3.161 \(\int \frac {x^2 \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx\)

Optimal. Leaf size=337 \[ \frac {i \text {Li}_3\left (-\frac {\left (\sqrt {a}-\sqrt {b}\right ) e^{2 i (c+d x)}}{\sqrt {a}+\sqrt {b}}\right )}{4 \sqrt {a} \sqrt {b} d^3}-\frac {i \text {Li}_3\left (-\frac {\left (\sqrt {a}+\sqrt {b}\right ) e^{2 i (c+d x)}}{\sqrt {a}-\sqrt {b}}\right )}{4 \sqrt {a} \sqrt {b} d^3}-\frac {x \text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d^2}+\frac {x \text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d^2}-\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d} \]

[Out]

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

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Rubi [A]  time = 0.90, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {4588, 3321, 2264, 2190, 2531, 2282, 6589} \[ -\frac {x \text {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d^2}+\frac {x \text {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d^2}+\frac {i \text {PolyLog}\left (3,-\frac {\left (\sqrt {a}-\sqrt {b}\right ) e^{2 i (c+d x)}}{\sqrt {a}+\sqrt {b}}\right )}{4 \sqrt {a} \sqrt {b} d^3}-\frac {i \text {PolyLog}\left (3,-\frac {\left (\sqrt {a}+\sqrt {b}\right ) e^{2 i (c+d x)}}{\sqrt {a}-\sqrt {b}}\right )}{4 \sqrt {a} \sqrt {b} d^3}-\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Sec[c + d*x]^2)/(a + b*Tan[c + d*x]^2),x]

[Out]

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

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 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && 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 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
+ d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 4588

Int[(((f_.) + (g_.)*(x_))^(m_.)*Sec[(d_.) + (e_.)*(x_)]^2)/((b_) + (c_.)*Tan[(d_.) + (e_.)*(x_)]^2), x_Symbol]
:> Dist[2, Int[(f + g*x)^m/(b + c + (b - c)*Cos[2*d + 2*e*x]), x], x] /; FreeQ[{b, c, d, e, f, g}, x] && IGtQ
[m, 0]

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]

Rubi steps

\begin {align*} \int \frac {x^2 \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx &=2 \int \frac {x^2}{a+b+(a-b) \cos (2 c+2 d x)} \, dx\\ &=4 \int \frac {e^{i (2 c+2 d x)} x^2}{a-b+2 (a+b) e^{i (2 c+2 d x)}+(a-b) e^{2 i (2 c+2 d x)}} \, dx\\ &=\frac {(2 (a-b)) \int \frac {e^{i (2 c+2 d x)} x^2}{-4 \sqrt {a} \sqrt {b}+2 (a+b)+2 (a-b) e^{i (2 c+2 d x)}} \, dx}{\sqrt {a} \sqrt {b}}-\frac {(2 (a-b)) \int \frac {e^{i (2 c+2 d x)} x^2}{4 \sqrt {a} \sqrt {b}+2 (a+b)+2 (a-b) e^{i (2 c+2 d x)}} \, dx}{\sqrt {a} \sqrt {b}}\\ &=-\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i \int x \log \left (1+\frac {2 (a-b) e^{i (2 c+2 d x)}}{-4 \sqrt {a} \sqrt {b}+2 (a+b)}\right ) \, dx}{\sqrt {a} \sqrt {b} d}-\frac {i \int x \log \left (1+\frac {2 (a-b) e^{i (2 c+2 d x)}}{4 \sqrt {a} \sqrt {b}+2 (a+b)}\right ) \, dx}{\sqrt {a} \sqrt {b} d}\\ &=-\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}-\frac {x \text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d^2}+\frac {x \text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d^2}+\frac {\int \text {Li}_2\left (-\frac {2 (a-b) e^{i (2 c+2 d x)}}{-4 \sqrt {a} \sqrt {b}+2 (a+b)}\right ) \, dx}{2 \sqrt {a} \sqrt {b} d^2}-\frac {\int \text {Li}_2\left (-\frac {2 (a-b) e^{i (2 c+2 d x)}}{4 \sqrt {a} \sqrt {b}+2 (a+b)}\right ) \, dx}{2 \sqrt {a} \sqrt {b} d^2}\\ &=-\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}-\frac {x \text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d^2}+\frac {x \text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d^2}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {\left (-\sqrt {a}+\sqrt {b}\right ) x}{\sqrt {a}+\sqrt {b}}\right )}{x} \, dx,x,e^{i (2 c+2 d x)}\right )}{4 \sqrt {a} \sqrt {b} d^3}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {\left (\sqrt {a}+\sqrt {b}\right ) x}{\sqrt {a}-\sqrt {b}}\right )}{x} \, dx,x,e^{i (2 c+2 d x)}\right )}{4 \sqrt {a} \sqrt {b} d^3}\\ &=-\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}-\frac {x \text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d^2}+\frac {x \text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d^2}+\frac {i \text {Li}_3\left (-\frac {\left (\sqrt {a}-\sqrt {b}\right ) e^{2 i (c+d x)}}{\sqrt {a}+\sqrt {b}}\right )}{4 \sqrt {a} \sqrt {b} d^3}-\frac {i \text {Li}_3\left (-\frac {\left (\sqrt {a}+\sqrt {b}\right ) e^{2 i (c+d x)}}{\sqrt {a}-\sqrt {b}}\right )}{4 \sqrt {a} \sqrt {b} d^3}\\ \end {align*}

