\(\int x (a+b \tan (c+d \sqrt {x}))^2 \, dx\) [32]

Optimal result

Integrand size = 18, antiderivative size = 274 \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=-\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {6 i b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i a b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {3 b^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 a b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {3 i a b \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d} \]

[Out]

-2*I*b^2*x^(3/2)/d+1/2*a^2*x^2+I*a*b*x^2-1/2*b^2*x^2+6*b^2*x*ln(1+exp(2*I*(c+d*x^(1/2))))/d^2-4*a*b*x^(3/2)*ln
(1+exp(2*I*(c+d*x^(1/2))))/d+6*I*a*b*x*polylog(2,-exp(2*I*(c+d*x^(1/2))))/d^2+3*b^2*polylog(3,-exp(2*I*(c+d*x^
(1/2))))/d^4-3*I*a*b*polylog(4,-exp(2*I*(c+d*x^(1/2))))/d^4-6*I*b^2*polylog(2,-exp(2*I*(c+d*x^(1/2))))*x^(1/2)
/d^3-6*a*b*polylog(3,-exp(2*I*(c+d*x^(1/2))))*x^(1/2)/d^3+2*b^2*x^(3/2)*tan(c+d*x^(1/2))/d

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3832, 3803, 3800, 2221, 2611, 6744, 2320, 6724, 3801, 30} \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {a^2 x^2}{2}-\frac {3 i a b \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 a b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i a b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+i a b x^2+\frac {3 b^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 i b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{3/2}}{d}-\frac {1}{2} b^2 x^2 \]

[In]

Int[x*(a + b*Tan[c + d*Sqrt[x]])^2,x]

[Out]

((-2*I)*b^2*x^(3/2))/d + (a^2*x^2)/2 + I*a*b*x^2 - (b^2*x^2)/2 + (6*b^2*x*Log[1 + E^((2*I)*(c + d*Sqrt[x]))])/
d^2 - (4*a*b*x^(3/2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))])/d - ((6*I)*b^2*Sqrt[x]*PolyLog[2, -E^((2*I)*(c + d*Sq
rt[x]))])/d^3 + ((6*I)*a*b*x*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])/d^2 + (3*b^2*PolyLog[3, -E^((2*I)*(c + d*
Sqrt[x]))])/d^4 - (6*a*b*Sqrt[x]*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 - ((3*I)*a*b*PolyLog[4, -E^((2*I)
*(c + d*Sqrt[x]))])/d^4 + (2*b^2*x^(3/2)*Tan[c + d*Sqrt[x]])/d

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 3801

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

Rule 3803

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 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}& = 2 \text {Subst}\left (\int x^3 (a+b \tan (c+d x))^2 \, dx,x,\sqrt {x}\right ) \\ & = 2 \text {Subst}\left (\int \left (a^2 x^3+2 a b x^3 \tan (c+d x)+b^2 x^3 \tan ^2(c+d x)\right ) \, dx,x,\sqrt {x}\right ) \\ & = \frac {a^2 x^2}{2}+(4 a b) \text {Subst}\left (\int x^3 \tan (c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \text {Subst}\left (\int x^3 \tan ^2(c+d x) \, dx,x,\sqrt {x}\right ) \\ & = \frac {a^2 x^2}{2}+i a b x^2+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-(8 i a 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 )-\left (2 b^2\right ) \text {Subst}\left (\int x^3 \, dx,x,\sqrt {x}\right )-\frac {\left (6 b^2\right ) \text {Subst}\left (\int x^2 \tan (c+d x) \, dx,x,\sqrt {x}\right )}{d} \\ & = -\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}-\frac {4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(12 a b) \text {Subst}\left (\int x^2 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {\left (12 i b^2\right ) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x^2}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )}{d} \\ & = -\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 i a b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(12 i a 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 {\left (12 b^2\right ) \text {Subst}\left (\int x \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2} \\ & = -\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {6 i b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i a b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 a b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(6 a b) \text {Subst}\left (\int \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {\left (6 i b^2\right ) \text {Subst}\left (\int \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3} \\ & = -\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {6 i b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i a b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 a b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(3 i a b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4} \\ & = -\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {6 i b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i a b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {3 b^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 a b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {3 i a b \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.72 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.33 \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {b \left (4 i b d^3 x^{3/2}-2 i a d^4 x^2+6 b d^2 x \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )+6 b d^2 e^{2 i c} x \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )-4 a d^3 x^{3/2} \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )-4 a d^3 e^{2 i c} x^{3/2} \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )-6 i d \left (1+e^{2 i c}\right ) \left (-b+a d \sqrt {x}\right ) \sqrt {x} \operatorname {PolyLog}\left (2,-e^{-2 i \left (c+d \sqrt {x}\right )}\right )+3 \left (1+e^{2 i c}\right ) \left (b-2 a d \sqrt {x}\right ) \operatorname {PolyLog}\left (3,-e^{-2 i \left (c+d \sqrt {x}\right )}\right )+3 i a \operatorname {PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt {x}\right )}\right )+3 i a e^{2 i c} \operatorname {PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt {x}\right )}\right )\right )}{d^4 \left (1+e^{2 i c}\right )}+\frac {2 b^2 x^{3/2} \sec (c) \sec \left (c+d \sqrt {x}\right ) \sin \left (d \sqrt {x}\right )}{d}+\frac {1}{2} x^2 \left (a^2-b^2+2 a b \tan (c)\right ) \]

