3.300 \(\int f^{a+b \sqrt [3]{c+d x}} \, dx\)
Optimal. Leaf size=100 \[ -\frac{6 \sqrt [3]{c+d x} f^{a+b \sqrt [3]{c+d x}}}{b^2 d \log ^2(f)}+\frac{6 f^{a+b \sqrt [3]{c+d x}}}{b^3 d \log ^3(f)}+\frac{3 (c+d x)^{2/3} f^{a+b \sqrt [3]{c+d x}}}{b d \log (f)} \]
[Out]
(6*f^(a + b*(c + d*x)^(1/3)))/(b^3*d*Log[f]^3) - (6*f^(a + b*(c + d*x)^(1/3))*(c + d*x)^(1/3))/(b^2*d*Log[f]^2
) + (3*f^(a + b*(c + d*x)^(1/3))*(c + d*x)^(2/3))/(b*d*Log[f])
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Rubi [A] time = 0.0640819, antiderivative size = 100, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{2207, 2176, 2194} \[ -\frac{6 \sqrt [3]{c+d x} f^{a+b \sqrt [3]{c+d x}}}{b^2 d \log ^2(f)}+\frac{6 f^{a+b \sqrt [3]{c+d x}}}{b^3 d \log ^3(f)}+\frac{3 (c+d x)^{2/3} f^{a+b \sqrt [3]{c+d x}}}{b d \log (f)} \]
Antiderivative was successfully verified.
[In]
Int[f^(a + b*(c + d*x)^(1/3)),x]
[Out]
(6*f^(a + b*(c + d*x)^(1/3)))/(b^3*d*Log[f]^3) - (6*f^(a + b*(c + d*x)^(1/3))*(c + d*x)^(1/3))/(b^2*d*Log[f]^2
) + (3*f^(a + b*(c + d*x)^(1/3))*(c + d*x)^(2/3))/(b*d*Log[f])
Rule 2207
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> With[{k = Denominator[n]}, Dist[k/d, Subst[In
t[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] &&
!IntegerQ[n]
Rule 2176
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] - Dist[(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] && !$UseGamma === True
Rule 2194
Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]
Rubi steps
\begin{align*} \int f^{a+b \sqrt [3]{c+d x}} \, dx &=\frac{3 \operatorname{Subst}\left (\int f^{a+b x} x^2 \, dx,x,\sqrt [3]{c+d x}\right )}{d}\\ &=\frac{3 f^{a+b \sqrt [3]{c+d x}} (c+d x)^{2/3}}{b d \log (f)}-\frac{6 \operatorname{Subst}\left (\int f^{a+b x} x \, dx,x,\sqrt [3]{c+d x}\right )}{b d \log (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)}+\frac{6 \operatorname{Subst}\left (\int f^{a+b x} \, dx,x,\sqrt [3]{c+d x}\right )}{b^2 d \log ^2(f)}\\ &=\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)}\\ \end{align*}
Mathematica [A] time = 0.0412831, size = 60, normalized size = 0.6 \[ \frac{3 f^{a+b \sqrt [3]{c+d x}} \left (b^2 \log ^2(f) (c+d x)^{2/3}-2 b \log (f) \sqrt [3]{c+d x}+2\right )}{b^3 d \log ^3(f)} \]
Antiderivative was successfully verified.
[In]
Integrate[f^(a + b*(c + d*x)^(1/3)),x]
[Out]
(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)
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Maple [F] time = 0.005, size = 0, normalized size = 0. \begin{align*} \int{f}^{a+b\sqrt [3]{dx+c}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(f^(a+b*(d*x+c)^(1/3)),x)
[Out]
int(f^(a+b*(d*x+c)^(1/3)),x)
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Maxima [A] time = 1.01445, size = 84, normalized size = 0.84 \begin{align*} \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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(f^(a+b*(d*x+c)^(1/3)),x, algorithm="maxima")
[Out]
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)
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Fricas [A] time = 1.57271, size = 167, normalized size = 1.67 \begin{align*} \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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(f^(a+b*(d*x+c)^(1/3)),x, algorithm="fricas")
[Out]
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)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{a + b \sqrt [3]{c + d x}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(f**(a+b*(d*x+c)**(1/3)),x)
[Out]
Integral(f**(a + b*(c + d*x)**(1/3)), x)
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Giac [B] time = 1.52837, size = 1804, normalized size = 18.04 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(f^(a+b*(d*x+c)^(1/3)),x, algorithm="giac")
[Out]
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^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*lo
g(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*log(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*l
og(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))*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(a
bs(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))*sin(-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))*e^((d*x + c)^(1/3)*b*log(abs(f)) + a*log(abs(f))) + ((pi^2*(d*x + c)^(2/3)*b^2*i*sgn(f) - pi^2*
(d*x + c)^(2/3)*b^2*i + 2*(d*x + c)^(2/3)*b^2*i*log(abs(f))^2 - 2*pi*(d*x + c)^(2/3)*b^2*log(abs(f))*sgn(f) +
2*pi*(d*x + c)^(2/3)*b^2*log(abs(f)) - 4*(d*x + c)^(1/3)*b*i*log(abs(f)) + 2*pi*(d*x + c)^(1/3)*b*sgn(f) - 2*p
i*(d*x + c)^(1/3)*b + 4*i)*e^(1/2*(pi*(d*x + c)^(1/3)*b*(sgn(f) - 1) + pi*a*(sgn(f) - 1))*i)/(pi^3*b^3*i*sgn(f
) - 3*pi*b^3*i*log(abs(f))^2*sgn(f) - pi^3*b^3*i + 3*pi*b^3*i*log(abs(f))^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^2*(d*x + c)^(2/3)*b^2*i*sgn(f) - pi^2*(d*x + c)^(2/3)*b^2*
i + 2*(d*x + c)^(2/3)*b^2*i*log(abs(f))^2 + 2*pi*(d*x + c)^(2/3)*b^2*log(abs(f))*sgn(f) - 2*pi*(d*x + c)^(2/3)
*b^2*log(abs(f)) - 4*(d*x + c)^(1/3)*b*i*log(abs(f)) - 2*pi*(d*x + c)^(1/3)*b*sgn(f) + 2*pi*(d*x + c)^(1/3)*b
+ 4*i)*e^(-1/2*(pi*(d*x + c)^(1/3)*b*(sgn(f) - 1) + pi*a*(sgn(f) - 1))*i)/(pi^3*b^3*i*sgn(f) - 3*pi*b^3*i*log(
abs(f))^2*sgn(f) - pi^3*b^3*i + 3*pi*b^3*i*log(abs(f))^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))*e^((d*x + c)^(1/3)*b*log(abs(f)) + a*log(abs(f)))/i)/d