\(\int (c x)^{7/3} (a+b x^2)^{4/3} \, dx\) [803]

Optimal result

Integrand size = 19, antiderivative size = 192 \[ \int (c x)^{7/3} \left (a+b x^2\right )^{4/3} \, dx=\frac {a^2 c (c x)^{4/3} \sqrt [3]{a+b x^2}}{27 b}+\frac {a (c x)^{10/3} \sqrt [3]{a+b x^2}}{9 c}+\frac {(c x)^{10/3} \left (a+b x^2\right )^{4/3}}{6 c}+\frac {2 a^3 c^{7/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+b x^2}}}{\sqrt {3}}\right )}{27 \sqrt {3} b^{5/3}}+\frac {a^3 c^{7/3} \log \left (\sqrt [3]{b} (c x)^{2/3}-c^{2/3} \sqrt [3]{a+b x^2}\right )}{27 b^{5/3}} \] Output:

1/27*a^2*c*(c*x)^(4/3)*(b*x^2+a)^(1/3)/b+1/9*a*(c*x)^(10/3)*(b*x^2+a)^(1/3 
)/c+1/6*(c*x)^(10/3)*(b*x^2+a)^(4/3)/c+2/81*a^3*c^(7/3)*arctan(1/3*(1+2*b^ 
(1/3)*(c*x)^(2/3)/c^(2/3)/(b*x^2+a)^(1/3))*3^(1/2))*3^(1/2)/b^(5/3)+1/27*a 
^3*c^(7/3)*ln(b^(1/3)*(c*x)^(2/3)-c^(2/3)*(b*x^2+a)^(1/3))/b^(5/3)
 

Mathematica [A] (verified)

Time = 1.56 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.20 \[ \int (c x)^{7/3} \left (a+b x^2\right )^{4/3} \, dx=\frac {(c x)^{7/3} \left (6 a^2 b^{2/3} x^{4/3} \sqrt [3]{a+b x^2}+45 a b^{5/3} x^{10/3} \sqrt [3]{a+b x^2}+27 b^{8/3} x^{16/3} \sqrt [3]{a+b x^2}+4 \sqrt {3} a^3 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x^{2/3}}{\sqrt [3]{b} x^{2/3}+2 \sqrt [3]{a+b x^2}}\right )+4 a^3 \log \left (-\sqrt [3]{b} x^{2/3}+\sqrt [3]{a+b x^2}\right )-2 a^3 \log \left (b^{2/3} x^{4/3}+\sqrt [3]{b} x^{2/3} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )\right )}{162 b^{5/3} x^{7/3}} \] Input:

Integrate[(c*x)^(7/3)*(a + b*x^2)^(4/3),x]
 

Output:

((c*x)^(7/3)*(6*a^2*b^(2/3)*x^(4/3)*(a + b*x^2)^(1/3) + 45*a*b^(5/3)*x^(10 
/3)*(a + b*x^2)^(1/3) + 27*b^(8/3)*x^(16/3)*(a + b*x^2)^(1/3) + 4*Sqrt[3]* 
a^3*ArcTan[(Sqrt[3]*b^(1/3)*x^(2/3))/(b^(1/3)*x^(2/3) + 2*(a + b*x^2)^(1/3 
))] + 4*a^3*Log[-(b^(1/3)*x^(2/3)) + (a + b*x^2)^(1/3)] - 2*a^3*Log[b^(2/3 
)*x^(4/3) + b^(1/3)*x^(2/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3)]))/(162* 
b^(5/3)*x^(7/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.29 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {248, 248, 262, 266, 807, 853}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (c x)^{7/3} \left (a+b x^2\right )^{4/3} \, dx\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {4}{9} a \int (c x)^{7/3} \sqrt [3]{b x^2+a}dx+\frac {(c x)^{10/3} \left (a+b x^2\right )^{4/3}}{6 c}\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {4}{9} a \left (\frac {1}{6} a \int \frac {(c x)^{7/3}}{\left (b x^2+a\right )^{2/3}}dx+\frac {(c x)^{10/3} \sqrt [3]{a+b x^2}}{4 c}\right )+\frac {(c x)^{10/3} \left (a+b x^2\right )^{4/3}}{6 c}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {4}{9} a \left (\frac {1}{6} a \left (\frac {c (c x)^{4/3} \sqrt [3]{a+b x^2}}{2 b}-\frac {2 a c^2 \int \frac {\sqrt [3]{c x}}{\left (b x^2+a\right )^{2/3}}dx}{3 b}\right )+\frac {(c x)^{10/3} \sqrt [3]{a+b x^2}}{4 c}\right )+\frac {(c x)^{10/3} \left (a+b x^2\right )^{4/3}}{6 c}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {4}{9} a \left (\frac {1}{6} a \left (\frac {c (c x)^{4/3} \sqrt [3]{a+b x^2}}{2 b}-\frac {2 a c \int \frac {c x}{\left (b x^2+a\right )^{2/3}}d\sqrt [3]{c x}}{b}\right )+\frac {(c x)^{10/3} \sqrt [3]{a+b x^2}}{4 c}\right )+\frac {(c x)^{10/3} \left (a+b x^2\right )^{4/3}}{6 c}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {4}{9} a \left (\frac {1}{6} a \left (\frac {c (c x)^{4/3} \sqrt [3]{a+b x^2}}{2 b}-\frac {a c \int \frac {(c x)^{2/3}}{\left (a+\frac {b x}{c}\right )^{2/3}}d(c x)^{2/3}}{b}\right )+\frac {(c x)^{10/3} \sqrt [3]{a+b x^2}}{4 c}\right )+\frac {(c x)^{10/3} \left (a+b x^2\right )^{4/3}}{6 c}\)

