\(\int (A+B x) (a+b x^2)^{4/3} \, dx\) [15]

Optimal result

Integrand size = 17, antiderivative size = 307 \[ \int (A+B x) \left (a+b x^2\right )^{4/3} \, dx=\frac {24}{55} a A x \sqrt [3]{a+b x^2}+\frac {3}{11} A x \left (a+b x^2\right )^{4/3}+\frac {3 B \left (a+b x^2\right )^{7/3}}{14 b}-\frac {16\ 3^{3/4} \sqrt {2-\sqrt {3}} a^2 A \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{55 b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:

24/55*a*A*x*(b*x^2+a)^(1/3)+3/11*A*x*(b*x^2+a)^(4/3)+3/14*B*(b*x^2+a)^(7/3 
)/b-16/55*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*a^2*A*(a^(1/3)-(b*x^2+a)^(1/3) 
)*((a^(2/3)+a^(1/3)*(b*x^2+a)^(1/3)+(b*x^2+a)^(2/3))/((1-3^(1/2))*a^(1/3)- 
(b*x^2+a)^(1/3))^2)^(1/2)*EllipticF(((1+3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))/ 
((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3)),2*I-I*3^(1/2))/b/x/(-a^(1/3)*(a^(1/3 
)-(b*x^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.17 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.34 \[ \int (A+B x) \left (a+b x^2\right )^{4/3} \, dx=\frac {3 \left (a+b x^2\right ) \left (55 a^2 B+5 b^2 x^3 (14 A+11 B x)+2 a b x (91 A+55 B x)\right )+224 a^2 A b x \left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},-\frac {b x^2}{a}\right )}{770 b \left (a+b x^2\right )^{2/3}} \] Input:

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

Output:

(3*(a + b*x^2)*(55*a^2*B + 5*b^2*x^3*(14*A + 11*B*x) + 2*a*b*x*(91*A + 55* 
B*x)) + 224*a^2*A*b*x*(1 + (b*x^2)/a)^(2/3)*Hypergeometric2F1[1/2, 2/3, 3/ 
2, -((b*x^2)/a)])/(770*b*(a + b*x^2)^(2/3))
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {455, 211, 211, 234, 760}

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.

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

\(\Big \downarrow \) 455

\(\displaystyle A \int \left (b x^2+a\right )^{4/3}dx+\frac {3 B \left (a+b x^2\right )^{7/3}}{14 b}\)

\(\Big \downarrow \) 211

\(\displaystyle A \left (\frac {8}{11} a \int \sqrt [3]{b x^2+a}dx+\frac {3}{11} x \left (a+b x^2\right )^{4/3}\right )+\frac {3 B \left (a+b x^2\right )^{7/3}}{14 b}\)

\(\Big \downarrow \) 211

\(\displaystyle A \left (\frac {8}{11} a \left (\frac {2}{5} a \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )+\frac {3}{11} x \left (a+b x^2\right )^{4/3}\right )+\frac {3 B \left (a+b x^2\right )^{7/3}}{14 b}\)

\(\Big \downarrow \) 234

\(\displaystyle A \left (\frac {8}{11} a \left (\frac {3 a \sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{5 b x}+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )+\frac {3}{11} x \left (a+b x^2\right )^{4/3}\right )+\frac {3 B \left (a+b x^2\right )^{7/3}}{14 b}\)

\(\Big \downarrow \) 760

\(\displaystyle A \left (\frac {8}{11} a \left (\frac {3}{5} x \sqrt [3]{a+b x^2}-\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{5 b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\right )+\frac {3}{11} x \left (a+b x^2\right )^{4/3}\right )+\frac {3 B \left (a+b x^2\right )^{7/3}}{14 b}\)

Input:

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

Output:

(3*B*(a + b*x^2)^(7/3))/(14*b) + A*((3*x*(a + b*x^2)^(4/3))/11 + (8*a*((3* 
x*(a + b*x^2)^(1/3))/5 - (2*3^(3/4)*Sqrt[2 - Sqrt[3]]*a*(a^(1/3) - (a + b* 
x^2)^(1/3))*Sqrt[(a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3)) 
/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqr 
t[3])*a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1 
/3))], -7 + 4*Sqrt[3]])/(5*b*x*Sqrt[-((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3 
)))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2)])))/11)
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 234