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Mathematica [A]  time = 1.09, size = 294, normalized size = 0.87 \[ \frac {i \left (2 d^2 x^2 \log \left (1+\frac {\left (\sqrt {a}-\sqrt {b}\right ) e^{2 i (c+d x)}}{\sqrt {a}+\sqrt {b}}\right )-2 d^2 x^2 \log \left (1+\frac {\left (\sqrt {a}+\sqrt {b}\right ) e^{2 i (c+d x)}}{\sqrt {a}-\sqrt {b}}\right )-2 i d x \text {Li}_2\left (\frac {\left (\sqrt {b}-\sqrt {a}\right ) e^{2 i (c+d x)}}{\sqrt {a}+\sqrt {b}}\right )+2 i d x \text {Li}_2\left (-\frac {\left (\sqrt {a}+\sqrt {b}\right ) e^{2 i (c+d x)}}{\sqrt {a}-\sqrt {b}}\right )+\text {Li}_3\left (\frac {\left (\sqrt {b}-\sqrt {a}\right ) e^{2 i (c+d x)}}{\sqrt {a}+\sqrt {b}}\right )-\text {Li}_3\left (-\frac {\left (\sqrt {a}+\sqrt {b}\right ) e^{2 i (c+d x)}}{\sqrt {a}-\sqrt {b}}\right )\right )}{4 \sqrt {a} \sqrt {b} d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 2.28, size = 4644, normalized size = 13.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(8*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*d*x*dilog(1/2*((2*(a + b)*cos(d*x + c) + (2*I*a + 2*I*b)*sin(d*x
+ c) - 4*((a - b)*cos(d*x + c) - (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*s
qrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2*a + 2*b)/(a - b) + 1) + 8*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b
^2))*d*x*dilog(-1/2*((2*(a + b)*cos(d*x + c) - (2*I*a + 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x + c) + (-I*a
+ I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a
- b)) + 2*a - 2*b)/(a - b) + 1) + 8*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*d*x*dilog(1/2*((2*(a + b)*cos(d*x +
c) + (-2*I*a - 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x + c) - (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*
b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2*a + 2*b)/(a - b) + 1) + 8*(a -
b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*d*x*dilog(-1/2*((2*(a + b)*cos(d*x + c) - (-2*I*a - 2*I*b)*sin(d*x + c) - 4*
((a - b)*cos(d*x + c) + (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^
2 - 2*a*b + b^2)) + a + b)/(a - b)) + 2*a - 2*b)/(a - b) + 1) - 8*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*d*x*di
log(1/2*((2*(a + b)*cos(d*x + c) + (2*I*a + 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) + (I*a - I*b)*sin(d*
x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) - 2*a +
2*b)/(a - b) + 1) - 8*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*d*x*dilog(-1/2*((2*(a + b)*cos(d*x + c) - (2*I*a
+ 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) - (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqr
t((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) + 2*a - 2*b)/(a - b) + 1) - 8*(a - b)*sqrt(a*b/(a
^2 - 2*a*b + b^2))*d*x*dilog(1/2*((2*(a + b)*cos(d*x + c) + (-2*I*a - 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x
+ c) + (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)
) - a - b)/(a - b)) - 2*a + 2*b)/(a - b) + 1) - 8*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*d*x*dilog(-1/2*((2*(a
+ b)*cos(d*x + c) - (-2*I*a - 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) - (-I*a + I*b)*sin(d*x + c))*sqrt(
a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) + 2*a - 2*b)/(a - b)
+ 1) + 4*I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c^2*log(2*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a
+ b)/(a - b)) + 2*cos(d*x + c) + 2*I*sin(d*x + c)) - 4*I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c^2*log(2*sqrt
(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) + 2*cos(d*x + c) - 2*I*sin(d*x + c)) - 4*I*(a - b
)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c^2*log(2*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2
*cos(d*x + c) + 2*I*sin(d*x + c)) + 4*I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c^2*log(2*sqrt(-(2*(a - b)*sqrt(
a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2*cos(d*x + c) - 2*I*sin(d*x + c)) - 4*I*(a - b)*sqrt(a*b/(a^2 -
2*a*b + b^2))*c^2*log(2*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) + 2*cos(d*x + c) + 2*I
*sin(d*x + c)) + 4*I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c^2*log(2*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b
^2)) - a - b)/(a - b)) + 2*cos(d*x + c) - 2*I*sin(d*x + c)) + 4*I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c^2*lo
g(2*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) - 2*cos(d*x + c) + 2*I*sin(d*x + c)) - 4*I
*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c^2*log(2*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b
)) - 2*cos(d*x + c) - 2*I*sin(d*x + c)) + 4*(I*(a - b)*d^2*x^2 - I*(a - b)*c^2)*sqrt(a*b/(a^2 - 2*a*b + b^2))*
