Integrand size = 33, antiderivative size = 187 \[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\frac {87 (b c-a d)^2 (c+d x)^{2/3} (b c+a d+2 b d x)^{2/3}}{112 d^4}-\frac {39 b (b c-a d) (c+d x)^{5/3} (b c+a d+2 b d x)^{2/3}}{70 d^4}+\frac {3 b^2 (c+d x)^{8/3} (b c+a d+2 b d x)^{2/3}}{20 d^4}+\frac {81 (b c-a d)^2 (c+d x)^{2/3} (b c+a d+2 b d x)^{2/3} \operatorname {Hypergeometric2F1}\left (1,\frac {4}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{112 d^4} \] Output:
87/112*(-a*d+b*c)^2*(d*x+c)^(2/3)*(2*b*d*x+a*d+b*c)^(2/3)/d^4-39/70*b*(-a* d+b*c)*(d*x+c)^(5/3)*(2*b*d*x+a*d+b*c)^(2/3)/d^4+3/20*b^2*(d*x+c)^(8/3)*(2 *b*d*x+a*d+b*c)^(2/3)/d^4+81/112*(-a*d+b*c)^2*(d*x+c)^(2/3)*(2*b*d*x+a*d+b *c)^(2/3)*hypergeom([1, 4/3],[5/3],2*b*(d*x+c)/(-a*d+b*c))/d^4
Time = 10.42 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\frac {3 (c+d x)^{2/3} (a d+b (c+2 d x))^{2/3} \left (2 (a+b x)^2+\frac {3 (b c-a d)^2 \left (2 \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {2}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )+3 \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {2}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )+\operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )\right )}{d^2 \left (\frac {a d+b (c+2 d x)}{-b c+a d}\right )^{2/3}}\right )}{40 d^2} \] Input:
Integrate[(a + b*x)^3/((c + d*x)^(1/3)*(b*c + a*d + 2*b*d*x)^(1/3)),x]
Output:
(3*(c + d*x)^(2/3)*(a*d + b*(c + 2*d*x))^(2/3)*(2*(a + b*x)^2 + (3*(b*c - a*d)^2*(2*Hypergeometric2F1[-5/3, 2/3, 5/3, (2*b*(c + d*x))/(b*c - a*d)] + 3*Hypergeometric2F1[-2/3, 2/3, 5/3, (2*b*(c + d*x))/(b*c - a*d)] + Hyperg eometric2F1[1/3, 2/3, 5/3, (2*b*(c + d*x))/(b*c - a*d)]))/(d^2*((a*d + b*( c + 2*d*x))/(-(b*c) + a*d))^(2/3))))/(40*d^2)
Time = 0.34 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {111, 27, 164, 80, 79}
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.
\(\displaystyle \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} \sqrt [3]{a d+b c+2 b d x}} \, dx\) |
\(\Big \downarrow \) 111 |
\(\displaystyle \frac {3 \int -\frac {2 b (b c-a d) (a+b x) (b c+3 a d+4 b d x)}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx}{20 b d^2}+\frac {3 (a+b x)^2 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{20 d^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{20 d^2}-\frac {3 (b c-a d) \int \frac {(a+b x) (b c+3 a d+4 b d x)}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx}{10 d^2}\) |
\(\Big \downarrow \) 164 |
\(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{20 d^2}-\frac {3 (b c-a d) \left (\frac {45 (b c-a d)^2 \int \frac {1}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx}{28 d}-\frac {3 (c+d x)^{2/3} (-39 a d+23 b c-16 b d x) (a d+b c+2 b d x)^{2/3}}{56 d^2}\right )}{10 d^2}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{20 d^2}-\frac {3 (b c-a d) \left (\frac {45 (b c-a d)^2 \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \int \frac {1}{\sqrt [3]{c+d x} \sqrt [3]{-\frac {b c+a d}{b c-a d}-\frac {2 b d x}{b c-a d}}}dx}{28 d \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (c+d x)^{2/3} (-39 a d+23 b c-16 b d x) (a d+b c+2 b d x)^{2/3}}{56 d^2}\right )}{10 d^2}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{20 d^2}-\frac {3 (b c-a d) \left (\frac {135 (c+d x)^{2/3} (b c-a d)^2 \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{56 d^2 \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (c+d x)^{2/3} (-39 a d+23 b c-16 b d x) (a d+b c+2 b d x)^{2/3}}{56 d^2}\right )}{10 d^2}\) |
