\(\int x (a+b \tan (c+d \sqrt {x})) \, dx\) [27]

Optimal result

Integrand size = 16, antiderivative size = 135 \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {2 b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {3 i b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {3 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {3 i b \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4} \]

[Out]

1/2*a*x^2+1/2*I*b*x^2-2*b*x^(3/2)*ln(1+exp(2*I*(c+d*x^(1/2))))/d+3*I*b*x*polylog(2,-exp(2*I*(c+d*x^(1/2))))/d^
2-3/2*I*b*polylog(4,-exp(2*I*(c+d*x^(1/2))))/d^4-3*b*polylog(3,-exp(2*I*(c+d*x^(1/2))))*x^(1/2)/d^3

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {14, 3832, 3800, 2221, 2611, 6744, 2320, 6724} \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^2}{2}-\frac {3 i b \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}-\frac {3 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {3 i b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {1}{2} i b x^2 \]

[In]

Int[x*(a + b*Tan[c + d*Sqrt[x]]),x]

[Out]

(a*x^2)/2 + (I/2)*b*x^2 - (2*b*x^(3/2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))])/d + ((3*I)*b*x*PolyLog[2, -E^((2*I)
*(c + d*Sqrt[x]))])/d^2 - (3*b*Sqrt[x]*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 - (((3*I)/2)*b*PolyLog[4, -
E^((2*I)*(c + d*Sqrt[x]))])/d^4

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x
] - Dist[2*I, Int[(c + d*x)^m*(E^(2*I*(e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 3832

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Tan[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6744

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a x+b x \tan \left (c+d \sqrt {x}\right )\right ) \, dx \\ & = \frac {a x^2}{2}+b \int x \tan \left (c+d \sqrt {x}\right ) \, dx \\ & = \frac {a x^2}{2}+(2 b) \text {Subst}\left (\int x^3 \tan (c+d x) \, dx,x,\sqrt {x}\right ) \\ & = \frac {a x^2}{2}+\frac {1}{2} i b x^2-(4 i b) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x^3}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {2 b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {(6 b) \text {Subst}\left (\int x^2 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d} \\ & = \frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {2 b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {3 i b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(6 i b) \text {Subst}\left (\int x \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2} \\ & = \frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {2 b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {3 i b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {3 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(3 b) \text {Subst}\left (\int \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3} \\ & = \frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {2 b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {3 i b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {3 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {(3 i b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4} \\ & = \frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {2 b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {3 i b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {3 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {3 i b \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00 \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {2 b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {3 i b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {3 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {3 i b \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4} \]

[In]

Integrate[x*(a + b*Tan[c + d*Sqrt[x]]),x]

[Out]

(a*x^2)/2 + (I/2)*b*x^2 - (2*b*x^(3/2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))])/d + ((3*I)*b*x*PolyLog[2, -E^((2*I)
*(c + d*Sqrt[x]))])/d^2 - (3*b*Sqrt[x]*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 - (((3*I)/2)*b*PolyLog[4, -
E^((2*I)*(c + d*Sqrt[x]))])/d^4

Maple [F]

\[\int x \left (a +b \tan \left (c +d \sqrt {x}\right )\right )d x\]

[In]

int(x*(a+b*tan(c+d*x^(1/2))),x)

[Out]

int(x*(a+b*tan(c+d*x^(1/2))),x)

Fricas [F]

\[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \tan \left (d \sqrt {x} + c\right ) + a\right )} x \,d x } \]

[In]

integrate(x*(a+b*tan(c+d*x^(1/2))),x, algorithm="fricas")

[Out]

integral(b*x*tan(d*sqrt(x) + c) + a*x, x)

Sympy [F]

\[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\int x \left (a + b \tan {\left (c + d \sqrt {x} \right )}\right )\, dx \]

[In]

integrate(x*(a+b*tan(c+d*x**(1/2))),x)

[Out]

Integral(x*(a + b*tan(c + d*sqrt(x))), x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 359 vs. \(2 (102) = 204\).

