\(\int \frac {A+B x}{(c+d x)^{7/3} \sqrt [3]{A^2-B^2 x^2}} \, dx\) [8]

Optimal result

Integrand size = 31, antiderivative size = 112 \[ \int \frac {A+B x}{(c+d x)^{7/3} \sqrt [3]{A^2-B^2 x^2}} \, dx=-\frac {3 \left (A^2-B^2 x^2\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {2}{3},\frac {5}{3},\frac {(B c-A d) (A-B x)}{2 A B (c+d x)}\right )}{\sqrt [3]{2} (B c+A d) \left (\frac {(B c+A d) (A+B x)}{A B (c+d x)}\right )^{2/3} (c+d x)^{4/3}} \] Output:

-3/2*(-B^2*x^2+A^2)^(2/3)*hypergeom([-2/3, 2/3],[5/3],1/2*(-A*d+B*c)*(-B*x 
+A)/A/B/(d*x+c))*2^(2/3)/(A*d+B*c)/((A*d+B*c)*(B*x+A)/A/B/(d*x+c))^(2/3)/( 
d*x+c)^(4/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 20.87 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.65 \[ \int \frac {A+B x}{(c+d x)^{7/3} \sqrt [3]{A^2-B^2 x^2}} \, dx=\frac {3 \left (-B^2 c^2+A^2 d^2\right ) \left (A^2-B^2 x^2\right )+4 A B \left (A^2 d+B^2 c x\right ) (c+d x) \sqrt [3]{-\frac {\left (-B^2 c^2+A^2 d^2\right ) \left (A^2-B^2 x^2\right )}{A^2 B^2 (c+d x)^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {3}{2},\frac {\left (A^2 d+B^2 c x\right )^2}{A^2 B^2 (c+d x)^2}\right )}{4 (B c-A d) (B c+A d)^2 (c+d x)^{4/3} \sqrt [3]{A^2-B^2 x^2}} \] Input:

Integrate[(A + B*x)/((c + d*x)^(7/3)*(A^2 - B^2*x^2)^(1/3)),x]
 

Output:

(3*(-(B^2*c^2) + A^2*d^2)*(A^2 - B^2*x^2) + 4*A*B*(A^2*d + B^2*c*x)*(c + d 
*x)*(-(((-(B^2*c^2) + A^2*d^2)*(A^2 - B^2*x^2))/(A^2*B^2*(c + d*x)^2)))^(1 
/3)*Hypergeometric2F1[1/3, 1/2, 3/2, (A^2*d + B^2*c*x)^2/(A^2*B^2*(c + d*x 
)^2)])/(4*(B*c - A*d)*(B*c + A*d)^2*(c + d*x)^(4/3)*(A^2 - B^2*x^2)^(1/3))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.46, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {679, 489}

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.

\(\displaystyle \int \frac {A+B x}{\sqrt [3]{A^2-B^2 x^2} (c+d x)^{7/3}} \, dx\)

\(\Big \downarrow \) 679

\(\displaystyle \frac {A B \int \frac {1}{(c+d x)^{4/3} \sqrt [3]{A^2-B^2 x^2}}dx}{A d+B c}-\frac {3 \left (A^2-B^2 x^2\right )^{2/3}}{4 (c+d x)^{4/3} (A d+B c)}\)

\(\Big \downarrow \) 489

\(\displaystyle \frac {3 A B (A+B x) \sqrt [3]{-\frac {(A-B x) (B c-A d)}{(A+B x) (A d+B c)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {2}{3},\frac {2 A B (c+d x)}{(B c+A d) (A+B x)}\right )}{\sqrt [3]{A^2-B^2 x^2} \sqrt [3]{c+d x} (B c-A d) (A d+B c)}-\frac {3 \left (A^2-B^2 x^2\right )^{2/3}}{4 (c+d x)^{4/3} (A d+B c)}\)

Input:

Int[(A + B*x)/((c + d*x)^(7/3)*(A^2 - B^2*x^2)^(1/3)),x]
 

Output:

(-3*(A^2 - B^2*x^2)^(2/3))/(4*(B*c + A*d)*(c + d*x)^(4/3)) + (3*A*B*(-(((B 
*c - A*d)*(A - B*x))/((B*c + A*d)*(A + B*x))))^(1/3)*(A + B*x)*Hypergeomet 
ric2F1[-1/3, 1/3, 2/3, (2*A*B*(c + d*x))/((B*c + A*d)*(A + B*x))])/((B*c - 
 A*d)*(B*c + A*d)*(c + d*x)^(1/3)*(A^2 - B^2*x^2)^(1/3))
 

Defintions of rubi rules used

rule 489

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Rt[(-a)*b, 2]}, Simp[(q - b*x)*(c + d*x)^(n + 1)*((a + b*x^2)^p/((n + 
1)*(b*c + d*q)*((b*c + d*q)*((q + b*x)/((b*c - d*q)*(-q + b*x))))^p))*Hyper 
geometric2F1[n + 1, -p, n + 2, 2*b*q*((c + d*x)/((b*c - d*q)*(q - b*x)))], 
x]] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[n + 2*p + 2, 0]
 

rule 679

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 
)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) 
 Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, 
 p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
 

