\(\int \frac {(A+B x) \sqrt [3]{c+d x}}{(A^2-B^2 x^2)^{5/3}} \, dx\) [11]

Optimal result

Integrand size = 31, antiderivative size = 114 \[ \int \frac {(A+B x) \sqrt [3]{c+d x}}{\left (A^2-B^2 x^2\right )^{5/3}} \, dx=\frac {3 \left (\frac {(B c+A d) (A+B x)}{A B (c+d x)}\right )^{2/3} (c+d x)^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {2}{3},\frac {1}{3},\frac {(B c-A d) (A-B x)}{2 A B (c+d x)}\right )}{2\ 2^{2/3} (B c+A d) \left (A^2-B^2 x^2\right )^{2/3}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 13.65 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.70 \[ \int \frac {(A+B x) \sqrt [3]{c+d x}}{\left (A^2-B^2 x^2\right )^{5/3}} \, dx=\frac {3 \sqrt [3]{c+d x} \left (d (A+B x)+(B c-A d) \sqrt [3]{\frac {d \left (\sqrt {\frac {A^2}{B^2}}-x\right )}{c+\sqrt {\frac {A^2}{B^2}} d}} \left (\frac {d \left (\sqrt {\frac {A^2}{B^2}}+x\right )}{-c+\sqrt {\frac {A^2}{B^2}} d}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},\frac {2 \sqrt {\frac {A^2}{B^2}} (c+d x)}{\left (c-\sqrt {\frac {A^2}{B^2}} d\right ) \left (\sqrt {\frac {A^2}{B^2}}-x\right )}\right )\right )}{4 A B d \left (A^2-B^2 x^2\right )^{2/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.33, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {678, 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.

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

\(\Big \downarrow \) 678

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

\(\Big \downarrow \) 489

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

Input:

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

Output:

(3*(A + B*x)*(c + d*x)^(1/3))/(4*A*B*(A^2 - B^2*x^2)^(2/3)) - (3*(-(((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 
))])/(4*A*B*(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 678

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

Maple [F]

\[\int \frac {\left (B x +A \right ) \left (d x +c \right )^{\frac {1}{3}}}{\left (-B^{2} x^{2}+A^{2}\right )^{\frac {5}{3}}}d x\]

Input:

int((B*x+A)*(d*x+c)^(1/3)/(-B^2*x^2+A^2)^(5/3),x)
 

Output:

int((B*x+A)*(d*x+c)^(1/3)/(-B^2*x^2+A^2)^(5/3),x)
 

Fricas [F]

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

integrate((B*x+A)*(d*x+c)^(1/3)/(-B^2*x^2+A^2)^(5/3),x, algorithm="fricas" 
)
 

Output:

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

Sympy [F]

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

integrate((B*x+A)*(d*x+c)**(1/3)/(-B**2*x**2+A**2)**(5/3),x)
 

Output:

Integral((A + B*x)*(c + d*x)**(1/3)/(-(-A + B*x)*(A + B*x))**(5/3), x)
 

Maxima [F]

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

integrate((B*x+A)*(d*x+c)^(1/3)/(-B^2*x^2+A^2)^(5/3),x, algorithm="maxima" 
)
 

Output:

integrate((B*x + A)*(d*x + c)^(1/3)/(-B^2*x^2 + A^2)^(5/3), x)
 

Giac [F]

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

integrate((B*x+A)*(d*x+c)^(1/3)/(-B^2*x^2+A^2)^(5/3),x, algorithm="giac")
 

Output:

integrate((B*x + A)*(d*x + c)^(1/3)/(-B^2*x^2 + A^2)^(5/3), x)
 

Mupad [F(-1)]

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

int(((A + B*x)*(c + d*x)^(1/3))/(A^2 - B^2*x^2)^(5/3),x)
 

Output:

int(((A + B*x)*(c + d*x)^(1/3))/(A^2 - B^2*x^2)^(5/3), x)
 

Reduce [F]

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

int((B*x+A)*(d*x+c)^(1/3)/(-B^2*x^2+A^2)^(5/3),x)
 

Output:

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