[In]

Integrate[x*(a + b*Tan[c + d*Sqrt[x]])^2,x]

[Out]

(b*((4*I)*b*d^3*x^(3/2) - (2*I)*a*d^4*x^2 + 6*b*d^2*x*Log[1 + E^((-2*I)*(c + d*Sqrt[x]))] + 6*b*d^2*E^((2*I)*c
)*x*Log[1 + E^((-2*I)*(c + d*Sqrt[x]))] - 4*a*d^3*x^(3/2)*Log[1 + E^((-2*I)*(c + d*Sqrt[x]))] - 4*a*d^3*E^((2*
I)*c)*x^(3/2)*Log[1 + E^((-2*I)*(c + d*Sqrt[x]))] - (6*I)*d*(1 + E^((2*I)*c))*(-b + a*d*Sqrt[x])*Sqrt[x]*PolyL
og[2, -E^((-2*I)*(c + d*Sqrt[x]))] + 3*(1 + E^((2*I)*c))*(b - 2*a*d*Sqrt[x])*PolyLog[3, -E^((-2*I)*(c + d*Sqrt
[x]))] + (3*I)*a*PolyLog[4, -E^((-2*I)*(c + d*Sqrt[x]))] + (3*I)*a*E^((2*I)*c)*PolyLog[4, -E^((-2*I)*(c + d*Sq
rt[x]))]))/(d^4*(1 + E^((2*I)*c))) + (2*b^2*x^(3/2)*Sec[c]*Sec[c + d*Sqrt[x]]*Sin[d*Sqrt[x]])/d + (x^2*(a^2 -
b^2 + 2*a*b*Tan[c]))/2

Maple [F]

\[\int x \left (a +b \tan \left (c +d \sqrt {x}\right )\right )^{2}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \tan \left (d \sqrt {x} + c\right ) + a\right )}^{2} x \,d x } \]

[In]

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

[Out]

integral(b^2*x*tan(d*sqrt(x) + c)^2 + 2*a*b*x*tan(d*sqrt(x) + c) + a^2*x, x)

Sympy [F]

\[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x \left (a + b \tan {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]

[In]

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

[Out]

Integral(x*(a + b*tan(c + d*sqrt(x)))**2, 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. 1290 vs. \(2 (218) = 436\).