Maple [F]

\[\int \frac {B x +A}{\left (d x +c \right )^{\frac {7}{3}} \left (-B^{2} x^{2}+A^{2}\right )^{\frac {1}{3}}}d x\]

Input:

int((B*x+A)/(d*x+c)^(7/3)/(-B^2*x^2+A^2)^(1/3),x)
 

Output:

int((B*x+A)/(d*x+c)^(7/3)/(-B^2*x^2+A^2)^(1/3),x)
 

Fricas [F]

\[ \int \frac {A+B x}{(c+d x)^{7/3} \sqrt [3]{A^2-B^2 x^2}} \, dx=\int { \frac {B x + A}{{\left (-B^{2} x^{2} + A^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {7}{3}}} \,d x } \] Input:

integrate((B*x+A)/(d*x+c)^(7/3)/(-B^2*x^2+A^2)^(1/3),x, algorithm="fricas" 
)
 

Output:

integral(-(-B^2*x^2 + A^2)^(2/3)*(d*x + c)^(2/3)/(B*d^3*x^4 - A*c^3 + (3*B 
*c*d^2 - A*d^3)*x^3 + 3*(B*c^2*d - A*c*d^2)*x^2 + (B*c^3 - 3*A*c^2*d)*x), 
x)
 

Sympy [F]

\[ \int \frac {A+B x}{(c+d x)^{7/3} \sqrt [3]{A^2-B^2 x^2}} \, dx=\int \frac {A + B x}{\sqrt [3]{- \left (- A + B x\right ) \left (A + B x\right )} \left (c + d x\right )^{\frac {7}{3}}}\, dx \] Input:

integrate((B*x+A)/(d*x+c)**(7/3)/(-B**2*x**2+A**2)**(1/3),x)
 

Output:

Integral((A + B*x)/((-(-A + B*x)*(A + B*x))**(1/3)*(c + d*x)**(7/3)), x)
 

Maxima [F]

\[ \int \frac {A+B x}{(c+d x)^{7/3} \sqrt [3]{A^2-B^2 x^2}} \, dx=\int { \frac {B x + A}{{\left (-B^{2} x^{2} + A^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {7}{3}}} \,d x } \] Input:

integrate((B*x+A)/(d*x+c)^(7/3)/(-B^2*x^2+A^2)^(1/3),x, algorithm="maxima" 
)
 

Output:

integrate((B*x + A)/((-B^2*x^2 + A^2)^(1/3)*(d*x + c)^(7/3)), x)
 

Giac [F]

\[ \int \frac {A+B x}{(c+d x)^{7/3} \sqrt [3]{A^2-B^2 x^2}} \, dx=\int { \frac {B x + A}{{\left (-B^{2} x^{2} + A^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {7}{3}}} \,d x } \] Input:

integrate((B*x+A)/(d*x+c)^(7/3)/(-B^2*x^2+A^2)^(1/3),x, algorithm="giac")
 

Output:

integrate((B*x + A)/((-B^2*x^2 + A^2)^(1/3)*(d*x + c)^(7/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{(c+d x)^{7/3} \sqrt [3]{A^2-B^2 x^2}} \, dx=\int \frac {A+B\,x}{{\left (A^2-B^2\,x^2\right )}^{1/3}\,{\left (c+d\,x\right )}^{7/3}} \,d x \] Input:

int((A + B*x)/((A^2 - B^2*x^2)^(1/3)*(c + d*x)^(7/3)),x)
 

Output:

int((A + B*x)/((A^2 - B^2*x^2)^(1/3)*(c + d*x)^(7/3)), x)
 

Reduce [F]

\[ \int \frac {A+B x}{(c+d x)^{7/3} \sqrt [3]{A^2-B^2 x^2}} \, dx=\left (\int \frac {x}{\left (d x +c \right )^{\frac {1}{3}} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {1}{3}} c^{2}+2 \left (d x +c \right )^{\frac {1}{3}} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {1}{3}} c d x +\left (d x +c \right )^{\frac {1}{3}} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {1}{3}} d^{2} x^{2}}d x \right ) b +\left (\int \frac {1}{\left (d x +c \right )^{\frac {1}{3}} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {1}{3}} c^{2}+2 \left (d x +c \right )^{\frac {1}{3}} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {1}{3}} c d x +\left (d x +c \right )^{\frac {1}{3}} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {1}{3}} d^{2} x^{2}}d x \right ) a \] Input:

int((B*x+A)/(d*x+c)^(7/3)/(-B^2*x^2+A^2)^(1/3),x)
 

Output:

int(x/((c + d*x)**(1/3)*(a**2 - b**2*x**2)**(1/3)*c**2 + 2*(c + d*x)**(1/3 
)*(a**2 - b**2*x**2)**(1/3)*c*d*x + (c + d*x)**(1/3)*(a**2 - b**2*x**2)**( 
1/3)*d**2*x**2),x)*b + int(1/((c + d*x)**(1/3)*(a**2 - b**2*x**2)**(1/3)*c 
**2 + 2*(c + d*x)**(1/3)*(a**2 - b**2*x**2)**(1/3)*c*d*x + (c + d*x)**(1/3 
)*(a**2 - b**2*x**2)**(1/3)*d**2*x**2),x)*a