3.308 \(\int \frac{(e+f x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=413 \[ \frac{i a f \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac{i a f \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}+\frac{i b f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )}+\frac{i b f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )}-\frac{i b f \text{PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d^2 \left (a^2-b^2\right )}-\frac{b (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )}-\frac{b (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )}+\frac{b (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{d \left (a^2-b^2\right )}-\frac{2 i a (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{d \left (a^2-b^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.635319, antiderivative size = 413, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4533, 4519, 2190, 2279, 2391, 6742, 4181, 3719} \[ \frac{i a f \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac{i a f \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}+\frac{i b f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )}+\frac{i b f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )}-\frac{i b f \text{PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d^2 \left (a^2-b^2\right )}-\frac{b (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )}-\frac{b (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )}+\frac{b (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{d \left (a^2-b^2\right )}-\frac{2 i a (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Sec[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

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

Rule 4533

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> -Dist[b^2/(a^2 - b^2), Int[((e + f*x)^m*Sec[c + d*x]^(n - 2))/(a + b*Sin[c + d*x]), x], x] + Dist[1/(a^2
- b^2), Int[(e + f*x)^m*Sec[c + d*x]^n*(a - b*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m
, 0] && NeQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

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 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 4181

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

Rubi steps

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

Mathematica [B]  time = 16.59, size = 2743, normalized size = 6.64 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)*Sec[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

((d*e + d*f*x)*(((-I)*b*(d*e + d*f*x)^2)/f + 2*(a - b)*(d*e - c*f)*Log[1 - Tan[(c + d*x)/2]] - 4*b*(d*e + d*f*
x)*Log[(-2*I)/(-I + Tan[(c + d*x)/2])] - 2*(a + b)*(d*e - c*f)*Log[1 + Tan[(c + d*x)/2]] - (4*I)*b*f*PolyLog[2
, -Cos[c + d*x] + I*Sin[c + d*x]] + (2*I)*(a + b)*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(1/2 - I/2)*(1 + Tan[(c +
d*x)/2])] + PolyLog[2, ((1 + I) - (1 - I)*Tan[(c + d*x)/2])/2]) - (2*I)*(a + b)*f*(Log[1 - I*Tan[(c + d*x)/2]
]*Log[(1/2 + I/2)*(1 + Tan[(c + d*x)/2])] + PolyLog[2, (-1/2 - I/2)*(I + Tan[(c + d*x)/2])]) + (2*I)*(a - b)*f
*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(-1/2 + I/2)*(-1 + Tan[(c + d*x)/2])] + PolyLog[2, ((1 + I) + (1 - I)*Tan[(c
+ d*x)/2])/2]) - (2*I)*(a - b)*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(-1/2 - I/2)*(-1 + Tan[(c + d*x)/2])] + Pol
yLog[2, ((1 - I) + (1 + I)*Tan[(c + d*x)/2])/2]))*(a - b*Sin[c + d*x]))/((a^2 - b^2)*d^2*(-2*a*d*e + 2*a*c*f -
(2*I)*a*f*Log[1 - I*Tan[(c + d*x)/2]] + (2*I)*a*f*Log[1 + I*Tan[(c + d*x)/2]] + 4*b*f*Cos[c + d*x]*(Log[1 + C
os[c + d*x] - I*Sin[c + d*x]] - Log[(-2*I)/(-I + Tan[(c + d*x)/2])]) + b*(d*e - c*f + f*(c + d*x))*Sec[(c + d*
x)/2]*Sin[(3*(c + d*x))/2] + b*d*e*Tan[(c + d*x)/2] - b*c*f*Tan[(c + d*x)/2] - b*f*(c + d*x)*Tan[(c + d*x)/2]
+ (2*I)*b*f*Log[1 - I*Tan[(c + d*x)/2]]*Tan[(c + d*x)/2] - (2*I)*b*f*Log[1 + I*Tan[(c + d*x)/2]]*Tan[(c + d*x)
/2])) + ((f*(c + d*x)^2 + (2*I)*d*e*Log[Sec[(c + d*x)/2]^2] - (2*I)*c*f*Log[Sec[(c + d*x)/2]^2] - (2*I)*d*e*Lo
g[Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])] + (2*I)*c*f*Log[Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])] - (4*I)*f*
(c + d*x)*Log[(-2*I)/(-I + Tan[(c + d*x)/2])] - 2*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b - Sqrt[-a^2 + b^2] + a*
Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])] + 2*f*Log[1 - I*Tan[(c + d*x)/2]]*Log[-((b - Sqrt[-a^2 + b^2]
+ a*Tan[(c + d*x)/2])/(I*a - b + Sqrt[-a^2 + b^2]))] + 2*f*Log[1 - I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 + b^
2] + a*Tan[(c + d*x)/2])/((-I)*a + b + Sqrt[-a^2 + b^2])] - 2*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2
+ b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])] + 4*f*PolyLog[2, -Cos[c + d*x] + I*Sin[c + d*x]] +
2*f*PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[-a^2 + b^2]))] - 2*f*PolyLog[2, (a*(1 + I*Tan[(c
+ d*x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))] + 2*f*PolyLog[2, (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2
+ b^2])] - 2*f*PolyLog[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]))])*(-((b^2*e*Cos[c + d*x])
/((a^2 - b^2)*(a + b*Sin[c + d*x]))) + (b^2*c*f*Cos[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x])) - (b^2*f*(c
+ d*x)*Cos[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x]))))/(d*(2*f*(c + d*x) - (4*I)*f*Log[(-2*I)/(-I + Tan[
(c + d*x)/2])] - (4*f*Log[1 + Cos[c + d*x] - I*Sin[c + d*x]]*(I*Cos[c + d*x] + Sin[c + d*x]))/(-Cos[c + d*x] +
I*Sin[c + d*x]) + (I*f*Log[1 - (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^
2)/(1 - I*Tan[(c + d*x)/2]) - (I*f*Log[-((b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a - b + Sqrt[-a^2 + b^
2]))]*Sec[(c + d*x)/2]^2)/(1 - I*Tan[(c + d*x)/2]) - (I*f*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/((-I
)*a + b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(1 - I*Tan[(c + d*x)/2]) + (I*f*Log[1 - (a*(1 + I*Tan[(c + d*
x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^2)/(1 + I*Tan[(c + d*x)/2]) - (I*f*Log[(b - Sqrt[-a^2
+ b^2] + a*Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(1 + I*Tan[(c + d*x)/2]) - (I*
f*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(1 + I*Tan
[(c + d*x)/2]) + (2*I)*d*e*Tan[(c + d*x)/2] - (2*I)*c*f*Tan[(c + d*x)/2] + ((2*I)*f*(c + d*x)*Sec[(c + d*x)/2]
^2)/(-I + Tan[(c + d*x)/2]) - (f*Log[1 - (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x
)/2]^2)/(I + Tan[(c + d*x)/2]) + (I*a*f*Log[1 - (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]))]*Se
c[(c + d*x)/2]^2)/(a + I*a*Tan[(c + d*x)/2]) + (a*f*Log[1 - I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(b - Sqrt[
-a^2 + b^2] + a*Tan[(c + d*x)/2]) - (a*f*Log[1 + I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(b - Sqrt[-a^2 + b^2]
+ a*Tan[(c + d*x)/2]) + (a*f*Log[1 - I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(b + Sqrt[-a^2 + b^2] + a*Tan[(c
+ d*x)/2]) - (a*f*Log[1 + I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])
- ((2*I)*d*e*Cos[(c + d*x)/2]^2*(b*Cos[c + d*x]*Sec[(c + d*x)/2]^2 + Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])*
Tan[(c + d*x)/2]))/(a + b*Sin[c + d*x]) + ((2*I)*c*f*Cos[(c + d*x)/2]^2*(b*Cos[c + d*x]*Sec[(c + d*x)/2]^2 + S
ec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])*Tan[(c + d*x)/2]))/(a + b*Sin[c + d*x])))

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Maple [B]  time = 0.254, size = 861, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*sec(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

4/d*e/(4*a-4*b)*ln(exp(I*(d*x+c))+I)-1/d*e*b/(a-b)/(a+b)*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-4/d*e
/(4*a+4*b)*ln(exp(I*(d*x+c))-I)-1/d*f*b/(a-b)/(a+b)*ln((I*b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1
/2)))*x-1/d^2*f*b/(a-b)/(a+b)*ln((I*b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*c-4*I/d^2*f/(4*a
+4*b)*dilog(-I*exp(I*(d*x+c)))-4/d*f/(4*a+4*b)*ln(-I*(I-exp(I*(d*x+c))))*x-4/d^2*f/(4*a+4*b)*ln(-I*(I-exp(I*(d
*x+c))))*c-4*I/d^2*f/(4*a-4*b)*dilog(-I*(exp(I*(d*x+c))+I))+4/d*f/(4*a-4*b)*ln(-I*(exp(I*(d*x+c))+I))*x+4/d^2*
f/(4*a-4*b)*ln(-I*(exp(I*(d*x+c))+I))*c+I/d^2*f*b/(a-b)/(a+b)*dilog((I*b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a
+(a^2-b^2)^(1/2)))+I/d^2*f*b/(a-b)/(a+b)*dilog(-(I*b*exp(I*(d*x+c))-(a^2-b^2)^(1/2)-a)/(a+(a^2-b^2)^(1/2)))-1/
d*f*b/(a-b)/(a+b)*ln(-(I*b*exp(I*(d*x+c))-(a^2-b^2)^(1/2)-a)/(a+(a^2-b^2)^(1/2)))*x-1/d^2*f*b/(a-b)/(a+b)*ln(-
(I*b*exp(I*(d*x+c))-(a^2-b^2)^(1/2)-a)/(a+(a^2-b^2)^(1/2)))*c-4*I/d^2*f/(4*a+4*b)*ln(-I*(I-exp(I*(d*x+c))))*ln
(-I*exp(I*(d*x+c)))-4/d^2*f*c/(4*a-4*b)*ln(exp(I*(d*x+c))+I)+1/d^2*f*c*b/(a-b)/(a+b)*ln(I*b*exp(2*I*(d*x+c))-2
*a*exp(I*(d*x+c))-I*b)+4/d^2*f*c/(4*a+4*b)*ln(exp(I*(d*x+c))-I)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.77881, size = 3094, normalized size = 7.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(I*b*f*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a
^2 - b^2)/b^2) + 2*b)/b + 1) + I*b*f*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I
*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - I*b*f*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x +
c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - I*b*f*dilog(-1/2*(-2*I*a*cos
(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + I*
(a - b)*f*dilog(I*cos(d*x + c) + sin(d*x + c)) + I*(a + b)*f*dilog(I*cos(d*x + c) - sin(d*x + c)) - I*(a - b)*
f*dilog(-I*cos(d*x + c) + sin(d*x + c)) - I*(a + b)*f*dilog(-I*cos(d*x + c) - sin(d*x + c)) + (b*d*e - b*c*f)*
log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + (b*d*e - b*c*f)*log(2*b*cos(
d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + (b*d*e - b*c*f)*log(-2*b*cos(d*x + c) +
2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + (b*d*e - b*c*f)*log(-2*b*cos(d*x + c) - 2*I*b*sin(d
*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + (b*d*f*x + b*c*f)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x +
c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + (b*d*f*x + b*c*f)*log(1/2*(2*I*a
*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + (b
*d*f*x + b*c*f)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-
(a^2 - b^2)/b^2) + 2*b)/b) + (b*d*f*x + b*c*f)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x
+ c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - ((a + b)*d*e - (a + b)*c*f)*log(cos(d*x + c) + I*s
in(d*x + c) + I) + ((a - b)*d*e - (a - b)*c*f)*log(cos(d*x + c) - I*sin(d*x + c) + I) - ((a + b)*d*f*x + (a +
b)*c*f)*log(I*cos(d*x + c) + sin(d*x + c) + 1) + ((a - b)*d*f*x + (a - b)*c*f)*log(I*cos(d*x + c) - sin(d*x +
c) + 1) - ((a + b)*d*f*x + (a + b)*c*f)*log(-I*cos(d*x + c) + sin(d*x + c) + 1) + ((a - b)*d*f*x + (a - b)*c*f
)*log(-I*cos(d*x + c) - sin(d*x + c) + 1) - ((a + b)*d*e - (a + b)*c*f)*log(-cos(d*x + c) + I*sin(d*x + c) + I
) + ((a - b)*d*e - (a - b)*c*f)*log(-cos(d*x + c) - I*sin(d*x + c) + I))/((a^2 - b^2)*d^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e + f x\right ) \sec{\left (c + d x \right )}}{a + b \sin{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sec(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

Integral((e + f*x)*sec(c + d*x)/(a + b*sin(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )} \sec \left (d x + c\right )}{b \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)*sec(d*x + c)/(b*sin(d*x + c) + a), x)