\(\int \frac {(a+b x^2)^3 (A+B x^2)}{x^{7/2}} \, dx\) [366]

Optimal result

Integrand size = 22, antiderivative size = 81 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x^2\right )}{x^{7/2}} \, dx=-\frac {2 a^3 A}{5 x^{5/2}}-\frac {2 a^2 (3 A b+a B)}{\sqrt {x}}+2 a b (A b+a B) x^{3/2}+\frac {2}{7} b^2 (A b+3 a B) x^{7/2}+\frac {2}{11} b^3 B x^{11/2} \]

[Out]

-2/5*a^3*A/x^(5/2)+2*a*b*(A*b+B*a)*x^(3/2)+2/7*b^2*(A*b+3*B*a)*x^(7/2)+2/11*b^3*B*x^(11/2)-2*a^2*(3*A*b+B*a)/x
^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 81, 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} \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x^2\right )}{x^{7/2}} \, dx=-\frac {2 a^3 A}{5 x^{5/2}}-\frac {2 a^2 (a B+3 A b)}{\sqrt {x}}+\frac {2}{7} b^2 x^{7/2} (3 a B+A b)+2 a b x^{3/2} (a B+A b)+\frac {2}{11} b^3 B x^{11/2} \]

[In]

Int[((a + b*x^2)^3*(A + B*x^2))/x^(7/2),x]

[Out]

(-2*a^3*A)/(5*x^(5/2)) - (2*a^2*(3*A*b + a*B))/Sqrt[x] + 2*a*b*(A*b + a*B)*x^(3/2) + (2*b^2*(A*b + 3*a*B)*x^(7
/2))/7 + (2*b^3*B*x^(11/2))/11

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 (\frac {a^3 A}{x^{7/2}}+\frac {a^2 (3 A b+a B)}{x^{3/2}}+3 a b (A b+a B) \sqrt {x}+b^2 (A b+3 a B) x^{5/2}+b^3 B x^{9/2}\right ) \, dx \\ & = -\frac {2 a^3 A}{5 x^{5/2}}-\frac {2 a^2 (3 A b+a B)}{\sqrt {x}}+2 a b (A b+a B) x^{3/2}+\frac {2}{7} b^2 (A b+3 a B) x^{7/2}+\frac {2}{11} b^3 B x^{11/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x^2\right )}{x^{7/2}} \, dx=\frac {2 \left (385 a^2 b x^2 \left (-3 A+B x^2\right )+55 a b^2 x^4 \left (7 A+3 B x^2\right )-77 a^3 \left (A+5 B x^2\right )+5 b^3 x^6 \left (11 A+7 B x^2\right )\right )}{385 x^{5/2}} \]

[In]

Integrate[((a + b*x^2)^3*(A + B*x^2))/x^(7/2),x]

[Out]

(2*(385*a^2*b*x^2*(-3*A + B*x^2) + 55*a*b^2*x^4*(7*A + 3*B*x^2) - 77*a^3*(A + 5*B*x^2) + 5*b^3*x^6*(11*A + 7*B
*x^2)))/(385*x^(5/2))

Maple [A] (verified)

Time = 2.65 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {2 b^{3} B \,x^{\frac {11}{2}}}{11}+\frac {2 A \,b^{3} x^{\frac {7}{2}}}{7}+\frac {6 B a \,b^{2} x^{\frac {7}{2}}}{7}+2 A a \,b^{2} x^{\frac {3}{2}}+2 B \,a^{2} b \,x^{\frac {3}{2}}-\frac {2 a^{3} A}{5 x^{\frac {5}{2}}}-\frac {2 a^{2} \left (3 A b +B a \right )}{\sqrt {x}}\)	\(75\)
default	\(\frac {2 b^{3} B \,x^{\frac {11}{2}}}{11}+\frac {2 A \,b^{3} x^{\frac {7}{2}}}{7}+\frac {6 B a \,b^{2} x^{\frac {7}{2}}}{7}+2 A a \,b^{2} x^{\frac {3}{2}}+2 B \,a^{2} b \,x^{\frac {3}{2}}-\frac {2 a^{3} A}{5 x^{\frac {5}{2}}}-\frac {2 a^{2} \left (3 A b +B a \right )}{\sqrt {x}}\)	\(75\)
gosper	\(-\frac {2 \left (-35 b^{3} B \,x^{8}-55 x^{6} b^{3} A -165 x^{6} a \,b^{2} B -385 A a \,b^{2} x^{4}-385 B \,a^{2} b \,x^{4}+1155 x^{2} a^{2} b A +385 B \,a^{3} x^{2}+77 a^{3} A \right )}{385 x^{\frac {5}{2}}}\)	\(80\)
trager	\(-\frac {2 \left (-35 b^{3} B \,x^{8}-55 x^{6} b^{3} A -165 x^{6} a \,b^{2} B -385 A a \,b^{2} x^{4}-385 B \,a^{2} b \,x^{4}+1155 x^{2} a^{2} b A +385 B \,a^{3} x^{2}+77 a^{3} A \right )}{385 x^{\frac {5}{2}}}\)	\(80\)
risch	\(-\frac {2 \left (-35 b^{3} B \,x^{8}-55 x^{6} b^{3} A -165 x^{6} a \,b^{2} B -385 A a \,b^{2} x^{4}-385 B \,a^{2} b \,x^{4}+1155 x^{2} a^{2} b A +385 B \,a^{3} x^{2}+77 a^{3} A \right )}{385 x^{\frac {5}{2}}}\)	\(80\)

