Integrand size = 18, antiderivative size = 305 \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=-\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {d \log (a+b \sin (e+f x))}{\left (a^2-b^2\right ) f^2}-\frac {a d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {a d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))} \]
-d*ln(a+b*sin(f*x+e))/(a^2-b^2)/f^2-I*a*(d*x+c)*ln(1-I*b*exp(I*(f*x+e))/(a -(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f+I*a*(d*x+c)*ln(1-I*b*exp(I*(f*x+e))/( a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f-a*d*polylog(2,I*b*exp(I*(f*x+e))/(a- (a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f^2+a*d*polylog(2,I*b*exp(I*(f*x+e))/(a+ (a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f^2+b*(d*x+c)*cos(f*x+e)/(a^2-b^2)/f/(a+ b*sin(f*x+e))
Time = 0.72 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.77 \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\frac {-d \log (a+b \sin (e+f x))+\frac {a \left (-i f (c+d x) \left (\log \left (1+\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )-\log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )-d \operatorname {PolyLog}\left (2,-\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )+d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )}{\sqrt {a^2-b^2}}+\frac {b f (c+d x) \cos (e+f x)}{a+b \sin (e+f x)}}{\left (a^2-b^2\right ) f^2} \]
(-(d*Log[a + b*Sin[e + f*x]]) + (a*((-I)*f*(c + d*x)*(Log[1 + (I*b*E^(I*(e + f*x)))/(-a + Sqrt[a^2 - b^2])] - Log[1 - (I*b*E^(I*(e + f*x)))/(a + Sqr t[a^2 - b^2])]) - d*PolyLog[2, ((-I)*b*E^(I*(e + f*x)))/(-a + Sqrt[a^2 - b ^2])] + d*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])]))/Sqrt[a ^2 - b^2] + (b*f*(c + d*x)*Cos[e + f*x])/(a + b*Sin[e + f*x]))/((a^2 - b^2 )*f^2)
Time = 1.13 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {3042, 3805, 3042, 3147, 16, 3804, 2694, 27, 2620, 2715, 2838}
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.
\(\displaystyle \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {c+d x}{(a+b \sin (e+f x))^2}dx\) |
\(\Big \downarrow \) 3805 |
\(\displaystyle \frac {a \int \frac {c+d x}{a+b \sin (e+f x)}dx}{a^2-b^2}-\frac {b d \int \frac {\cos (e+f x)}{a+b \sin (e+f x)}dx}{f \left (a^2-b^2\right )}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a \int \frac {c+d x}{a+b \sin (e+f x)}dx}{a^2-b^2}-\frac {b d \int \frac {\cos (e+f x)}{a+b \sin (e+f x)}dx}{f \left (a^2-b^2\right )}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {a \int \frac {c+d x}{a+b \sin (e+f x)}dx}{a^2-b^2}-\frac {d \int \frac {1}{a+b \sin (e+f x)}d(b \sin (e+f x))}{f^2 \left (a^2-b^2\right )}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {a \int \frac {c+d x}{a+b \sin (e+f x)}dx}{a^2-b^2}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3804 |
\(\displaystyle \frac {2 a \int \frac {e^{i (e+f x)} (c+d x)}{2 e^{i (e+f x)} a-i b e^{2 i (e+f x)}+i b}dx}{a^2-b^2}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 2694 |
\(\displaystyle \frac {2 a \left (\frac {i b \int \frac {e^{i (e+f x)} (c+d x)}{2 \left (a-i b e^{i (e+f x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (e+f x)} (c+d x)}{2 \left (a-i b e^{i (e+f x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )}{a^2-b^2}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 a \left (\frac {i b \int \frac {e^{i (e+f x)} (c+d x)}{a-i b e^{i (e+f x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (e+f x)} (c+d x)}{a-i b e^{i (e+f x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {2 a \left (\frac {i b \left (\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{b f}-\frac {d \int \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b f}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{b f}-\frac {d \int \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b f}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {2 a \left (\frac {i b \left (\frac {i d \int e^{-i (e+f x)} \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (e+f x)}}{b f^2}+\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{b f}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {i d \int e^{-i (e+f x)} \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (e+f x)}}{b f^2}+\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{b f}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {2 a \left (\frac {i b \left (\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{b f}-\frac {i d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{b f^2}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{b f}-\frac {i d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{b f^2}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )}\) |
-((d*Log[a + b*Sin[e + f*x]])/((a^2 - b^2)*f^2)) + (2*a*(((-1/2*I)*b*(((c + d*x)*Log[1 - (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/(b*f) - (I*d* PolyLog[2, (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/(b*f^2)))/Sqrt[a^ 2 - b^2] + ((I/2)*b*(((c + d*x)*Log[1 - (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^ 2 - b^2])])/(b*f) - (I*d*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/(b*f^2)))/Sqrt[a^2 - b^2]))/(a^2 - b^2) + (b*(c + d*x)*Cos[e + f*x ])/((a^2 - b^2)*f*(a + b*Sin[e + f*x]))
3.2.70.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
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[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_ Symbol] :> Simp[b*(c + d*x)^m*(Cos[e + f*x]/(f*(a^2 - b^2)*(a + b*Sin[e + f *x]))), x] + (Simp[a/(a^2 - b^2) Int[(c + d*x)^m/(a + b*Sin[e + f*x]), x] , x] - Simp[b*d*(m/(f*(a^2 - b^2))) Int[(c + d*x)^(m - 1)*(Cos[e + f*x]/( a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 649 vs. \(2 (275 ) = 550\).
