3.312 \(\int \frac{(e+f x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=349 \[ \frac{b^2 f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}-\frac{b^2 f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac{b f \tanh ^{-1}(\sin (c+d x))}{d^2 \left (a^2-b^2\right )}+\frac{a f \log (\cos (c+d x))}{d^2 \left (a^2-b^2\right )}+\frac{i b^2 (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac{i b^2 (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac{a (e+f x) \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac{b (e+f x) \sec (c+d x)}{d \left (a^2-b^2\right )} \]

[Out]

(b*f*ArcTanh[Sin[c + d*x]])/((a^2 - b^2)*d^2) + (I*b^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 -
b^2])])/((a^2 - b^2)^(3/2)*d) - (I*b^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/((a^2
- b^2)^(3/2)*d) + (a*f*Log[Cos[c + d*x]])/((a^2 - b^2)*d^2) + (b^2*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqr
t[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^2) - (b^2*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/((a^
2 - b^2)^(3/2)*d^2) - (b*(e + f*x)*Sec[c + d*x])/((a^2 - b^2)*d) + (a*(e + f*x)*Tan[c + d*x])/((a^2 - b^2)*d)

________________________________________________________________________________________

Rubi [A]  time = 0.794791, antiderivative size = 349, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.423, Rules used = {4533, 3323, 2264, 2190, 2279, 2391, 6742, 4184, 3475, 4409, 3770} \[ \frac{b^2 f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}-\frac{b^2 f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac{b f \tanh ^{-1}(\sin (c+d x))}{d^2 \left (a^2-b^2\right )}+\frac{a f \log (\cos (c+d x))}{d^2 \left (a^2-b^2\right )}+\frac{i b^2 (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac{i b^2 (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac{a (e+f x) \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac{b (e+f x) \sec (c+d x)}{d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*f*ArcTanh[Sin[c + d*x]])/((a^2 - b^2)*d^2) + (I*b^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 -
b^2])])/((a^2 - b^2)^(3/2)*d) - (I*b^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/((a^2
- b^2)^(3/2)*d) + (a*f*Log[Cos[c + d*x]])/((a^2 - b^2)*d^2) + (b^2*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqr
t[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^2) - (b^2*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/((a^
2 - b^2)^(3/2)*d^2) - (b*(e + f*x)*Sec[c + d*x])/((a^2 - b^2)*d) + (a*(e + f*x)*Tan[c + d*x])/((a^2 - b^2)*d)

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 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && 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 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 4184

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4409

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
((c + d*x)^m*Sec[a + b*x]^n)/(b*n), x] - Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 9.7094, size = 842, normalized size = 2.41 \[ \frac{\frac{d (e+f x) \left (\frac{2 (d e-c f) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{i f \left (\log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt{b^2-a^2}}{-i a+b+\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\frac{a \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a+i \left (b+\sqrt{b^2-a^2}\right )}\right )\right )}{\sqrt{b^2-a^2}}+\frac{i f \left (\log \left (i \tan \left (\frac{1}{2} (c+d x)\right )+1\right ) \log \left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt{b^2-a^2}}{i a+b+\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\frac{a \left (i \tan \left (\frac{1}{2} (c+d x)\right )+1\right )}{a-i \left (b+\sqrt{b^2-a^2}\right )}\right )\right )}{\sqrt{b^2-a^2}}+\frac{i f \left (\log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (\frac{-b-a \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt{b^2-a^2}}{i a-b+\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\frac{a \left (\tan \left (\frac{1}{2} (c+d x)\right )+i\right )}{i a-b+\sqrt{b^2-a^2}}\right )\right )}{\sqrt{b^2-a^2}}-\frac{i f \left (\log \left (i \tan \left (\frac{1}{2} (c+d x)\right )+1\right ) \log \left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )-\sqrt{b^2-a^2}}{i a+b-\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\frac{i \tan \left (\frac{1}{2} (c+d x)\right ) a+a}{a+i \left (\sqrt{b^2-a^2}-b\right )}\right )\right )}{\sqrt{b^2-a^2}}\right ) b^2}{\left (b^2-a^2\right ) \left (d e-c f+i f \log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right )-i f \log \left (i \tan \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}+\frac{d (e+f x) b}{b^2-a^2}+\frac{f \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a+b}+\frac{f \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a-b}+\frac{d (e+f x) \sin \left (\frac{1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{d (e+f x) \sin \left (\frac{1}{2} (c+d x)\right )}{(a-b) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}}{d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.359, size = 1542, normalized size = 4.4 \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)^2/(a+b*sin(d*x+c)),x)