[In]

int((b*x^2+a)^3*(B*x^2+A)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/11*b^3*B*x^(11/2)+2/7*A*b^3*x^(7/2)+6/7*B*a*b^2*x^(7/2)+2*A*a*b^2*x^(3/2)+2*B*a^2*b*x^(3/2)-2/5*a^3*A/x^(5/2
)-2*a^2*(3*A*b+B*a)/x^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x^2\right )}{x^{7/2}} \, dx=\frac {2 \, {\left (35 \, B b^{3} x^{8} + 55 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{6} + 385 \, {\left (B a^{2} b + A a b^{2}\right )} x^{4} - 77 \, A a^{3} - 385 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2}\right )}}{385 \, x^{\frac {5}{2}}} \]

[In]

integrate((b*x^2+a)^3*(B*x^2+A)/x^(7/2),x, algorithm="fricas")

[Out]

2/385*(35*B*b^3*x^8 + 55*(3*B*a*b^2 + A*b^3)*x^6 + 385*(B*a^2*b + A*a*b^2)*x^4 - 77*A*a^3 - 385*(B*a^3 + 3*A*a
^2*b)*x^2)/x^(5/2)

Sympy [A] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x^2\right )}{x^{7/2}} \, dx=- \frac {2 A a^{3}}{5 x^{\frac {5}{2}}} - \frac {6 A a^{2} b}{\sqrt {x}} + 2 A a b^{2} x^{\frac {3}{2}} + \frac {2 A b^{3} x^{\frac {7}{2}}}{7} - \frac {2 B a^{3}}{\sqrt {x}} + 2 B a^{2} b x^{\frac {3}{2}} + \frac {6 B a b^{2} x^{\frac {7}{2}}}{7} + \frac {2 B b^{3} x^{\frac {11}{2}}}{11} \]

[In]

integrate((b*x**2+a)**3*(B*x**2+A)/x**(7/2),x)

[Out]

-2*A*a**3/(5*x**(5/2)) - 6*A*a**2*b/sqrt(x) + 2*A*a*b**2*x**(3/2) + 2*A*b**3*x**(7/2)/7 - 2*B*a**3/sqrt(x) + 2
*B*a**2*b*x**(3/2) + 6*B*a*b**2*x**(7/2)/7 + 2*B*b**3*x**(11/2)/11

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x^2\right )}{x^{7/2}} \, dx=\frac {2}{11} \, B b^{3} x^{\frac {11}{2}} + \frac {2}{7} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{\frac {7}{2}} + 2 \, {\left (B a^{2} b + A a b^{2}\right )} x^{\frac {3}{2}} - \frac {2 \, {\left (A a^{3} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2}\right )}}{5 \, x^{\frac {5}{2}}} \]

[In]

integrate((b*x^2+a)^3*(B*x^2+A)/x^(7/2),x, algorithm="maxima")

[Out]

2/11*B*b^3*x^(11/2) + 2/7*(3*B*a*b^2 + A*b^3)*x^(7/2) + 2*(B*a^2*b + A*a*b^2)*x^(3/2) - 2/5*(A*a^3 + 5*(B*a^3
+ 3*A*a^2*b)*x^2)/x^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x^2\right )}{x^{7/2}} \, dx=\frac {2}{11} \, B b^{3} x^{\frac {11}{2}} + \frac {6}{7} \, B a b^{2} x^{\frac {7}{2}} + \frac {2}{7} \, A b^{3} x^{\frac {7}{2}} + 2 \, B a^{2} b x^{\frac {3}{2}} + 2 \, A a b^{2} x^{\frac {3}{2}} - \frac {2 \, {\left (5 \, B a^{3} x^{2} + 15 \, A a^{2} b x^{2} + A a^{3}\right )}}{5 \, x^{\frac {5}{2}}} \]

[In]

integrate((b*x^2+a)^3*(B*x^2+A)/x^(7/2),x, algorithm="giac")

[Out]

2/11*B*b^3*x^(11/2) + 6/7*B*a*b^2*x^(7/2) + 2/7*A*b^3*x^(7/2) + 2*B*a^2*b*x^(3/2) + 2*A*a*b^2*x^(3/2) - 2/5*(5
*B*a^3*x^2 + 15*A*a^2*b*x^2 + A*a^3)/x^(5/2)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x^2\right )}{x^{7/2}} \, dx=x^{7/2}\,\left (\frac {2\,A\,b^3}{7}+\frac {6\,B\,a\,b^2}{7}\right )-\frac {\frac {2\,A\,a^3}{5}+x^2\,\left (2\,B\,a^3+6\,A\,b\,a^2\right )}{x^{5/2}}+\frac {2\,B\,b^3\,x^{11/2}}{11}+2\,a\,b\,x^{3/2}\,\left (A\,b+B\,a\right ) \]

[In]

int(((A + B*x^2)*(a + b*x^2)^3)/x^(7/2),x)

[Out]

x^(7/2)*((2*A*b^3)/7 + (6*B*a*b^2)/7) - ((2*A*a^3)/5 + x^2*(2*B*a^3 + 6*A*a^2*b))/x^(5/2) + (2*B*b^3*x^(11/2))
/11 + 2*a*b*x^(3/2)*(A*b + B*a)