Time = 1.25 (sec) , antiderivative size = 650, normalized size of antiderivative = 2.13
method | result | size |
risch | \(\frac {2 \left (d x +c \right ) \left (i b +a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (a^{2}-b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (f x +e \right )}-b +2 i a \,{\mathrm e}^{i \left (f x +e \right )}\right )}+\frac {d \ln \left (i b \,{\mathrm e}^{2 i \left (f x +e \right )}-i b -2 a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{\left (-a^{2}+b^{2}\right ) f^{2}}-\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{\left (-a^{2}+b^{2}\right ) f^{2}}+\frac {2 i a d e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}-\frac {2 i a c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f}-\frac {d a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f}+\frac {d a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f}-\frac {d a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) e}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}+\frac {d a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) e}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}+\frac {i d a \operatorname {dilog}\left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}-\frac {i d a \operatorname {dilog}\left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}\) | \(650\) |
2*(d*x+c)*(I*b+a*exp(I*(f*x+e)))/f/(a^2-b^2)/(b*exp(2*I*(f*x+e))-b+2*I*a*e xp(I*(f*x+e)))+1/(-a^2+b^2)/f^2*d*ln(I*b*exp(2*I*(f*x+e))-I*b-2*a*exp(I*(f *x+e)))-2/(-a^2+b^2)/f^2*d*ln(exp(I*(f*x+e)))+2*I/(-a^2+b^2)^(3/2)/f^2*a*d *e*arctan(1/2*(2*I*b*exp(I*(f*x+e))-2*a)/(-a^2+b^2)^(1/2))-2*I/(-a^2+b^2)^ (3/2)/f*a*c*arctan(1/2*(2*I*b*exp(I*(f*x+e))-2*a)/(-a^2+b^2)^(1/2))-1/(-a^ 2+b^2)^(3/2)/f*d*a*ln((I*a+b*exp(I*(f*x+e))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b ^2)^(1/2)))*x+1/(-a^2+b^2)^(3/2)/f*d*a*ln((I*a+b*exp(I*(f*x+e))+(-a^2+b^2) ^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*x-1/(-a^2+b^2)^(3/2)/f^2*d*a*ln((I*a+b*exp (I*(f*x+e))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*e+1/(-a^2+b^2)^(3/2) /f^2*d*a*ln((I*a+b*exp(I*(f*x+e))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)) )*e+I/(-a^2+b^2)^(3/2)/f^2*d*a*dilog((I*a+b*exp(I*(f*x+e))-(-a^2+b^2)^(1/2 ))/(I*a-(-a^2+b^2)^(1/2)))-I/(-a^2+b^2)^(3/2)/f^2*d*a*dilog((I*a+b*exp(I*( f*x+e))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1512 vs. \(2 (267) = 534\).
Time = 0.49 (sec) , antiderivative size = 1512, normalized size of antiderivative = 4.96 \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\text {Too large to display} \]
1/2*((I*a*b^2*d*sin(f*x + e) + I*a^2*b*d)*sqrt(-(a^2 - b^2)/b^2)*dilog((I* a*cos(f*x + e) - a*sin(f*x + e) + (b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt (-(a^2 - b^2)/b^2) - b)/b + 1) + (-I*a*b^2*d*sin(f*x + e) - I*a^2*b*d)*sqr t(-(a^2 - b^2)/b^2)*dilog((I*a*cos(f*x + e) - a*sin(f*x + e) - (b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + (-I*a*b^2*d* sin(f*x + e) - I*a^2*b*d)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(f*x + e) - a*sin(f*x + e) + (b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b ^2) - b)/b + 1) + (I*a*b^2*d*sin(f*x + e) + I*a^2*b*d)*sqrt(-(a^2 - b^2)/b ^2)*dilog((-I*a*cos(f*x + e) - a*sin(f*x + e) - (b*cos(f*x + e) - I*b*sin( f*x + e))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - (a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*c os(f*x + e) - a*sin(f*x + e) + (b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-( a^2 - b^2)/b^2) - b)/b) + (a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2* d*e)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(f*x + e) - a*sin(f *x + e) - (b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) - b)/ b) - (a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*sin(f*x + e))*sq rt(-(a^2 - b^2)/b^2)*log(-(-I*a*cos(f*x + e) - a*sin(f*x + e) + (b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + (a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) *log(-(-I*a*cos(f*x + e) - a*sin(f*x + e) - (b*cos(f*x + e) - I*b*sin(f...
Timed out. \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\int { \frac {d x + c}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\text {Hanged} \]