[Out]

2*(f*x+e)*(-I*a+b*exp(I*(d*x+c)))/d/(-a^2+b^2)/(1+exp(2*I*(d*x+c)))-I/d/(a^2-b^2)^(3/2)*b^2*f/(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)))*a^2*x-2*I/d^2/(a^2-b^2)*b^4*c*f/(a-b)/(a+b)/(-a^2+
b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))+I/d/(a^2-b^2)^(3/2)*b^4*f/(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-I/d^2/(a^2-b^2)^(3/2)*b^4*f/(a-b)/(a+b)*ln((I*b*e
xp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*c-I/d/(a^2-b^2)^(3/2)*b^4*f/(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+2*I/d^2/(a^2-b^2)*b^2*c*f/(a-b)/(a+b)/(-a^2+b^2)^(1/2)*arct
an(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))*a^2-4/d^2/(a^2-b^2)*b^2*f/(4*a-4*b)*ln(exp(I*(d*x+c))+I)-4
/d^2/(a^2-b^2)*b^2*f/(4*a+4*b)*ln(exp(I*(d*x+c))-I)+4/d^2/(a^2-b^2)*a^2*f/(4*a-4*b)*ln(exp(I*(d*x+c))+I)+4/d^2
/(a^2-b^2)*a^2*f/(4*a+4*b)*ln(exp(I*(d*x+c))-I)-2/d^2/(a^2-b^2)*a*f*ln(exp(I*(d*x+c)))+1/d^2/(a^2-b^2)^(3/2)*b
^4*f/(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^2/(a^2-b^2)^(3/2)*b^4*
f/(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)))+2*I/d/(a^2-b^2)*b^4*e/(a-b)/(
a+b)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))+I/d^2/(a^2-b^2)^(3/2)*b^4*f/(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-I/d^2/(a^2-b^2)^(3/2)*b^2*f/(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)))*a^2*c-2*I/d/(a^2-b^2)*b^2*e/(a-b)/(a+b)/(-
a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))*a^2+I/d/(a^2-b^2)^(3/2)*b^2*f/(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)))*a^2*x+I/d^2/(a^2-b^2)^(3/2)*b^2*f/(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)))*a^2*c-1/d^2/(a^2-b^2)^(3/2)*b^2*f/(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)))*a^2+1/d^2/(a^2-b^2)^(3/2)*b^2*f/(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)))*a^2

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.95523, size = 3087, normalized size = 8.85 \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)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(-2*I*b^3*f*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*c
os(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*I*b^3*f*sqrt(-(a^2 - b^2)/b^2)*cos(d*
x + c)*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) + 2*I*b^3*f*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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) - 2*I*b^3*f*sqrt(-(a
^2 - b^2)/b^2)*cos(d*x + c)*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) - 4*(a^2*b - b^3)*d*f*x + 2*(a^3 + a^2*b - a*b^2 - b^3)*f*cos(d*
x + c)*log(sin(d*x + c) + 1) + 2*(a^3 - a^2*b - a*b^2 + b^3)*f*cos(d*x + c)*log(-sin(d*x + c) + 1) - 2*(b^3*d*
e - b^3*c*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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) - 2*(b^3*d*e - b^3*c*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(2*b*cos(d*x + c) - 2*I*b*si
n(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + 2*(b^3*d*e - b^3*c*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*l
og(-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) + 2*(b^3*d*e - b^3*c*f)*sqrt(-
(a^2 - b^2)/b^2)*cos(d*x + c)*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)
- 2*(b^3*d*f*x + b^3*c*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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) + 2*(b^3*d*f*x + b^3*c*f)*sqrt(-(a^2
- b^2)/b^2)*cos(d*x + c)*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) - 2*(b^3*d*f*x + b^3*c*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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) +
2*(b^3*d*f*x + b^3*c*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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) - 4*(a^2*b - b^3)*d*e + 4*((a^3 - a*b^2
)*d*f*x + (a^3 - a*b^2)*d*e)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*d^2*cos(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e + f x\right ) \sec ^{2}{\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)**2/(a+b*sin(d*x+c)),x)

[Out]

Integral((e + f*x)*sec(c + d*x)**2/(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 )^{2}}{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)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)*sec(d*x + c)^2/(b*sin(d*x + c) + a), x)