\(\int \frac {(3 a+b x^2)^2}{(a-b x^2)^{7/3}} \, dx\) [137]

Optimal result

Integrand size = 24, antiderivative size = 44 \[ \int \frac {\left (3 a+b x^2\right )^2}{\left (a-b x^2\right )^{7/3}} \, dx=\frac {9 x}{2 \sqrt [3]{a-b x^2}}+\frac {3 x \left (3 a+b x^2\right )}{2 \left (a-b x^2\right )^{4/3}} \]

[Out]

9/2*x/(-b*x^2+a)^(1/3)+3/2*x*(b*x^2+3*a)/(-b*x^2+a)^(4/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {424, 391} \[ \int \frac {\left (3 a+b x^2\right )^2}{\left (a-b x^2\right )^{7/3}} \, dx=\frac {3 x \left (3 a+b x^2\right )}{2 \left (a-b x^2\right )^{4/3}}+\frac {9 x}{2 \sqrt [3]{a-b x^2}} \]

[In]

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

[Out]

(9*x)/(2*(a - b*x^2)^(1/3)) + (3*x*(3*a + b*x^2))/(2*(a - b*x^2)^(4/3))

Rule 391

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

Rule 424

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a*d - c*b)*x*(a + b*x^n)^(
p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 1))), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rubi steps \begin{align*} \text {integral}& = \frac {3 x \left (3 a+b x^2\right )}{2 \left (a-b x^2\right )^{4/3}}-\frac {3 \int \frac {-12 a^2 b+4 a b^2 x^2}{\left (a-b x^2\right )^{4/3}} \, dx}{8 a b} \\ & = \frac {9 x}{2 \sqrt [3]{a-b x^2}}+\frac {3 x \left (3 a+b x^2\right )}{2 \left (a-b x^2\right )^{4/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 15.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.55 \[ \int \frac {\left (3 a+b x^2\right )^2}{\left (a-b x^2\right )^{7/3}} \, dx=\frac {9 a x-3 b x^3}{\left (a-b x^2\right )^{4/3}} \]

[In]

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

[Out]

(9*a*x - 3*b*x^3)/(a - b*x^2)^(4/3)

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.55

method	result	size
gosper	\(\frac {3 x \left (-b \,x^{2}+3 a \right )}{\left (-b \,x^{2}+a \right )^{\frac {4}{3}}}\)	\(24\)
trager	\(\frac {3 x \left (-b \,x^{2}+3 a \right )}{\left (-b \,x^{2}+a \right )^{\frac {4}{3}}}\)	\(24\)

[In]

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

[Out]

3/(-b*x^2+a)^(4/3)*x*(-b*x^2+3*a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {\left (3 a+b x^2\right )^2}{\left (a-b x^2\right )^{7/3}} \, dx=-\frac {3 \, {\left (b x^{3} - 3 \, a x\right )} {\left (-b x^{2} + a\right )}^{\frac {2}{3}}}{b^{2} x^{4} - 2 \, a b x^{2} + a^{2}} \]

[In]

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

[Out]

-3*(b*x^3 - 3*a*x)*(-b*x^2 + a)^(2/3)/(b^2*x^4 - 2*a*b*x^2 + a^2)

Sympy [F]

\[ \int \frac {\left (3 a+b x^2\right )^2}{\left (a-b x^2\right )^{7/3}} \, dx=\int \frac {\left (3 a + b x^{2}\right )^{2}}{\left (a - b x^{2}\right )^{\frac {7}{3}}}\, dx \]

[In]

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

[Out]

Integral((3*a + b*x**2)**2/(a - b*x**2)**(7/3), x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {\left (3 a+b x^2\right )^2}{\left (a-b x^2\right )^{7/3}} \, dx=\frac {3 \, {\left (b x^{3} - 3 \, a x\right )}}{{\left (b x^{2} - a\right )} {\left (-b x^{2} + a\right )}^{\frac {1}{3}}} \]

[In]

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

[Out]

3*(b*x^3 - 3*a*x)/((b*x^2 - a)*(-b*x^2 + a)^(1/3))

Giac [F]

\[ \int \frac {\left (3 a+b x^2\right )^2}{\left (a-b x^2\right )^{7/3}} \, dx=\int { \frac {{\left (b x^{2} + 3 \, a\right )}^{2}}{{\left (-b x^{2} + a\right )}^{\frac {7}{3}}} \,d x } \]

[In]

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

[Out]

integrate((b*x^2 + 3*a)^2/(-b*x^2 + a)^(7/3), x)

Mupad [B] (verification not implemented)

Time = 4.88 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.61 \[ \int \frac {\left (3 a+b x^2\right )^2}{\left (a-b x^2\right )^{7/3}} \, dx=\frac {3\,x\,\left (a-b\,x^2\right )+6\,a\,x}{{\left (a-b\,x^2\right )}^{4/3}} \]

[In]

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

[Out]

(3*x*(a - b*x^2) + 6*a*x)/(a - b*x^2)^(4/3)