\(\Big \downarrow \) 853

\(\displaystyle \frac {4}{9} a \left (\frac {1}{6} a \left (\frac {c (c x)^{4/3} \sqrt [3]{a+b x^2}}{2 b}-\frac {a c \left (-\frac {c^{4/3} \arctan \left (\frac {\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+\frac {b x}{c}}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {c^{4/3} \log \left (\frac {\sqrt [3]{b} (c x)^{2/3}}{c^{2/3}}-\sqrt [3]{a+\frac {b x}{c}}\right )}{2 b^{2/3}}\right )}{b}\right )+\frac {(c x)^{10/3} \sqrt [3]{a+b x^2}}{4 c}\right )+\frac {(c x)^{10/3} \left (a+b x^2\right )^{4/3}}{6 c}\)

Input:

Int[(c*x)^(7/3)*(a + b*x^2)^(4/3),x]
 

Output:

((c*x)^(10/3)*(a + b*x^2)^(4/3))/(6*c) + (4*a*(((c*x)^(10/3)*(a + b*x^2)^( 
1/3))/(4*c) + (a*((c*(c*x)^(4/3)*(a + b*x^2)^(1/3))/(2*b) - (a*c*(-((c^(4/ 
3)*ArcTan[(1 + (2*b^(1/3)*(c*x)^(2/3))/(c^(2/3)*(a + (b*x)/c)^(1/3)))/Sqrt 
[3]])/(Sqrt[3]*b^(2/3))) - (c^(4/3)*Log[(b^(1/3)*(c*x)^(2/3))/c^(2/3) - (a 
 + (b*x)/c)^(1/3)])/(2*b^(2/3))))/b))/6))/9
 

Defintions of rubi rules used

rule 248

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) 
  Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ 
p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 853

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim 
p[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp 
[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]
 

Maple [F]

\[\int \left (c x \right )^{\frac {7}{3}} \left (b \,x^{2}+a \right )^{\frac {4}{3}}d x\]

Input:

int((c*x)^(7/3)*(b*x^2+a)^(4/3),x)
 

Output:

int((c*x)^(7/3)*(b*x^2+a)^(4/3),x)
 

Fricas [F(-1)]

Timed out. \[ \int (c x)^{7/3} \left (a+b x^2\right )^{4/3} \, dx=\text {Timed out} \] Input:

integrate((c*x)^(7/3)*(b*x^2+a)^(4/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 37.36 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.24 \[ \int (c x)^{7/3} \left (a+b x^2\right )^{4/3} \, dx=\frac {a^{\frac {4}{3}} c^{\frac {7}{3}} x^{\frac {10}{3}} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {8}{3}\right )} \] Input:

integrate((c*x)**(7/3)*(b*x**2+a)**(4/3),x)
 

Output:

a**(4/3)*c**(7/3)*x**(10/3)*gamma(5/3)*hyper((-4/3, 5/3), (8/3,), b*x**2*e 
xp_polar(I*pi)/a)/(2*gamma(8/3))
 

Maxima [F]

\[ \int (c x)^{7/3} \left (a+b x^2\right )^{4/3} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {4}{3}} \left (c x\right )^{\frac {7}{3}} \,d x } \] Input:

integrate((c*x)^(7/3)*(b*x^2+a)^(4/3),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(4/3)*(c*x)^(7/3), x)
 

Giac [F]

\[ \int (c x)^{7/3} \left (a+b x^2\right )^{4/3} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {4}{3}} \left (c x\right )^{\frac {7}{3}} \,d x } \] Input:

integrate((c*x)^(7/3)*(b*x^2+a)^(4/3),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((b*x^2 + a)^(4/3)*(c*x)^(7/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int (c x)^{7/3} \left (a+b x^2\right )^{4/3} \, dx=\int {\left (c\,x\right )}^{7/3}\,{\left (b\,x^2+a\right )}^{4/3} \,d x \] Input:

int((c*x)^(7/3)*(a + b*x^2)^(4/3),x)
 

Output:

int((c*x)^(7/3)*(a + b*x^2)^(4/3), x)
 

Reduce [F]

\[ \int (c x)^{7/3} \left (a+b x^2\right )^{4/3} \, dx=\frac {c^{\frac {7}{3}} \left (6 x^{\frac {4}{3}} \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{2}+45 x^{\frac {10}{3}} \left (b \,x^{2}+a \right )^{\frac {1}{3}} a b +27 x^{\frac {16}{3}} \left (b \,x^{2}+a \right )^{\frac {1}{3}} b^{2}-8 \left (\int \frac {x^{\frac {1}{3}}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a^{3}\right )}{162 b} \] Input:

int((c*x)^(7/3)*(b*x^2+a)^(4/3),x)
 

Output:

(c**(1/3)*c**2*(6*x**(1/3)*(a + b*x**2)**(1/3)*a**2*x + 45*x**(1/3)*(a + b 
*x**2)**(1/3)*a*b*x**3 + 27*x**(1/3)*(a + b*x**2)**(1/3)*b**2*x**5 - 8*int 
((x**(1/3)*(a + b*x**2)**(1/3))/(a + b*x**2),x)*a**3))/(162*b)