Input:
Int[(a + b*x)^3/((c + d*x)^(1/3)*(b*c + a*d + 2*b*d*x)^(1/3)),x]
Output:
(3*(a + b*x)^2*(c + d*x)^(2/3)*(b*c + a*d + 2*b*d*x)^(2/3))/(20*d^2) - (3* (b*c - a*d)*((-3*(c + d*x)^(2/3)*(23*b*c - 39*a*d - 16*b*d*x)*(b*c + a*d + 2*b*d*x)^(2/3))/(56*d^2) + (135*(b*c - a*d)^2*(c + d*x)^(2/3)*(-((b*c + a *d + 2*b*d*x)/(b*c - a*d)))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, (2*b*(c + d*x))/(b*c - a*d)])/(56*d^2*(b*c + a*d + 2*b*d*x)^(1/3))))/(10*d^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
\[\int \frac {\left (b x +a \right )^{3}}{\left (x d +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}}}d x\]
Input:
int((b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x)
Output:
int((b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x)
\[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{3}}{{\left (2 \, b d x + b c + a d\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x, algorithm="fr icas")
Output:
integral((b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*(2*b*d*x + b*c + a*d)^( 2/3)*(d*x + c)^(2/3)/(2*b*d^2*x^2 + b*c^2 + a*c*d + (3*b*c*d + a*d^2)*x), x)
\[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int \frac {\left (a + b x\right )^{3}}{\sqrt [3]{c + d x} \sqrt [3]{a d + b c + 2 b d x}}\, dx \] Input:
integrate((b*x+a)**3/(d*x+c)**(1/3)/(2*b*d*x+a*d+b*c)**(1/3),x)
Output:
Integral((a + b*x)**3/((c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)), x)
\[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{3}}{{\left (2 \, b d x + b c + a d\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x, algorithm="ma xima")
Output:
integrate((b*x + a)^3/((2*b*d*x + b*c + a*d)^(1/3)*(d*x + c)^(1/3)), x)
\[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{3}}{{\left (2 \, b d x + b c + a d\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x, algorithm="gi ac")
Output:
integrate((b*x + a)^3/((2*b*d*x + b*c + a*d)^(1/3)*(d*x + c)^(1/3)), x)
Timed out. \[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int \frac {{\left (a+b\,x\right )}^3}{{\left (c+d\,x\right )}^{1/3}\,{\left (a\,d+b\,c+2\,b\,d\,x\right )}^{1/3}} \,d x \] Input:
int((a + b*x)^3/((c + d*x)^(1/3)*(a*d + b*c + 2*b*d*x)^(1/3)),x)
Output:
int((a + b*x)^3/((c + d*x)^(1/3)*(a*d + b*c + 2*b*d*x)^(1/3)), x)
\[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\left (\int \frac {x^{3}}{\left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}}}d x \right ) b^{3}+3 \left (\int \frac {x^{2}}{\left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}}}d x \right ) a \,b^{2}+3 \left (\int \frac {x}{\left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}}}d x \right ) a^{2} b +\left (\int \frac {1}{\left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}}}d x \right ) a^{3} \] Input:
int((b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x)
Output:
int(x**3/((c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)),x)*b**3 + 3*int(x **2/((c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)),x)*a*b**2 + 3*int(x/(( c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)),x)*a**2*b + int(1/((c + d*x) **(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)),x)*a**3