Time = 0.51 (sec) , antiderivative size = 1290, normalized size of antiderivative = 4.71 \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/2*((d*sqrt(x) + c)^4*a^2 - 4*(d*sqrt(x) + c)^3*a^2*c + 6*(d*sqrt(x) + c)^2*a^2*c^2 - 4*(d*sqrt(x) + c)*a^2*c
^3 - 8*a*b*c^3*log(sec(d*sqrt(x) + c)) - 4*(12*I*(d*sqrt(x) + c)*b^2*c^3 - 3*(2*a*b + I*b^2)*(d*sqrt(x) + c)^4
+ 12*(2*a*b + I*b^2)*(d*sqrt(x) + c)^3*c - 18*(2*a*b + I*b^2)*(d*sqrt(x) + c)^2*c^2 + 24*b^2*c^3 + 4*(8*(d*sq
rt(x) + c)^3*a*b - 9*b^2*c^2 - 9*(2*a*b*c + b^2)*(d*sqrt(x) + c)^2 + 18*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c) + (8
*(d*sqrt(x) + c)^3*a*b - 9*b^2*c^2 - 9*(2*a*b*c + b^2)*(d*sqrt(x) + c)^2 + 18*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c
))*cos(2*d*sqrt(x) + 2*c) - (-8*I*(d*sqrt(x) + c)^3*a*b + 9*I*b^2*c^2 + 9*(2*I*a*b*c + I*b^2)*(d*sqrt(x) + c)^
2 + 18*(-I*a*b*c^2 - I*b^2*c)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*arctan2(sin(2*d*sqrt(x) + 2*c), cos(2*d
*sqrt(x) + 2*c) + 1) - 3*((2*a*b + I*b^2)*(d*sqrt(x) + c)^4 - 4*(2*b^2 + (2*a*b + I*b^2)*c)*(d*sqrt(x) + c)^3
+ 6*(4*b^2*c + (2*a*b + I*b^2)*c^2)*(d*sqrt(x) + c)^2 + 4*(-I*b^2*c^3 - 6*b^2*c^2)*(d*sqrt(x) + c))*cos(2*d*sq
rt(x) + 2*c) - 12*(4*(d*sqrt(x) + c)^2*a*b + 3*a*b*c^2 + 3*b^2*c - 3*(2*a*b*c + b^2)*(d*sqrt(x) + c) + (4*(d*s
qrt(x) + c)^2*a*b + 3*a*b*c^2 + 3*b^2*c - 3*(2*a*b*c + b^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + (4*I*(d*
sqrt(x) + c)^2*a*b + 3*I*a*b*c^2 + 3*I*b^2*c + 3*(-2*I*a*b*c - I*b^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))
*dilog(-e^(2*I*d*sqrt(x) + 2*I*c)) - 2*(8*I*(d*sqrt(x) + c)^3*a*b - 9*I*b^2*c^2 + 9*(-2*I*a*b*c - I*b^2)*(d*sq
rt(x) + c)^2 + 18*(I*a*b*c^2 + I*b^2*c)*(d*sqrt(x) + c) + (8*I*(d*sqrt(x) + c)^3*a*b - 9*I*b^2*c^2 + 9*(-2*I*a
*b*c - I*b^2)*(d*sqrt(x) + c)^2 + 18*(I*a*b*c^2 + I*b^2*c)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - (8*(d*sqr
t(x) + c)^3*a*b - 9*b^2*c^2 - 9*(2*a*b*c + b^2)*(d*sqrt(x) + c)^2 + 18*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c))*sin(
2*d*sqrt(x) + 2*c))*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) +
24*(a*b*cos(2*d*sqrt(x) + 2*c) + I*a*b*sin(2*d*sqrt(x) + 2*c) + a*b)*polylog(4, -e^(2*I*d*sqrt(x) + 2*I*c)) -
6*(8*I*(d*sqrt(x) + c)*a*b - 6*I*a*b*c - 3*I*b^2 + (8*I*(d*sqrt(x) + c)*a*b - 6*I*a*b*c - 3*I*b^2)*cos(2*d*sqr
t(x) + 2*c) - (8*(d*sqrt(x) + c)*a*b - 6*a*b*c - 3*b^2)*sin(2*d*sqrt(x) + 2*c))*polylog(3, -e^(2*I*d*sqrt(x) +
2*I*c)) - 3*((2*I*a*b - b^2)*(d*sqrt(x) + c)^4 + 4*(-2*I*b^2 + (-2*I*a*b + b^2)*c)*(d*sqrt(x) + c)^3 + 6*(4*I
*b^2*c + (2*I*a*b - b^2)*c^2)*(d*sqrt(x) + c)^2 + 4*(b^2*c^3 - 6*I*b^2*c^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) +
2*c))/(-12*I*cos(2*d*sqrt(x) + 2*c) + 12*sin(2*d*sqrt(x) + 2*c) - 12*I))/d^4

Giac [F]

\[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \tan \left (d \sqrt {x} + c\right ) + a\right )}^{2} x \,d x } \]

[In]

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

[Out]

integrate((b*tan(d*sqrt(x) + c) + a)^2*x, x)

Mupad [F(-1)]

Timed out. \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x\,{\left (a+b\,\mathrm {tan}\left (c+d\,\sqrt {x}\right )\right )}^2 \,d x \]

[In]

int(x*(a + b*tan(c + d*x^(1/2)))^2,x)

[Out]

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