Time = 0.39 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.66 \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {3 \, {\left (d \sqrt {x} + c\right )}^{4} a + 3 i \, {\left (d \sqrt {x} + c\right )}^{4} b - 12 \, {\left (d \sqrt {x} + c\right )}^{3} a c - 12 i \, {\left (d \sqrt {x} + c\right )}^{3} b c + 18 \, {\left (d \sqrt {x} + c\right )}^{2} a c^{2} + 18 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} - 12 \, {\left (d \sqrt {x} + c\right )} a c^{3} - 12 \, b c^{3} \log \left (\sec \left (d \sqrt {x} + c\right )\right ) + 4 \, {\left (-4 i \, {\left (d \sqrt {x} + c\right )}^{3} b + 9 i \, {\left (d \sqrt {x} + c\right )}^{2} b c - 9 i \, {\left (d \sqrt {x} + c\right )} b c^{2}\right )} \arctan \left (\sin \left (2 \, d \sqrt {x} + 2 \, c\right ), \cos \left (2 \, d \sqrt {x} + 2 \, c\right ) + 1\right ) + 6 \, {\left (4 i \, {\left (d \sqrt {x} + c\right )}^{2} b - 6 i \, {\left (d \sqrt {x} + c\right )} b c + 3 i \, b c^{2}\right )} {\rm Li}_2\left (-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}\right ) - 2 \, {\left (4 \, {\left (d \sqrt {x} + c\right )}^{3} b - 9 \, {\left (d \sqrt {x} + c\right )}^{2} b c + 9 \, {\left (d \sqrt {x} + c\right )} b c^{2}\right )} \log \left (\cos \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + \sin \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d \sqrt {x} + 2 \, c\right ) + 1\right ) - 12 i \, b {\rm Li}_{4}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}) - 6 \, {\left (4 \, {\left (d \sqrt {x} + c\right )} b - 3 \, b c\right )} {\rm Li}_{3}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )})}{6 \, d^{4}} \]

[In]

integrate(x*(a+b*tan(c+d*x^(1/2))),x, algorithm="maxima")

[Out]

1/6*(3*(d*sqrt(x) + c)^4*a + 3*I*(d*sqrt(x) + c)^4*b - 12*(d*sqrt(x) + c)^3*a*c - 12*I*(d*sqrt(x) + c)^3*b*c +
18*(d*sqrt(x) + c)^2*a*c^2 + 18*I*(d*sqrt(x) + c)^2*b*c^2 - 12*(d*sqrt(x) + c)*a*c^3 - 12*b*c^3*log(sec(d*sqr
t(x) + c)) + 4*(-4*I*(d*sqrt(x) + c)^3*b + 9*I*(d*sqrt(x) + c)^2*b*c - 9*I*(d*sqrt(x) + c)*b*c^2)*arctan2(sin(
2*d*sqrt(x) + 2*c), cos(2*d*sqrt(x) + 2*c) + 1) + 6*(4*I*(d*sqrt(x) + c)^2*b - 6*I*(d*sqrt(x) + c)*b*c + 3*I*b
*c^2)*dilog(-e^(2*I*d*sqrt(x) + 2*I*c)) - 2*(4*(d*sqrt(x) + c)^3*b - 9*(d*sqrt(x) + c)^2*b*c + 9*(d*sqrt(x) +
c)*b*c^2)*log(cos(2*d*sqrt(x) + 2*c)^2 + sin(2*d*sqrt(x) + 2*c)^2 + 2*cos(2*d*sqrt(x) + 2*c) + 1) - 12*I*b*pol
ylog(4, -e^(2*I*d*sqrt(x) + 2*I*c)) - 6*(4*(d*sqrt(x) + c)*b - 3*b*c)*polylog(3, -e^(2*I*d*sqrt(x) + 2*I*c)))/
d^4

Giac [F]

\[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \tan \left (d \sqrt {x} + c\right ) + a\right )} x \,d x } \]

[In]

integrate(x*(a+b*tan(c+d*x^(1/2))),x, algorithm="giac")

[Out]

integrate((b*tan(d*sqrt(x) + c) + a)*x, x)

Mupad [F(-1)]

Timed out. \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\int x\,\left (a+b\,\mathrm {tan}\left (c+d\,\sqrt {x}\right )\right ) \,d x \]

[In]

int(x*(a + b*tan(c + d*x^(1/2))),x)

[Out]

int(x*(a + b*tan(c + d*x^(1/2))), x)