Integrand size = 15, antiderivative size = 100 \[ \int f^{a+b \sqrt [3]{c+d x}} \, dx=\frac {6 f^{a+b \sqrt [3]{c+d x}}}{b^3 d \log ^3(f)}-\frac {6 f^{a+b \sqrt [3]{c+d x}} \sqrt [3]{c+d x}}{b^2 d \log ^2(f)}+\frac {3 f^{a+b \sqrt [3]{c+d x}} (c+d x)^{2/3}}{b d \log (f)} \] Output:
6*f^(a+b*(d*x+c)^(1/3))/b^3/d/ln(f)^3-6*f^(a+b*(d*x+c)^(1/3))*(d*x+c)^(1/3 )/b^2/d/ln(f)^2+3*f^(a+b*(d*x+c)^(1/3))*(d*x+c)^(2/3)/b/d/ln(f)
Time = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.60 \[ \int f^{a+b \sqrt [3]{c+d x}} \, dx=\frac {3 f^{a+b \sqrt [3]{c+d x}} \left (2-2 b \sqrt [3]{c+d x} \log (f)+b^2 (c+d x)^{2/3} \log ^2(f)\right )}{b^3 d \log ^3(f)} \] Input:
Integrate[f^(a + b*(c + d*x)^(1/3)),x]
Output:
(3*f^(a + b*(c + d*x)^(1/3))*(2 - 2*b*(c + d*x)^(1/3)*Log[f] + b^2*(c + d* x)^(2/3)*Log[f]^2))/(b^3*d*Log[f]^3)
Time = 0.45 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2636, 2607, 2607, 2624}
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.
\(\displaystyle \int f^{a+b \sqrt [3]{c+d x}} \, dx\) |
\(\Big \downarrow \) 2636 |
\(\displaystyle \frac {3 \int f^{a+b \sqrt [3]{c+d x}} (c+d x)^{2/3}d\sqrt [3]{c+d x}}{d}\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle \frac {3 \left (\frac {(c+d x)^{2/3} f^{a+b \sqrt [3]{c+d x}}}{b \log (f)}-\frac {2 \int f^{a+b \sqrt [3]{c+d x}} \sqrt [3]{c+d x}d\sqrt [3]{c+d x}}{b \log (f)}\right )}{d}\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle \frac {3 \left (\frac {(c+d x)^{2/3} f^{a+b \sqrt [3]{c+d x}}}{b \log (f)}-\frac {2 \left (\frac {\sqrt [3]{c+d x} f^{a+b \sqrt [3]{c+d x}}}{b \log (f)}-\frac {\int f^{a+b \sqrt [3]{c+d x}}d\sqrt [3]{c+d x}}{b \log (f)}\right )}{b \log (f)}\right )}{d}\) |
\(\Big \downarrow \) 2624 |
\(\displaystyle \frac {3 \left (\frac {(c+d x)^{2/3} f^{a+b \sqrt [3]{c+d x}}}{b \log (f)}-\frac {2 \left (\frac {\sqrt [3]{c+d x} f^{a+b \sqrt [3]{c+d x}}}{b \log (f)}-\frac {f^{a+b \sqrt [3]{c+d x}}}{b^2 \log ^2(f)}\right )}{b \log (f)}\right )}{d}\) |
Input:
Int[f^(a + b*(c + d*x)^(1/3)),x]
Output:
(3*((f^(a + b*(c + d*x)^(1/3))*(c + d*x)^(2/3))/(b*Log[f]) - (2*(-(f^(a + b*(c + d*x)^(1/3))/(b^2*Log[f]^2)) + (f^(a + b*(c + d*x)^(1/3))*(c + d*x)^ (1/3))/(b*Log[f])))/(b*Log[f])))/d
Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m _.), x_Symbol] :> Simp[(c + d*x)^m*((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Simp[d*(m/(f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x)))^ n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2* m] && !TrueQ[$UseGamma]
Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; FreeQ[{F, n}, x] && LinearQ[v, x]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> With[{k = Denominator[n]}, Simp[k/d Subst[Int[x^(k - 1)*F^(a + b*x^(k*n)), x], x, ( c + d*x)^(1/k)], x]] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n] && !Int egerQ[n]
\[\int f^{a +b \left (d x +c \right )^{\frac {1}{3}}}d x\]
Input:
int(f^(a+b*(d*x+c)^(1/3)),x)
Output:
int(f^(a+b*(d*x+c)^(1/3)),x)
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.58 \[ \int f^{a+b \sqrt [3]{c+d x}} \, dx=\frac {3 \, {\left ({\left (d x + c\right )}^{\frac {2}{3}} b^{2} \log \left (f\right )^{2} - 2 \, {\left (d x + c\right )}^{\frac {1}{3}} b \log \left (f\right ) + 2\right )} e^{\left ({\left (d x + c\right )}^{\frac {1}{3}} b \log \left (f\right ) + a \log \left (f\right )\right )}}{b^{3} d \log \left (f\right )^{3}} \] Input:
integrate(f^(a+b*(d*x+c)^(1/3)),x, algorithm="fricas")
Output:
3*((d*x + c)^(2/3)*b^2*log(f)^2 - 2*(d*x + c)^(1/3)*b*log(f) + 2)*e^((d*x + c)^(1/3)*b*log(f) + a*log(f))/(b^3*d*log(f)^3)
\[ \int f^{a+b \sqrt [3]{c+d x}} \, dx=\int f^{a + b \sqrt [3]{c + d x}}\, dx \] Input:
integrate(f**(a+b*(d*x+c)**(1/3)),x)
Output:
Integral(f**(a + b*(c + d*x)**(1/3)), x)
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.62 \[ \int f^{a+b \sqrt [3]{c+d x}} \, dx=\frac {3 \, {\left ({\left (d x + c\right )}^{\frac {2}{3}} b^{2} f^{a} \log \left (f\right )^{2} - 2 \, {\left (d x + c\right )}^{\frac {1}{3}} b f^{a} \log \left (f\right ) + 2 \, f^{a}\right )} f^{{\left (d x + c\right )}^{\frac {1}{3}} b}}{b^{3} d \log \left (f\right )^{3}} \] Input:
integrate(f^(a+b*(d*x+c)^(1/3)),x, algorithm="maxima")
Output:
3*((d*x + c)^(2/3)*b^2*f^a*log(f)^2 - 2*(d*x + c)^(1/3)*b*f^a*log(f) + 2*f ^a)*f^((d*x + c)^(1/3)*b)/(b^3*d*log(f)^3)
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 1338, normalized size of antiderivative = 13.38 \[ \int f^{a+b \sqrt [3]{c+d x}} \, dx=\text {Too large to display} \] Input:
integrate(f^(a+b*(d*x+c)^(1/3)),x, algorithm="giac")
Output:
