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

Optimal. Leaf size=274 \[ \frac {a^2 x^2}{2}-\frac {3 i a b \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 a b \sqrt {x} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i a b x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+i a b x^2+\frac {3 b^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 i b^2 \sqrt {x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{3/2}}{d}-\frac {1}{2} b^2 x^2 \]

[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

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Rubi [A]  time = 0.47, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3747, 3722, 3719, 2190, 2531, 6609, 2282, 6589, 3720, 30} \[ \frac {a^2 x^2}{2}+\frac {6 i a b x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 a b \sqrt {x} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {3 i a b \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+i a b x^2-\frac {6 i b^2 \sqrt {x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {3 b^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{3/2}}{d}-\frac {1}{2} b^2 x^2 \]

Antiderivative was successfully verified.

[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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2282

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 2531

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 3719

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 3720

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 3722

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 3747

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 6589

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 6609

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*} \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \operatorname {Subst}\left (\int x^3 (a+b \tan (c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {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) \operatorname {Subst}\left (\int x^3 \tan (c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \operatorname {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) \operatorname {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 ) \operatorname {Subst}\left (\int x^3 \, dx,x,\sqrt {x}\right )-\frac {\left (6 b^2\right ) \operatorname {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) \operatorname {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 ) \operatorname {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 \text {Li}_2\left (-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) \operatorname {Subst}\left (\int x \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {\left (12 b^2\right ) \operatorname {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} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i a b x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 a b \sqrt {x} \text {Li}_3\left (-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) \operatorname {Subst}\left (\int \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {\left (6 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-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} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i a b x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 a b \sqrt {x} \text {Li}_3\left (-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) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_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} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i a b x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {3 b^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 a b \sqrt {x} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {3 i a b \text {Li}_4\left (-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*}

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Mathematica [A]  time = 2.32, size = 365, normalized size = 1.33 \[ \frac {1}{2} x^2 \left (a^2+2 a b \tan (c)-b^2\right )+\frac {b \left (-6 i \left (1+e^{2 i c}\right ) d \sqrt {x} \left (a d \sqrt {x}-b\right ) \text {Li}_2\left (-e^{-2 i \left (c+d \sqrt {x}\right )}\right )+3 \left (1+e^{2 i c}\right ) \left (b-2 a d \sqrt {x}\right ) \text {Li}_3\left (-e^{-2 i \left (c+d \sqrt {x}\right )}\right )-4 a e^{2 i c} d^3 x^{3/2} \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )-4 a d^3 x^{3/2} \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )+3 i a e^{2 i c} \text {Li}_4\left (-e^{-2 i \left (c+d \sqrt {x}\right )}\right )+3 i a \text {Li}_4\left (-e^{-2 i \left (c+d \sqrt {x}\right )}\right )-2 i a d^4 x^2+6 b e^{2 i c} d^2 x \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )+6 b d^2 x \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )+4 i b d^3 x^{3/2}\right )}{\left (1+e^{2 i c}\right ) d^4}+\frac {2 b^2 x^{3/2} \sec (c) \sin \left (d \sqrt {x}\right ) \sec \left (c+d \sqrt {x}\right )}{d} \]

Antiderivative was successfully verified.

[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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x \tan \left (d \sqrt {x} + c\right )^{2} + 2 \, a b x \tan \left (d \sqrt {x} + c\right ) + a^{2} x, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (d \sqrt {x} + c\right ) + a\right )}^{2} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 1.39, size = 0, normalized size = 0.00 \[ \int x \left (a +b \tan \left (c +d \sqrt {x}\right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [B]  time = 1.13, size = 1282, normalized size = 4.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 + (32*(d*sqr
t(x) + c)^3*a*b - 36*b^2*c^2 - 36*(2*a*b*c + b^2)*(d*sqrt(x) + c)^2 + 72*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c) + 4
*(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) + (32*I*(d*sqrt(x) + c)^3*a*b - 36*I*b^2*c^2 + (-72*I*a*b*c - 36*I*b^2)*(d*sqrt(x
) + c)^2 + (72*I*a*b*c^2 + 72*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))*c
os(2*d*sqrt(x) + 2*c) - (48*(d*sqrt(x) + c)^2*a*b + 36*a*b*c^2 + 36*b^2*c - 36*(2*a*b*c + b^2)*(d*sqrt(x) + 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))*cos(2*d*sqrt(x) + 2*
c) - (-48*I*(d*sqrt(x) + c)^2*a*b - 36*I*a*b*c^2 - 36*I*b^2*c + (72*I*a*b*c + 36*I*b^2)*(d*sqrt(x) + c))*sin(2
*d*sqrt(x) + 2*c))*dilog(-e^(2*I*d*sqrt(x) + 2*I*c)) + (-16*I*(d*sqrt(x) + c)^3*a*b + 18*I*b^2*c^2 + (36*I*a*b
*c + 18*I*b^2)*(d*sqrt(x) + c)^2 + (-36*I*a*b*c^2 - 36*I*b^2*c)*(d*sqrt(x) + c) + (-16*I*(d*sqrt(x) + c)^3*a*b
+ 18*I*b^2*c^2 + (36*I*a*b*c + 18*I*b^2)*(d*sqrt(x) + c)^2 + (-36*I*a*b*c^2 - 36*I*b^2*c)*(d*sqrt(x) + c))*co
s(2*d*sqrt(x) + 2*c) + 2*(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))*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)) + (-48*I*(d*sqrt(x) + c)*a*b + 36*I*a*b*c + 18*I*b^2 + (-48*I*(d*sqrt(x) + c)*
a*b + 36*I*a*b*c + 18*I*b^2)*cos(2*d*sqrt(x) + 2*c) + 6*(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)) + ((-6*I*a*b + 3*b^2)*(d*sqrt(x) + c)^4 + (24*I*b^2 + (24*I
*a*b - 12*b^2)*c)*(d*sqrt(x) + c)^3 + (-72*I*b^2*c + (-36*I*a*b + 18*b^2)*c^2)*(d*sqrt(x) + c)^2 - (12*b^2*c^3
- 72*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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,{\left (a+b\,\mathrm {tan}\left (c+d\,\sqrt {x}\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (a + b \tan {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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