Integrand size = 33, antiderivative size = 1388 \[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\frac {3 (a+b x)^2 (c+d x)^{2/3}}{14 d^2 \sqrt [3]{b c+a d+2 b d x}}+\frac {9 (b c-a d) (c+d x)^{2/3} (b c-7 a d-6 b d x)}{112 d^4 \sqrt [3]{b c+a d+2 b d x}}+\frac {81 (b c-a d)^2 \sqrt [3]{(c+d x) (b c+a d+2 b d x)} \sqrt {d^2 (3 b c+a d+4 b d x)^2} \sqrt {\left (d (3 b c+a d)+4 b d^2 x\right )^2}}{112 b^{2/3} d^6 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x} (3 b c+a d+4 b d x) \left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )}-\frac {81 \sqrt [4]{3} \sqrt {2-\sqrt {3}} (b c-a d)^{8/3} \sqrt [3]{(c+d x) (b c+a d+2 b d x)} \sqrt {\left (d (3 b c+a d)+4 b d^2 x\right )^2} \left ((b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right ) \sqrt {\frac {(b c-a d)^{4/3}-2 \sqrt [3]{b} (b c-a d)^{2/3} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}+4 b^{2/3} ((c+d x) (a d+b (c+2 d x)))^{2/3}}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}}{\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}}\right )|-7-4 \sqrt {3}\right )}{224 b^{2/3} d^4 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x} (3 b c+a d+4 b d x) \sqrt {d^2 (3 b c+a d+4 b d x)^2} \sqrt {\frac {(b c-a d)^{2/3} \left ((b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )^2}}}+\frac {27\ 3^{3/4} (b c-a d)^{8/3} \sqrt [3]{(c+d x) (b c+a d+2 b d x)} \sqrt {\left (d (3 b c+a d)+4 b d^2 x\right )^2} \left ((b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right ) \sqrt {\frac {(b c-a d)^{4/3}-2 \sqrt [3]{b} (b c-a d)^{2/3} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}+4 b^{2/3} ((c+d x) (a d+b (c+2 d x)))^{2/3}}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}}{\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}}\right ),-7-4 \sqrt {3}\right )}{56 \sqrt {2} b^{2/3} d^4 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x} (3 b c+a d+4 b d x) \sqrt {d^2 (3 b c+a d+4 b d x)^2} \sqrt {\frac {(b c-a d)^{2/3} \left ((b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )^2}}} \]
3/14*(b*x+a)^2*(d*x+c)^(2/3)/d^2/(2*b*d*x+a*d+b*c)^(1/3)+9/112*(-a*d+b*c)* (d*x+c)^(2/3)*(-6*b*d*x-7*a*d+b*c)/d^4/(2*b*d*x+a*d+b*c)^(1/3)+81/112*(-a* d+b*c)^2*((d*x+c)*(2*b*d*x+a*d+b*c))^(1/3)*(d^2*(4*b*d*x+a*d+3*b*c)^2)^(1/ 2)*((d*(a*d+3*b*c)+4*b*d^2*x)^2)^(1/2)/b^(2/3)/d^6/(d*x+c)^(1/3)/(2*b*d*x+ a*d+b*c)^(1/3)/(4*b*d*x+a*d+3*b*c)/(2*b^(1/3)*((d*x+c)*(a*d+b*(2*d*x+c)))^ (1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2)))+27/112*3^(3/4)*(-a*d+b*c)^(8/3)*((d*x+ c)*(2*b*d*x+a*d+b*c))^(1/3)*((-a*d+b*c)^(2/3)+2*b^(1/3)*((d*x+c)*(a*d+b*(2 *d*x+c)))^(1/3))*EllipticF((2*b^(1/3)*((d*x+c)*(a*d+b*(2*d*x+c)))^(1/3)+(- a*d+b*c)^(2/3)*(1-3^(1/2)))/(2*b^(1/3)*((d*x+c)*(a*d+b*(2*d*x+c)))^(1/3)+( -a*d+b*c)^(2/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*((d*(a*d+3*b*c)+4*b*d^2*x)^2)^ (1/2)*(((-a*d+b*c)^(4/3)-2*b^(1/3)*(-a*d+b*c)^(2/3)*((d*x+c)*(a*d+b*(2*d*x +c)))^(1/3)+4*b^(2/3)*((d*x+c)*(a*d+b*(2*d*x+c)))^(2/3))/(2*b^(1/3)*((d*x+ c)*(a*d+b*(2*d*x+c)))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2)))^2)^(1/2)/b^(2/3) /d^4/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3)/(4*b*d*x+a*d+3*b*c)*2^(1/2)/(d^ 2*(4*b*d*x+a*d+3*b*c)^2)^(1/2)/((-a*d+b*c)^(2/3)*((-a*d+b*c)^(2/3)+2*b^(1/ 3)*((d*x+c)*(a*d+b*(2*d*x+c)))^(1/3))/(2*b^(1/3)*((d*x+c)*(a*d+b*(2*d*x+c) ))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2)))^2)^(1/2)-81/224*3^(1/4)*(-a*d+b*c)^ (8/3)*((d*x+c)*(2*b*d*x+a*d+b*c))^(1/3)*((-a*d+b*c)^(2/3)+2*b^(1/3)*((d*x+ c)*(a*d+b*(2*d*x+c)))^(1/3))*EllipticE((2*b^(1/3)*((d*x+c)*(a*d+b*(2*d*x+c )))^(1/3)+(-a*d+b*c)^(2/3)*(1-3^(1/2)))/(2*b^(1/3)*((d*x+c)*(a*d+b*(2*d...
