\(\int \sqrt {x} (a+b x^2)^2 (c+d x^2) \, dx\) [394]

Optimal result

Integrand size = 22, antiderivative size = 63 \[ \int \sqrt {x} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{3} a^2 c x^{3/2}+\frac {2}{7} a (2 b c+a d) x^{7/2}+\frac {2}{11} b (b c+2 a d) x^{11/2}+\frac {2}{15} b^2 d x^{15/2} \]

[Out]

2/3*a^2*c*x^(3/2)+2/7*a*(a*d+2*b*c)*x^(7/2)+2/11*b*(2*a*d+b*c)*x^(11/2)+2/15*b^2*d*x^(15/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {459} \[ \int \sqrt {x} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{3} a^2 c x^{3/2}+\frac {2}{11} b x^{11/2} (2 a d+b c)+\frac {2}{7} a x^{7/2} (a d+2 b c)+\frac {2}{15} b^2 d x^{15/2} \]

[In]

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

[Out]

(2*a^2*c*x^(3/2))/3 + (2*a*(2*b*c + a*d)*x^(7/2))/7 + (2*b*(b*c + 2*a*d)*x^(11/2))/11 + (2*b^2*d*x^(15/2))/15

Rule 459

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a^2 c \sqrt {x}+a (2 b c+a d) x^{5/2}+b (b c+2 a d) x^{9/2}+b^2 d x^{13/2}\right ) \, dx \\ & = \frac {2}{3} a^2 c x^{3/2}+\frac {2}{7} a (2 b c+a d) x^{7/2}+\frac {2}{11} b (b c+2 a d) x^{11/2}+\frac {2}{15} b^2 d x^{15/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int \sqrt {x} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2 x^{3/2} \left (385 a^2 c+330 a b c x^2+165 a^2 d x^2+105 b^2 c x^4+210 a b d x^4+77 b^2 d x^6\right )}{1155} \]

[In]

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

[Out]

(2*x^(3/2)*(385*a^2*c + 330*a*b*c*x^2 + 165*a^2*d*x^2 + 105*b^2*c*x^4 + 210*a*b*d*x^4 + 77*b^2*d*x^6))/1155

Maple [A] (verified)

Time = 2.67 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {2 b^{2} d \,x^{\frac {15}{2}}}{15}+\frac {2 \left (2 a b d +b^{2} c \right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (a^{2} d +2 a b c \right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{2} c \,x^{\frac {3}{2}}}{3}\)	\(52\)
default	\(\frac {2 b^{2} d \,x^{\frac {15}{2}}}{15}+\frac {2 \left (2 a b d +b^{2} c \right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (a^{2} d +2 a b c \right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{2} c \,x^{\frac {3}{2}}}{3}\)	\(52\)
gosper	\(\frac {2 x^{\frac {3}{2}} \left (77 b^{2} d \,x^{6}+210 a b d \,x^{4}+105 b^{2} c \,x^{4}+165 a^{2} d \,x^{2}+330 a b c \,x^{2}+385 a^{2} c \right )}{1155}\)	\(56\)
trager	\(\frac {2 x^{\frac {3}{2}} \left (77 b^{2} d \,x^{6}+210 a b d \,x^{4}+105 b^{2} c \,x^{4}+165 a^{2} d \,x^{2}+330 a b c \,x^{2}+385 a^{2} c \right )}{1155}\)	\(56\)
risch	\(\frac {2 x^{\frac {3}{2}} \left (77 b^{2} d \,x^{6}+210 a b d \,x^{4}+105 b^{2} c \,x^{4}+165 a^{2} d \,x^{2}+330 a b c \,x^{2}+385 a^{2} c \right )}{1155}\)	\(56\)

[In]

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

[Out]

2/15*b^2*d*x^(15/2)+2/11*(2*a*b*d+b^2*c)*x^(11/2)+2/7*(a^2*d+2*a*b*c)*x^(7/2)+2/3*a^2*c*x^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86 \[ \int \sqrt {x} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{1155} \, {\left (77 \, b^{2} d x^{7} + 105 \, {\left (b^{2} c + 2 \, a b d\right )} x^{5} + 385 \, a^{2} c x + 165 \, {\left (2 \, a b c + a^{2} d\right )} x^{3}\right )} \sqrt {x} \]

[In]

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

[Out]

2/1155*(77*b^2*d*x^7 + 105*(b^2*c + 2*a*b*d)*x^5 + 385*a^2*c*x + 165*(2*a*b*c + a^2*d)*x^3)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.05 \[ \int \sqrt {x} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2 a^{2} c x^{\frac {3}{2}}}{3} + \frac {2 b^{2} d x^{\frac {15}{2}}}{15} + \frac {2 x^{\frac {11}{2}} \cdot \left (2 a b d + b^{2} c\right )}{11} + \frac {2 x^{\frac {7}{2}} \left (a^{2} d + 2 a b c\right )}{7} \]

[In]

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

[Out]

2*a**2*c*x**(3/2)/3 + 2*b**2*d*x**(15/2)/15 + 2*x**(11/2)*(2*a*b*d + b**2*c)/11 + 2*x**(7/2)*(a**2*d + 2*a*b*c
)/7

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int \sqrt {x} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{15} \, b^{2} d x^{\frac {15}{2}} + \frac {2}{11} \, {\left (b^{2} c + 2 \, a b d\right )} x^{\frac {11}{2}} + \frac {2}{3} \, a^{2} c x^{\frac {3}{2}} + \frac {2}{7} \, {\left (2 \, a b c + a^{2} d\right )} x^{\frac {7}{2}} \]

[In]

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

[Out]

2/15*b^2*d*x^(15/2) + 2/11*(b^2*c + 2*a*b*d)*x^(11/2) + 2/3*a^2*c*x^(3/2) + 2/7*(2*a*b*c + a^2*d)*x^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84 \[ \int \sqrt {x} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{15} \, b^{2} d x^{\frac {15}{2}} + \frac {2}{11} \, b^{2} c x^{\frac {11}{2}} + \frac {4}{11} \, a b d x^{\frac {11}{2}} + \frac {4}{7} \, a b c x^{\frac {7}{2}} + \frac {2}{7} \, a^{2} d x^{\frac {7}{2}} + \frac {2}{3} \, a^{2} c x^{\frac {3}{2}} \]

[In]

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

[Out]

2/15*b^2*d*x^(15/2) + 2/11*b^2*c*x^(11/2) + 4/11*a*b*d*x^(11/2) + 4/7*a*b*c*x^(7/2) + 2/7*a^2*d*x^(7/2) + 2/3*
a^2*c*x^(3/2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int \sqrt {x} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=x^{7/2}\,\left (\frac {2\,d\,a^2}{7}+\frac {4\,b\,c\,a}{7}\right )+x^{11/2}\,\left (\frac {2\,c\,b^2}{11}+\frac {4\,a\,d\,b}{11}\right )+\frac {2\,a^2\,c\,x^{3/2}}{3}+\frac {2\,b^2\,d\,x^{15/2}}{15} \]

[In]

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

[Out]

x^(7/2)*((2*a^2*d)/7 + (4*a*b*c)/7) + x^(11/2)*((2*b^2*c)/11 + (4*a*b*d)/11) + (2*a^2*c*x^(3/2))/3 + (2*b^2*d*
x^(15/2))/15