^2)*(d*x^2+c)^(3/2)/d^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.34 \[ \int x \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {{\left (15 \, b^{2} d^{3} x^{6} + 8 \, b^{2} c^{3} - 28 \, a b c^{2} d + 35 \, a^{2} c d^{2} + 3 \, {\left (b^{2} c d^{2} + 14 \, a b d^{3}\right )} x^{4} - {\left (4 \, b^{2} c^{2} d - 14 \, a b c d^{2} - 35 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{105 \, d^{3}} \]

[In]

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

[Out]

1/105*(15*b^2*d^3*x^6 + 8*b^2*c^3 - 28*a*b*c^2*d + 35*a^2*c*d^2 + 3*(b^2*c*d^2 + 14*a*b*d^3)*x^4 - (4*b^2*c^2*
d - 14*a*b*c*d^2 - 35*a^2*d^3)*x^2)*sqrt(d*x^2 + c)/d^3

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (66) = 132\).

Time = 0.25 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.94 \[ \int x \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\begin {cases} \frac {a^{2} c \sqrt {c + d x^{2}}}{3 d} + \frac {a^{2} x^{2} \sqrt {c + d x^{2}}}{3} - \frac {4 a b c^{2} \sqrt {c + d x^{2}}}{15 d^{2}} + \frac {2 a b c x^{2} \sqrt {c + d x^{2}}}{15 d} + \frac {2 a b x^{4} \sqrt {c + d x^{2}}}{5} + \frac {8 b^{2} c^{3} \sqrt {c + d x^{2}}}{105 d^{3}} - \frac {4 b^{2} c^{2} x^{2} \sqrt {c + d x^{2}}}{105 d^{2}} + \frac {b^{2} c x^{4} \sqrt {c + d x^{2}}}{35 d} + \frac {b^{2} x^{6} \sqrt {c + d x^{2}}}{7} & \text {for}\: d \neq 0 \\\sqrt {c} \left (\frac {a^{2} x^{2}}{2} + \frac {a b x^{4}}{2} + \frac {b^{2} x^{6}}{6}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*c*sqrt(c + d*x**2)/(3*d) + a**2*x**2*sqrt(c + d*x**2)/3 - 4*a*b*c**2*sqrt(c + d*x**2)/(15*d**2
) + 2*a*b*c*x**2*sqrt(c + d*x**2)/(15*d) + 2*a*b*x**4*sqrt(c + d*x**2)/5 + 8*b**2*c**3*sqrt(c + d*x**2)/(105*d
**3) - 4*b**2*c**2*x**2*sqrt(c + d*x**2)/(105*d**2) + b**2*c*x**4*sqrt(c + d*x**2)/(35*d) + b**2*x**6*sqrt(c +
d*x**2)/7, Ne(d, 0)), (sqrt(c)*(a**2*x**2/2 + a*b*x**4/2 + b**2*x**6/6), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.49 \[ \int x \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x^{4}}{7 \, d} - \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c x^{2}}{35 \, d^{2}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b x^{2}}{5 \, d} + \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2}}{105 \, d^{3}} - \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c}{15 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{3 \, d} \]

[In]

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

[Out]

1/7*(d*x^2 + c)^(3/2)*b^2*x^4/d - 4/35*(d*x^2 + c)^(3/2)*b^2*c*x^2/d^2 + 2/5*(d*x^2 + c)^(3/2)*a*b*x^2/d + 8/1
05*(d*x^2 + c)^(3/2)*b^2*c^2/d^3 - 4/15*(d*x^2 + c)^(3/2)*a*b*c/d^2 + 1/3*(d*x^2 + c)^(3/2)*a^2/d

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.27 \[ \int x \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {15 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} - 42 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c + 35 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} + 42 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b d - 70 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d + 35 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2}}{105 \, d^{3}} \]

[In]

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

[Out]

1/105*(15*(d*x^2 + c)^(7/2)*b^2 - 42*(d*x^2 + c)^(5/2)*b^2*c + 35*(d*x^2 + c)^(3/2)*b^2*c^2 + 42*(d*x^2 + c)^(
5/2)*a*b*d - 70*(d*x^2 + c)^(3/2)*a*b*c*d + 35*(d*x^2 + c)^(3/2)*a^2*d^2)/d^3

Mupad [B] (verification not implemented)

Time = 5.15 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.31 \[ \int x \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\sqrt {d\,x^2+c}\,\left (\frac {35\,a^2\,c\,d^2-28\,a\,b\,c^2\,d+8\,b^2\,c^3}{105\,d^3}+\frac {b^2\,x^6}{7}+\frac {x^2\,\left (35\,a^2\,d^3+14\,a\,b\,c\,d^2-4\,b^2\,c^2\,d\right )}{105\,d^3}+\frac {b\,x^4\,\left (14\,a\,d+b\,c\right )}{35\,d}\right ) \]

[In]

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

[Out]

(c + d*x^2)^(1/2)*((8*b^2*c^3 + 35*a^2*c*d^2 - 28*a*b*c^2*d)/(105*d^3) + (b^2*x^6)/7 + (x^2*(35*a^2*d^3 - 4*b^
2*c^2*d + 14*a*b*c*d^2))/(105*d^3) + (b*x^4*(14*a*d + b*c))/(35*d))