\(\int \frac {(A+B x) \sqrt [3]{A^2-B^2 x^2}}{(c+d x)^{11/3}} \, dx\) [7]

Optimal result

Integrand size = 31, antiderivative size = 112 \[ \int \frac {(A+B x) \sqrt [3]{A^2-B^2 x^2}}{(c+d x)^{11/3}} \, dx=-\frac {3 \left (A^2-B^2 x^2\right )^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {4}{3},\frac {7}{3},\frac {(B c-A d) (A-B x)}{2 A B (c+d x)}\right )}{2^{2/3} (B c+A d) \left (\frac {(B c+A d) (A+B x)}{A B (c+d x)}\right )^{4/3} (c+d x)^{8/3}} \] Output:

-3/2*(-B^2*x^2+A^2)^(4/3)*hypergeom([-4/3, 4/3],[7/3],1/2*(-A*d+B*c)*(-B*x 
+A)/A/B/(d*x+c))*2^(1/3)/(A*d+B*c)/((A*d+B*c)*(B*x+A)/A/B/(d*x+c))^(4/3)/( 
d*x+c)^(8/3)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(284\) vs. \(2(112)=224\).

Time = 10.41 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.54 \[ \int \frac {(A+B x) \sqrt [3]{A^2-B^2 x^2}}{(c+d x)^{11/3}} \, dx=\frac {3 \left (d \left (A^2-B^2 x^2\right ) \left (5 A^4 d^2+5 B^4 c^2 x^2+8 A^3 B d (c+d x)+8 A B^3 c x (c+d x)-5 A^2 B^2 \left (c^2+d^2 x^2\right )\right )+16 A^3 B^3 \sqrt [3]{\frac {d \left (\sqrt {\frac {A^2}{B^2}}-x\right )}{c+\sqrt {\frac {A^2}{B^2}} d}} \left (\frac {d \left (\sqrt {\frac {A^2}{B^2}}+x\right )}{-c+\sqrt {\frac {A^2}{B^2}} d}\right )^{2/3} (c+d x)^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},\frac {2 \sqrt {\frac {A^2}{B^2}} (c+d x)}{\left (c-\sqrt {\frac {A^2}{B^2}} d\right ) \left (\sqrt {\frac {A^2}{B^2}}-x\right )}\right )\right )}{40 d (B c-A d) (B c+A d)^2 (c+d x)^{8/3} \left (A^2-B^2 x^2\right )^{2/3}} \] Input:

Integrate[((A + B*x)*(A^2 - B^2*x^2)^(1/3))/(c + d*x)^(11/3),x]
 

Output:

(3*(d*(A^2 - B^2*x^2)*(5*A^4*d^2 + 5*B^4*c^2*x^2 + 8*A^3*B*d*(c + d*x) + 8 
*A*B^3*c*x*(c + d*x) - 5*A^2*B^2*(c^2 + d^2*x^2)) + 16*A^3*B^3*((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)^3*Hypergeometric2F1[1/3, 2/3, 4/3, (2*Sq 
rt[A^2/B^2]*(c + d*x))/((c - Sqrt[A^2/B^2]*d)*(Sqrt[A^2/B^2] - x))]))/(40* 
d*(B*c - A*d)*(B*c + A*d)^2*(c + d*x)^(8/3)*(A^2 - B^2*x^2)^(2/3))
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(250\) vs. \(2(112)=224\).

Time = 0.33 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.23, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {679, 486, 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.

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

\(\Big \downarrow \) 679

\(\displaystyle \frac {A B \int \frac {\sqrt [3]{A^2-B^2 x^2}}{(c+d x)^{8/3}}dx}{A d+B c}-\frac {3 \left (A^2-B^2 x^2\right )^{4/3}}{8 (c+d x)^{8/3} (A d+B c)}\)

\(\Big \downarrow \) 486

\(\displaystyle \frac {A B \left (\frac {2 A^2 B^2 \int \frac {1}{(c+d x)^{2/3} \left (A^2-B^2 x^2\right )^{2/3}}dx}{5 \left (B^2 c^2-A^2 d^2\right )}+\frac {3 \sqrt [3]{A^2-B^2 x^2} \left (A^2 d+B^2 c x\right )}{5 (c+d x)^{5/3} \left (B^2 c^2-A^2 d^2\right )}\right )}{A d+B c}-\frac {3 \left (A^2-B^2 x^2\right )^{4/3}}{8 (c+d x)^{8/3} (A d+B c)}\)

\(\Big \downarrow \) 489

\(\displaystyle \frac {A B \left (\frac {3 \sqrt [3]{A^2-B^2 x^2} \left (A^2 d+B^2 c x\right )}{5 (c+d x)^{5/3} \left (B^2 c^2-A^2 d^2\right )}-\frac {6 A^2 B^2 (A+B x) \sqrt [3]{c+d x} \left (-\frac {(A-B x) (B c-A d)}{(A+B x) (A d+B c)}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},\frac {2 A B (c+d x)}{(B c+A d) (A+B x)}\right )}{5 \left (A^2-B^2 x^2\right )^{2/3} (B c-A d) \left (B^2 c^2-A^2 d^2\right )}\right )}{A d+B c}-\frac {3 \left (A^2-B^2 x^2\right )^{4/3}}{8 (c+d x)^{8/3} (A d+B c)}\)