3*((((3*pi^2*b^3*log(abs(f))*sgn(f) - 3*pi^2*b^3*log(abs(f)) + 2*b^3*log(a bs(f))^3)*(pi^2*(d*x + c)^(2/3)*b^2*sgn(f) - pi^2*(d*x + c)^(2/3)*b^2 + 2* (d*x + c)^(2/3)*b^2*log(abs(f))^2 - 4*(d*x + c)^(1/3)*b*log(abs(f)) + 4)/( (pi^3*b^3*sgn(f) - 3*pi*b^3*log(abs(f))^2*sgn(f) - pi^3*b^3 + 3*pi*b^3*log (abs(f))^2)^2 + (3*pi^2*b^3*log(abs(f))*sgn(f) - 3*pi^2*b^3*log(abs(f)) + 2*b^3*log(abs(f))^3)^2) - 2*(pi^3*b^3*sgn(f) - 3*pi*b^3*log(abs(f))^2*sgn( f) - pi^3*b^3 + 3*pi*b^3*log(abs(f))^2)*(pi*(d*x + c)^(2/3)*b^2*log(abs(f) )*sgn(f) - pi*(d*x + c)^(2/3)*b^2*log(abs(f)) - pi*(d*x + c)^(1/3)*b*sgn(f ) + pi*(d*x + c)^(1/3)*b)/((pi^3*b^3*sgn(f) - 3*pi*b^3*log(abs(f))^2*sgn(f ) - pi^3*b^3 + 3*pi*b^3*log(abs(f))^2)^2 + (3*pi^2*b^3*log(abs(f))*sgn(f) - 3*pi^2*b^3*log(abs(f)) + 2*b^3*log(abs(f))^3)^2))*cos(-1/2*pi*(d*x + c)^ (1/3)*b*sgn(f) - 1/2*pi*a*sgn(f) + 1/2*pi*(d*x + c)^(1/3)*b + 1/2*pi*a) + ((pi^3*b^3*sgn(f) - 3*pi*b^3*log(abs(f))^2*sgn(f) - pi^3*b^3 + 3*pi*b^3*lo g(abs(f))^2)*(pi^2*(d*x + c)^(2/3)*b^2*sgn(f) - pi^2*(d*x + c)^(2/3)*b^2 + 2*(d*x + c)^(2/3)*b^2*log(abs(f))^2 - 4*(d*x + c)^(1/3)*b*log(abs(f)) + 4 )/((pi^3*b^3*sgn(f) - 3*pi*b^3*log(abs(f))^2*sgn(f) - pi^3*b^3 + 3*pi*b^3* log(abs(f))^2)^2 + (3*pi^2*b^3*log(abs(f))*sgn(f) - 3*pi^2*b^3*log(abs(f)) + 2*b^3*log(abs(f))^3)^2) + 2*(3*pi^2*b^3*log(abs(f))*sgn(f) - 3*pi^2*b^3 *log(abs(f)) + 2*b^3*log(abs(f))^3)*(pi*(d*x + c)^(2/3)*b^2*log(abs(f))*sg n(f) - pi*(d*x + c)^(2/3)*b^2*log(abs(f)) - pi*(d*x + c)^(1/3)*b*sgn(f)...
Time = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.54 \[ \int f^{a+b \sqrt [3]{c+d x}} \, dx=\frac {f^{a+b\,{\left (c+d\,x\right )}^{1/3}}\,\left (3\,b^2\,{\ln \left (f\right )}^2\,{\left (c+d\,x\right )}^{2/3}-6\,b\,\ln \left (f\right )\,{\left (c+d\,x\right )}^{1/3}+6\right )}{b^3\,d\,{\ln \left (f\right )}^3} \] Input:
int(f^(a + b*(c + d*x)^(1/3)),x)
Output:
(f^(a + b*(c + d*x)^(1/3))*(3*b^2*log(f)^2*(c + d*x)^(2/3) - 6*b*log(f)*(c + d*x)^(1/3) + 6))/(b^3*d*log(f)^3)
Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.54 \[ \int f^{a+b \sqrt [3]{c+d x}} \, dx=\frac {3 f^{\left (d x +c \right )^{\frac {1}{3}} b +a} \left (\left (d x +c \right )^{\frac {2}{3}} \mathrm {log}\left (f \right )^{2} b^{2}-2 \left (d x +c \right )^{\frac {1}{3}} \mathrm {log}\left (f \right ) b +2\right )}{\mathrm {log}\left (f \right )^{3} b^{3} d} \] Input:
int(f^(a+b*(d*x+c)^(1/3)),x)
Output:
(3*f**((c + d*x)**(1/3)*b + a)*((c + d*x)**(2/3)*log(f)**2*b**2 - 2*(c + d *x)**(1/3)*log(f)*b + 2))/(log(f)**3*b**3*d)