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.20 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.10 \[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\frac {3 (c+d x)^{2/3} \left (29 a^2 d^2+2 a b d (-12 c+17 d x)+b^2 \left (3 c^2-18 c d x+8 d^2 x^2\right )+27 (b c-a d)^2 \sqrt [3]{\frac {a d+b (c+2 d x)}{-b c+a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )\right )}{112 d^4 \sqrt [3]{a d+b (c+2 d x)}} \]
(3*(c + d*x)^(2/3)*(29*a^2*d^2 + 2*a*b*d*(-12*c + 17*d*x) + b^2*(3*c^2 - 1 8*c*d*x + 8*d^2*x^2) + 27*(b*c - a*d)^2*((a*d + b*(c + 2*d*x))/(-(b*c) + a *d))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, (2*b*(c + d*x))/(b*c - a*d)])) /(112*d^4*(a*d + b*(c + 2*d*x))^(1/3))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.35 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.14, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {111, 27, 163, 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.
| \(\displaystyle \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (a d+b c+2 b d x)^{4/3}} \, dx\) |
| \(\Big \downarrow \) 111 |
| \(\displaystyle \frac {3 \int -\frac {2 b (b c-a d) (a+b x) (b c+2 a d+3 b d x)}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}}dx}{14 b d^2}+\frac {3 (a+b x)^2 (c+d x)^{2/3}}{14 d^2 \sqrt [3]{a d+b c+2 b d x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3}}{14 d^2 \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (b c-a d) \int \frac {(a+b x) (b c+2 a d+3 b d x)}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}}dx}{7 d^2}\) |
| \(\Big \downarrow \) 163 |
| \(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3}}{14 d^2 \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (b c-a d) \left (-\frac {9 (b c-a d) \int \frac {1}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx}{8 d}-\frac {3 (c+d x)^{2/3} (-7 a d+b c-6 b d x)}{16 d^2 \sqrt [3]{a d+b c+2 b d x}}\right )}{7 d^2}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3}}{14 d^2 \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (b c-a d) \left (-\frac {9 (b c-a d) \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \int \frac {1}{\sqrt [3]{c+d x} \sqrt [3]{-\frac {b c+a d}{b c-a d}-\frac {2 b d x}{b c-a d}}}dx}{8 d \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (c+d x)^{2/3} (-7 a d+b c-6 b d x)}{16 d^2 \sqrt [3]{a d+b c+2 b d x}}\right )}{7 d^2}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3}}{14 d^2 \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (b c-a d) \left (-\frac {27 (c+d x)^{2/3} (b c-a d) \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{16 d^2 \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (c+d x)^{2/3} (-7 a d+b c-6 b d x)}{16 d^2 \sqrt [3]{a d+b c+2 b d x}}\right )}{7 d^2}\) |
(3*(a + b*x)^2*(c + d*x)^(2/3))/(14*d^2*(b*c + a*d + 2*b*d*x)^(1/3)) - (3* (b*c - a*d)*((-3*(c + d*x)^(2/3)*(b*c - 7*a*d - 6*b*d*x))/(16*d^2*(b*c + a *d + 2*b*d*x)^(1/3)) - (27*(b*c - a*d)*(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)])/(16*d^2*(b*c + a*d + 2*b*d*x)^(1/3))))/(7*d^2)
3.31.29.3.1 Defintions of rubi rules used
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^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(b*c - a*d)* (m + 1)*x)/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), 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*(b*c - a*d)*(m + 1)*(m + n + 3)) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((GeQ[m, -2] && LtQ[m, - 1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]
\[\int \frac {\left (b x +a \right )^{3}}{\left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {4}{3}}}d x\]
\[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{3}}{{\left (2 \, b d x + b c + a d\right )}^{\frac {4}{3}} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \]
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)/(4*b^2*d^3*x^3 + b^2*c^3 + 2*a*b*c^2*d + a^2*c*d^2 + 4*(2*b^2*c*d^2 + a*b*d^3)*x^2 + (5*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x), x)
\[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\int \frac {\left (a + b x\right )^{3}}{\sqrt [3]{c + d x} \left (a d + b c + 2 b d x\right )^{\frac {4}{3}}}\, dx \]
\[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{3}}{{\left (2 \, b d x + b c + a d\right )}^{\frac {4}{3}} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{3}}{{\left (2 \, b d x + b c + a d\right )}^{\frac {4}{3}} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\int \frac {{\left (a+b\,x\right )}^3}{{\left (c+d\,x\right )}^{1/3}\,{\left (a\,d+b\,c+2\,b\,d\,x\right )}^{4/3}} \,d x \]