Input:

Int[((A + B*x)*(A^2 - B^2*x^2)^(1/3))/(c + d*x)^(11/3),x]
 

Output:

(-3*(A^2 - B^2*x^2)^(4/3))/(8*(B*c + A*d)*(c + d*x)^(8/3)) + (A*B*((3*(A^2 
*d + B^2*c*x)*(A^2 - B^2*x^2)^(1/3))/(5*(B^2*c^2 - A^2*d^2)*(c + d*x)^(5/3 
)) - (6*A^2*B^2*(-(((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))])/(5*(B*c - A*d)*(B^2*c^2 - A^2*d^2)*(A^2 - B^ 
2*x^2)^(2/3))))/(B*c + A*d)
 

Defintions of rubi rules used

rule 486

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

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]

\[\int \frac {\left (B x +A \right ) \left (-B^{2} x^{2}+A^{2}\right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {11}{3}}}d x\]

Input:

int((B*x+A)*(-B^2*x^2+A^2)^(1/3)/(d*x+c)^(11/3),x)
 

Output:

int((B*x+A)*(-B^2*x^2+A^2)^(1/3)/(d*x+c)^(11/3),x)
 

Fricas [F]

\[ \int \frac {(A+B x) \sqrt [3]{A^2-B^2 x^2}}{(c+d x)^{11/3}} \, dx=\int { \frac {{\left (-B^{2} x^{2} + A^{2}\right )}^{\frac {1}{3}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {11}{3}}} \,d x } \] Input:

integrate((B*x+A)*(-B^2*x^2+A^2)^(1/3)/(d*x+c)^(11/3),x, algorithm="fricas 
")
 

Output:

integral((-B^2*x^2 + A^2)^(1/3)*(B*x + A)*(d*x + c)^(1/3)/(d^4*x^4 + 4*c*d 
^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4), x)
 

Sympy [F]

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

integrate((B*x+A)*(-B**2*x**2+A**2)**(1/3)/(d*x+c)**(11/3),x)
 

Output:

Integral((-(-A + B*x)*(A + B*x))**(1/3)*(A + B*x)/(c + d*x)**(11/3), x)
 

Maxima [F]

\[ \int \frac {(A+B x) \sqrt [3]{A^2-B^2 x^2}}{(c+d x)^{11/3}} \, dx=\int { \frac {{\left (-B^{2} x^{2} + A^{2}\right )}^{\frac {1}{3}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {11}{3}}} \,d x } \] Input:

integrate((B*x+A)*(-B^2*x^2+A^2)^(1/3)/(d*x+c)^(11/3),x, algorithm="maxima 
")
 

Output:

integrate((-B^2*x^2 + A^2)^(1/3)*(B*x + A)/(d*x + c)^(11/3), x)
 

Giac [F]

\[ \int \frac {(A+B x) \sqrt [3]{A^2-B^2 x^2}}{(c+d x)^{11/3}} \, dx=\int { \frac {{\left (-B^{2} x^{2} + A^{2}\right )}^{\frac {1}{3}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {11}{3}}} \,d x } \] Input:

integrate((B*x+A)*(-B^2*x^2+A^2)^(1/3)/(d*x+c)^(11/3),x, algorithm="giac")
 

Output:

integrate((-B^2*x^2 + A^2)^(1/3)*(B*x + A)/(d*x + c)^(11/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \sqrt [3]{A^2-B^2 x^2}}{(c+d x)^{11/3}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (A^2-B^2\,x^2\right )}^{1/3}}{{\left (c+d\,x\right )}^{11/3}} \,d x \] Input:

int(((A + B*x)*(A^2 - B^2*x^2)^(1/3))/(c + d*x)^(11/3),x)
 

Output:

int(((A + B*x)*(A^2 - B^2*x^2)^(1/3))/(c + d*x)^(11/3), x)
 

Reduce [F]

\[ \int \frac {(A+B x) \sqrt [3]{A^2-B^2 x^2}}{(c+d x)^{11/3}} \, dx=\left (\int \frac {\left (-b^{2} x^{2}+a^{2}\right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {2}{3}} c^{3}+3 \left (d x +c \right )^{\frac {2}{3}} c^{2} d x +3 \left (d x +c \right )^{\frac {2}{3}} c \,d^{2} x^{2}+\left (d x +c \right )^{\frac {2}{3}} d^{3} x^{3}}d x \right ) a +\left (\int \frac {\left (-b^{2} x^{2}+a^{2}\right )^{\frac {1}{3}} x}{\left (d x +c \right )^{\frac {2}{3}} c^{3}+3 \left (d x +c \right )^{\frac {2}{3}} c^{2} d x +3 \left (d x +c \right )^{\frac {2}{3}} c \,d^{2} x^{2}+\left (d x +c \right )^{\frac {2}{3}} d^{3} x^{3}}d x \right ) b \] Input:

int((B*x+A)*(-B^2*x^2+A^2)^(1/3)/(d*x+c)^(11/3),x)
 

Output:

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