log(-1/2*((2*(a + b)*cos(d*x + c) + (2*I*a + 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x + c) - (-I*a + I*b)*sin(
d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2*
a + 2*b)/(a - b)) + 4*(-I*(a - b)*d^2*x^2 + I*(a - b)*c^2)*sqrt(a*b/(a^2 - 2*a*b + b^2))*log(1/2*((2*(a + b)*c
os(d*x + c) - (2*I*a + 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x + c) + (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^
2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) + 2*a - 2*b)/(a - b)) + 4*
(-I*(a - b)*d^2*x^2 + I*(a - b)*c^2)*sqrt(a*b/(a^2 - 2*a*b + b^2))*log(-1/2*((2*(a + b)*cos(d*x + c) + (-2*I*a
- 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x + c) - (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sq
rt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2*a + 2*b)/(a - b)) + 4*(I*(a - b)*d^2*x^2 -
I*(a - b)*c^2)*sqrt(a*b/(a^2 - 2*a*b + b^2))*log(1/2*((2*(a + b)*cos(d*x + c) - (-2*I*a - 2*I*b)*sin(d*x + c)
- 4*((a - b)*cos(d*x + c) + (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b
/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) + 2*a - 2*b)/(a - b)) + 4*(-I*(a - b)*d^2*x^2 + I*(a - b)*c^2)*sqrt(a*
b/(a^2 - 2*a*b + b^2))*log(-1/2*((2*(a + b)*cos(d*x + c) + (2*I*a + 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x +
c) + (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) -
a - b)/(a - b)) - 2*a + 2*b)/(a - b)) + 4*(I*(a - b)*d^2*x^2 - I*(a - b)*c^2)*sqrt(a*b/(a^2 - 2*a*b + b^2))*l
og(1/2*((2*(a + b)*cos(d*x + c) - (2*I*a + 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) - (I*a - I*b)*sin(d*x
+ c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) + 2*a -
2*b)/(a - b)) + 4*(I*(a - b)*d^2*x^2 - I*(a - b)*c^2)*sqrt(a*b/(a^2 - 2*a*b + b^2))*log(-1/2*((2*(a + b)*cos(d
*x + c) + (-2*I*a - 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) + (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 -
2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) - 2*a + 2*b)/(a - b)) + 4*(-I*
(a - b)*d^2*x^2 + I*(a - b)*c^2)*sqrt(a*b/(a^2 - 2*a*b + b^2))*log(1/2*((2*(a + b)*cos(d*x + c) - (-2*I*a - 2*
I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) - (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((
2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) + 2*a - 2*b)/(a - b)) + 4*(2*I*a - 2*I*b)*sqrt(a*b/(
a^2 - 2*a*b + b^2))*polylog(3, -1/2*(2*(a + b)*cos(d*x + c) + (2*I*a + 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*
x + c) - (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^
2)) + a + b)/(a - b))/(a - b)) + 4*(-2*I*a + 2*I*b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*polylog(3, 1/2*(2*(a + b)*co
s(d*x + c) - (2*I*a + 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x + c) + (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2
- 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b))/(a - b)) + 4*(-2*I*a + 2*I*
b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*polylog(3, -1/2*(2*(a + b)*cos(d*x + c) + (-2*I*a - 2*I*b)*sin(d*x + c) - 4*(
(a - b)*cos(d*x + c) - (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2
- 2*a*b + b^2)) + a + b)/(a - b))/(a - b)) + 4*(2*I*a - 2*I*b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*polylog(3, 1/2*(
2*(a + b)*cos(d*x + c) - (-2*I*a - 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x + c) + (I*a - I*b)*sin(d*x + c))*s
qrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b))/(a - b)) + 4*(-
2*I*a + 2*I*b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*polylog(3, -1/2*(2*(a + b)*cos(d*x + c) + (2*I*a + 2*I*b)*sin(d*x
+ c) + 4*((a - b)*cos(d*x + c) + (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqr
t(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b))/(a - b)) + 4*(2*I*a - 2*I*b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*polylo
g(3, 1/2*(2*(a + b)*cos(d*x + c) - (2*I*a + 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) - (I*a - I*b)*sin(d*
x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b))/(a - b)
) + 4*(2*I*a - 2*I*b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*polylog(3, -1/2*(2*(a + b)*cos(d*x + c) + (-2*I*a - 2*I*b)
*sin(d*x + c) + 4*((a - b)*cos(d*x + c) + (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a
- b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b))/(a - b)) + 4*(-2*I*a + 2*I*b)*sqrt(a*b/(a^2 - 2*a*b + b^
2))*polylog(3, 1/2*(2*(a + b)*cos(d*x + c) - (-2*I*a - 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) - (-I*a +
I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a -
b))/(a - b)))/(a*b*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.46, size = 1251, normalized size = 3.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sec(c + d*x)**2/(a + b*tan(c + d*x)**2), x)

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