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) 
   Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b 
}, x]
 

rule 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 760

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- 
s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) 
*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x 
] && NegQ[a]
 

Maple [F]

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

Input:

int((B*x+A)*(b*x^2+a)^(4/3),x)
 

Output:

int((B*x+A)*(b*x^2+a)^(4/3),x)
 

Fricas [F]

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

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

Output:

integral((B*b*x^3 + A*b*x^2 + B*a*x + A*a)*(b*x^2 + a)^(1/3), x)
 

Sympy [A] (verification not implemented)

Time = 1.65 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.08 \[ \int (A+B x) \left (a+b x^2\right )^{4/3} \, dx=A a^{\frac {4}{3}} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {A \sqrt [3]{a} b x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} - \frac {9 B a^{\frac {13}{3}} b \sqrt [3]{1 + \frac {b x^{2}}{a}}}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} + \frac {9 B a^{\frac {13}{3}} b}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} - \frac {6 B a^{\frac {10}{3}} b^{2} x^{2} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} + \frac {9 B a^{\frac {10}{3}} b^{2} x^{2}}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} + \frac {15 B a^{\frac {7}{3}} b^{3} x^{4} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} + \frac {12 B a^{\frac {4}{3}} b^{4} x^{6} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} + B a \left (\begin {cases} \frac {\sqrt [3]{a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {3 \left (a + b x^{2}\right )^{\frac {4}{3}}}{8 b} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((B*x+A)*(b*x**2+a)**(4/3),x)
 

Output:

A*a**(4/3)*x*hyper((-1/3, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a) + A*a**( 
1/3)*b*x**3*hyper((-1/3, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 - 9*B*a 
**(13/3)*b*(1 + b*x**2/a)**(1/3)/(56*a**2*b**2 + 56*a*b**3*x**2) + 9*B*a** 
(13/3)*b/(56*a**2*b**2 + 56*a*b**3*x**2) - 6*B*a**(10/3)*b**2*x**2*(1 + b* 
x**2/a)**(1/3)/(56*a**2*b**2 + 56*a*b**3*x**2) + 9*B*a**(10/3)*b**2*x**2/( 
56*a**2*b**2 + 56*a*b**3*x**2) + 15*B*a**(7/3)*b**3*x**4*(1 + b*x**2/a)**( 
1/3)/(56*a**2*b**2 + 56*a*b**3*x**2) + 12*B*a**(4/3)*b**4*x**6*(1 + b*x**2 
/a)**(1/3)/(56*a**2*b**2 + 56*a*b**3*x**2) + B*a*Piecewise((a**(1/3)*x**2/ 
2, Eq(b, 0)), (3*(a + b*x**2)**(4/3)/(8*b), True))
 

Maxima [F]

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

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

Output:

integrate((b*x^2 + a)^(4/3)*(B*x + A), x)
 

Giac [F]

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

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

Output:

integrate((b*x^2 + a)^(4/3)*(B*x + A), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.18 \[ \int (A+B x) \left (a+b x^2\right )^{4/3} \, dx=\frac {3\,B\,{\left (b\,x^2+a\right )}^{7/3}}{14\,b}+\frac {A\,x\,{\left (b\,x^2+a\right )}^{4/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {4}{3},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{4/3}} \] Input:

int((a + b*x^2)^(4/3)*(A + B*x),x)
 

Output:

(3*B*(a + b*x^2)^(7/3))/(14*b) + (A*x*(a + b*x^2)^(4/3)*hypergeom([-4/3, 1 
/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^(4/3)
 

Reduce [F]

\[ \int (A+B x) \left (a+b x^2\right )^{4/3} \, dx=\frac {39 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{2} x}{55}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{2}}{14}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a b \,x^{3}}{11}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a b \,x^{2}}{7}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b^{2} x^{4}}{14}+\frac {16 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a^{3}}{55} \] Input:

int((B*x+A)*(b*x^2+a)^(4/3),x)
 

Output:

(546*(a + b*x**2)**(1/3)*a**2*x + 165*(a + b*x**2)**(1/3)*a**2 + 210*(a + 
b*x**2)**(1/3)*a*b*x**3 + 330*(a + b*x**2)**(1/3)*a*b*x**2 + 165*(a + b*x* 
*2)**(1/3)*b**2*x**4 + 224*int((a + b*x**2)**(1/3)/(a + b*x**2),x)*a**3)/7 
70