\(\int t \sqrt [4]{1+t} \, dt\) [15]

Optimal result

Integrand size = 9, antiderivative size = 23 \[ \int t \sqrt [4]{1+t} \, dt=-\frac {4}{5} (1+t)^{5/4}+\frac {4}{9} (1+t)^{9/4} \]

[Out]

-4/5*(1+t)^(5/4)+4/9*(1+t)^(9/4)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {45} \[ \int t \sqrt [4]{1+t} \, dt=\frac {4}{9} (t+1)^{9/4}-\frac {4}{5} (t+1)^{5/4} \]

[In]

Int[t*(1 + t)^(1/4),t]

[Out]

(-4*(1 + t)^(5/4))/5 + (4*(1 + t)^(9/4))/9

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (-\sqrt [4]{1+t}+(1+t)^{5/4}\right ) \, dt \\ & = -\frac {4}{5} (1+t)^{5/4}+\frac {4}{9} (1+t)^{9/4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int t \sqrt [4]{1+t} \, dt=\frac {4}{45} (1+t)^{5/4} (-9+5 (1+t)) \]

[In]

Integrate[t*(1 + t)^(1/4),t]

[Out]

(4*(1 + t)^(5/4)*(-9 + 5*(1 + t)))/45

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57

method	result	size
gosper	\(\frac {4 \left (1+t \right )^{\frac {5}{4}} \left (5 t -4\right )}{45}\)	\(13\)
meijerg	\(\frac {t^{2} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{4},2;3;-t \right )}{2}\)	\(15\)
derivativedivides	\(-\frac {4 \left (1+t \right )^{\frac {5}{4}}}{5}+\frac {4 \left (1+t \right )^{\frac {9}{4}}}{9}\)	\(16\)
default	\(-\frac {4 \left (1+t \right )^{\frac {5}{4}}}{5}+\frac {4 \left (1+t \right )^{\frac {9}{4}}}{9}\)	\(16\)
risch	\(\frac {4 \left (1+t \right )^{\frac {1}{4}} \left (5 t^{2}+t -4\right )}{45}\)	\(16\)
trager	\(\left (\frac {4}{9} t^{2}+\frac {4}{45} t -\frac {16}{45}\right ) \left (1+t \right )^{\frac {1}{4}}\)	\(17\)

[In]

int(t*(1+t)^(1/4),t,method=_RETURNVERBOSE)

[Out]

4/45*(1+t)^(5/4)*(5*t-4)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int t \sqrt [4]{1+t} \, dt=\frac {4}{45} \, {\left (5 \, t^{2} + t - 4\right )} {\left (t + 1\right )}^{\frac {1}{4}} \]

[In]

integrate(t*(1+t)^(1/4),t, algorithm="fricas")

[Out]

4/45*(5*t^2 + t - 4)*(t + 1)^(1/4)

Sympy [A] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int t \sqrt [4]{1+t} \, dt=\frac {4 t^{2} \sqrt [4]{t + 1}}{9} + \frac {4 t \sqrt [4]{t + 1}}{45} - \frac {16 \sqrt [4]{t + 1}}{45} \]

[In]

integrate(t*(1+t)**(1/4),t)

[Out]

4*t**2*(t + 1)**(1/4)/9 + 4*t*(t + 1)**(1/4)/45 - 16*(t + 1)**(1/4)/45

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int t \sqrt [4]{1+t} \, dt=\frac {4}{9} \, {\left (t + 1\right )}^{\frac {9}{4}} - \frac {4}{5} \, {\left (t + 1\right )}^{\frac {5}{4}} \]

[In]

integrate(t*(1+t)^(1/4),t, algorithm="maxima")

[Out]

4/9*(t + 1)^(9/4) - 4/5*(t + 1)^(5/4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int t \sqrt [4]{1+t} \, dt=\frac {4}{9} \, {\left (t + 1\right )}^{\frac {9}{4}} - \frac {4}{5} \, {\left (t + 1\right )}^{\frac {5}{4}} \]

[In]

integrate(t*(1+t)^(1/4),t, algorithm="giac")

[Out]

4/9*(t + 1)^(9/4) - 4/5*(t + 1)^(5/4)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.52 \[ \int t \sqrt [4]{1+t} \, dt=\frac {4\,\left (5\,t-4\right )\,{\left (t+1\right )}^{5/4}}{45} \]

[In]

int(t*(t + 1)^(1/4),t)

[Out]

(4*(5*t - 4)*(t + 1)^(5/4))/45