\(\int \frac {A+B x}{(c+d x)^{5/3} (A^2-B^2 x^2)^{2/3}} \, dx\) [9]

Optimal result

Integrand size = 31, antiderivative size = 112 \[ \int \frac {A+B x}{(c+d x)^{5/3} \left (A^2-B^2 x^2\right )^{2/3}} \, dx=-\frac {3 \sqrt [3]{2} \sqrt [3]{A^2-B^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},\frac {(B c-A d) (A-B x)}{2 A B (c+d x)}\right )}{(B c+A d) \sqrt [3]{\frac {(B c+A d) (A+B x)}{A B (c+d x)}} (c+d x)^{2/3}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 15.85 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.83 \[ \int \frac {A+B x}{(c+d x)^{5/3} \left (A^2-B^2 x^2\right )^{2/3}} \, dx=\frac {-3 d \left (A^2-B^2 x^2\right )+6 A B \sqrt [3]{\frac {d \left (\sqrt {\frac {A^2}{B^2}}-x\right )}{c+\sqrt {\frac {A^2}{B^2}} d}} \left (\frac {d \left (\sqrt {\frac {A^2}{B^2}}+x\right )}{-c+\sqrt {\frac {A^2}{B^2}} d}\right )^{2/3} (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},\frac {2 \sqrt {\frac {A^2}{B^2}} (c+d x)}{\left (c-\sqrt {\frac {A^2}{B^2}} d\right ) \left (\sqrt {\frac {A^2}{B^2}}-x\right )}\right )}{2 d (B c+A d) (c+d x)^{2/3} \left (A^2-B^2 x^2\right )^{2/3}} \] Input:

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

Output:

(-3*d*(A^2 - B^2*x^2) + 6*A*B*((d*(Sqrt[A^2/B^2] - x))/(c + Sqrt[A^2/B^2]* 
d))^(1/3)*((d*(Sqrt[A^2/B^2] + x))/(-c + Sqrt[A^2/B^2]*d))^(2/3)*(c + d*x) 
*Hypergeometric2F1[1/3, 2/3, 4/3, (2*Sqrt[A^2/B^2]*(c + d*x))/((c - Sqrt[A 
^2/B^2]*d)*(Sqrt[A^2/B^2] - x))])/(2*d*(B*c + A*d)*(c + d*x)^(2/3)*(A^2 - 
B^2*x^2)^(2/3))
 

Rubi [A] (verified)

Time = 0.27 (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.

\(\displaystyle \int \frac {A+B x}{\left (A^2-B^2 x^2\right )^{2/3} (c+d x)^{5/3}} \, dx\)

\(\Big \downarrow \) 679

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

\(\Big \downarrow \) 489

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

Input:

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

Output:

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

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]

\[\int \frac {B x +A}{\left (d x +c \right )^{\frac {5}{3}} \left (-B^{2} x^{2}+A^{2}\right )^{\frac {2}{3}}}d x\]

Input:

int((B*x+A)/(d*x+c)^(5/3)/(-B^2*x^2+A^2)^(2/3),x)
 

Output:

int((B*x+A)/(d*x+c)^(5/3)/(-B^2*x^2+A^2)^(2/3),x)
 

Fricas [F]

\[ \int \frac {A+B x}{(c+d x)^{5/3} \left (A^2-B^2 x^2\right )^{2/3}} \, dx=\int { \frac {B x + A}{{\left (-B^{2} x^{2} + A^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate((B*x+A)/(d*x+c)^(5/3)/(-B^2*x^2+A^2)^(2/3),x, algorithm="fricas" 
)
 

Output:

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

Sympy [F]

\[ \int \frac {A+B x}{(c+d x)^{5/3} \left (A^2-B^2 x^2\right )^{2/3}} \, dx=\int \frac {A + B x}{\left (- \left (- A + B x\right ) \left (A + B x\right )\right )^{\frac {2}{3}} \left (c + d x\right )^{\frac {5}{3}}}\, dx \] Input:

integrate((B*x+A)/(d*x+c)**(5/3)/(-B**2*x**2+A**2)**(2/3),x)
 

Output:

Integral((A + B*x)/((-(-A + B*x)*(A + B*x))**(2/3)*(c + d*x)**(5/3)), x)
 

Maxima [F]

\[ \int \frac {A+B x}{(c+d x)^{5/3} \left (A^2-B^2 x^2\right )^{2/3}} \, dx=\int { \frac {B x + A}{{\left (-B^{2} x^{2} + A^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate((B*x+A)/(d*x+c)^(5/3)/(-B^2*x^2+A^2)^(2/3),x, algorithm="maxima" 
)
 

Output:

integrate((B*x + A)/((-B^2*x^2 + A^2)^(2/3)*(d*x + c)^(5/3)), x)
 

Giac [F]

\[ \int \frac {A+B x}{(c+d x)^{5/3} \left (A^2-B^2 x^2\right )^{2/3}} \, dx=\int { \frac {B x + A}{{\left (-B^{2} x^{2} + A^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate((B*x+A)/(d*x+c)^(5/3)/(-B^2*x^2+A^2)^(2/3),x, algorithm="giac")
 

Output:

integrate((B*x + A)/((-B^2*x^2 + A^2)^(2/3)*(d*x + c)^(5/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{(c+d x)^{5/3} \left (A^2-B^2 x^2\right )^{2/3}} \, dx=\int \frac {A+B\,x}{{\left (A^2-B^2\,x^2\right )}^{2/3}\,{\left (c+d\,x\right )}^{5/3}} \,d x \] Input:

int((A + B*x)/((A^2 - B^2*x^2)^(2/3)*(c + d*x)^(5/3)),x)
 

Output:

int((A + B*x)/((A^2 - B^2*x^2)^(2/3)*(c + d*x)^(5/3)), x)
 

Reduce [F]

\[ \int \frac {A+B x}{(c+d x)^{5/3} \left (A^2-B^2 x^2\right )^{2/3}} \, dx=\left (\int \frac {x}{\left (d x +c \right )^{\frac {2}{3}} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {2}{3}} c +\left (d x +c \right )^{\frac {2}{3}} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {2}{3}} d x}d x \right ) b +\left (\int \frac {1}{\left (d x +c \right )^{\frac {2}{3}} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {2}{3}} c +\left (d x +c \right )^{\frac {2}{3}} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {2}{3}} d x}d x \right ) a \] Input:

int((B*x+A)/(d*x+c)^(5/3)/(-B^2*x^2+A^2)^(2/3),x)
 

Output:

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