Integrand size = 26, antiderivative size = 349 \[ \int \frac {(e+f x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b f \text {arctanh}(\sin (c+d x))}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {a f \log (\cos (c+d x))}{\left (a^2-b^2\right ) d^2}+\frac {b^2 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {b^2 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {b (e+f x) \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x) \tan (c+d x)}{\left (a^2-b^2\right ) d} \]
b*f*arctanh(sin(d*x+c))/(a^2-b^2)/d^2+a*f*ln(cos(d*x+c))/(a^2-b^2)/d^2+I*b ^2*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d- I*b^2*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2) /d+b^2*f*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2) /d^2-b^2*f*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/ 2)/d^2-b*(f*x+e)*sec(d*x+c)/(a^2-b^2)/d+a*(f*x+e)*tan(d*x+c)/(a^2-b^2)/d
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(906\) vs. \(2(349)=698\).
Time = 8.06 (sec) , antiderivative size = 906, normalized size of antiderivative = 2.60 \[ \int \frac {(e+f x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {b d (e+f x)}{-a^2+b^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 {b^2 d (e+f x) \left (\frac {2 (d e-c f) \arctan \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 \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {-b+\sqrt {-a^2+b^2}-a \tan \left (\frac {1}{2} (c+d x)\right )}{i a-b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b-\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{i a+b-\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b+\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{-i a+b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b+\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{i a+b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \operatorname {PolyLog}\left (2,\frac {a \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a+i \left (b+\sqrt {-a^2+b^2}\right )}\right )}{\sqrt {-a^2+b^2}}+\frac {i f \operatorname {PolyLog}\left (2,\frac {a \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a-i \left (b+\sqrt {-a^2+b^2}\right )}\right )}{\sqrt {-a^2+b^2}}+\frac {i f \operatorname {PolyLog}\left (2,\frac {a \left (i+\tan \left (\frac {1}{2} (c+d x)\right )\right )}{i a-b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \operatorname {PolyLog}\left (2,\frac {a+i a \tan \left (\frac {1}{2} (c+d x)\right )}{a+i \left (-b+\sqrt {-a^2+b^2}\right )}\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^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 (1+i \tan \left (\frac {1}{2} (c+d x)\right )\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 )}+\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} \]
((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])])/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])])/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])])/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])])/Sqrt[-a^2 + b^2] - (I*f*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*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*PolyLog[2, (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2 + b^2]) ])/Sqrt[-a^2 + b^2] - (I*f*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)*(Cos[(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
Time = 1.52 (sec) , antiderivative size = 324, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5044, 3042, 3804, 2694, 27, 2620, 2715, 2838, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(e+f x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 5044 |
| \(\displaystyle \frac {\int (e+f x) \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \int \frac {e+f x}{a+b \sin (c+d x)}dx}{a^2-b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (e+f x) \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \int \frac {e+f x}{a+b \sin (c+d x)}dx}{a^2-b^2}\) |
| \(\Big \downarrow \) 3804 |
| \(\displaystyle \frac {\int (e+f x) \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \int \frac {e^{i (c+d x)} (e+f x)}{2 e^{i (c+d x)} a-i b e^{2 i (c+d x)}+i b}dx}{a^2-b^2}\) |
| \(\Big \downarrow \) 2694 |
| \(\displaystyle \frac {\int (e+f x) \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{2 \left (a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{2 \left (a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (e+f x) \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {\int (e+f x) \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {f \int \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {f \int \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle \frac {\int (e+f x) \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {i f \int e^{-i (c+d x)} \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{b d^2}+\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {i f \int e^{-i (c+d x)} \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{b d^2}+\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {\int (e+f x) \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {\int \left (a (e+f x) \sec ^2(c+d x)-b (e+f x) \sec (c+d x) \tan (c+d x)\right )dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a f \log (\cos (c+d x))}{d^2}+\frac {a (e+f x) \tan (c+d x)}{d}+\frac {b f \text {arctanh}(\sin (c+d x))}{d^2}-\frac {b (e+f x) \sec (c+d x)}{d}}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
(-2*b^2*(((-1/2*I)*b*(((e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a ^2 - b^2])])/(b*d) - (I*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d^2)))/Sqrt[a^2 - b^2] + ((I/2)*b*(((e + f*x)*Log[1 - (I*b*E^( I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*d) - (I*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*d^2)))/Sqrt[a^2 - b^2]))/(a^2 - b^2) + ((b*f*ArcTanh[Sin[c + d*x]])/d^2 + (a*f*Log[Cos[c + d*x]])/d^2 - (b*(e + f*x)*Sec[c + d*x])/d + (a*(e + f*x)*Tan[c + d*x])/d)/(a^2 - b^2)
3.4.12.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] - Si mp[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]
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]}, Simp[2*(c/q) Int [(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[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]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[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]
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]
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy mbol] :> Simp[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]
Int[(((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_. )*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[-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] + Simp[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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2034 vs. \(2 (319 ) = 638\).
Time = 0.73 (sec) , antiderivative size = 2035, normalized size of antiderivative = 5.83
I/(a^2-b^2)^(3/2)/d*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/(a^2-b^2)^(3/2)/d^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+I/(a ^2-b^2)/d^2*f*c*b^2/(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^2-4/(a^2-b^2)/d^2*b^2*f/(4*a-4*b)*ln(exp( I*(d*x+c))+I)-4/(a^2-b^2)/d^2*b^2*f/(4*a+4*b)*ln(-I+exp(I*(d*x+c)))+4/(a^2 -b^2)/d^2*a^2*f/(4*a-4*b)*ln(exp(I*(d*x+c))+I)+4/(a^2-b^2)/d^2*a^2*f/(4*a+ 4*b)*ln(-I+exp(I*(d*x+c)))-1/(a^2-b^2)^(3/2)/d^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/(a^2-b^2 )^(3/2)/d^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+I/(a^2-b^2)/d*e*b^2/(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/(a^2-b^2)/d*e* b^4/(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))+3/(a^2-b^2)/d^2*b^2*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))*a^2-I/(a^2-b^2)^(3/2)/d*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/(a^2-b^2)^(3/2)/d^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/(a^2-b^2)/d*e*b^2/(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/(a^2-b^2)^(3/2)/d*b^4*f/(a-b)/(a+b)*ln((-I*b*exp(I*(d*x+c))+(a^2-b^2...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1267 vs. \(2 (311) = 622\).
Time = 0.46 (sec) , antiderivative size = 1267, normalized size of antiderivative = 3.63 \[ \int \frac {(e+f x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]
-1/2*(I*b^3*f*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b ^2) - b)/b + 1) - I*b^3*f*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog((I*a*c os(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-( a^2 - b^2)/b^2) - b)/b + 1) - I*b^3*f*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)* dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + I*b^3*f*sqrt(-(a^2 - b^2)/b^2)* cos(d*x + c)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 2*(a^2*b - b^3)*d* f*x - (a^3 + a^2*b - a*b^2 - b^3)*f*cos(d*x + c)*log(sin(d*x + c) + 1) - ( a^3 - a^2*b - a*b^2 + b^3)*f*cos(d*x + c)*log(-sin(d*x + c) + 1) + (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) + (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) - (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) - (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) - (b^3*d*f*x + b^3*c*f)*sqrt(-(a^2 - b^2)/b^2)* cos(d*x + c)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) ...
\[ \int \frac {(e+f x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\left (e + f x\right ) \sec ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {(e+f x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
\[ \int \frac {(e+f x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )} \sec \left (d x + c\right )^{2}}{b \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e+f x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \]