\(\int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\) [255]

Optimal result

Integrand size = 28, antiderivative size = 94 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\log (1-\cos (c+d x))}{2 (a+b) d}-\frac {\log (\cos (c+d x))}{b d}-\frac {\log (1+\cos (c+d x))}{2 (a-b) d}+\frac {a^2 \log (b+a \cos (c+d x))}{b \left (a^2-b^2\right ) d} \]

[Out]

1/2*ln(1-cos(d*x+c))/(a+b)/d-ln(cos(d*x+c))/b/d-1/2*ln(1+cos(d*x+c))/(a-b)/d+a^2*ln(b+a*cos(d*x+c))/b/(a^2-b^2
)/d

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4482, 2916, 12, 908} \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {a^2 \log (a \cos (c+d x)+b)}{b d \left (a^2-b^2\right )}+\frac {\log (1-\cos (c+d x))}{2 d (a+b)}-\frac {\log (\cos (c+d x)+1)}{2 d (a-b)}-\frac {\log (\cos (c+d x))}{b d} \]

[In]

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

[Out]

Log[1 - Cos[c + d*x]]/(2*(a + b)*d) - Log[Cos[c + d*x]]/(b*d) - Log[1 + Cos[c + d*x]]/(2*(a - b)*d) + (a^2*Log
[b + a*Cos[c + d*x]])/(b*(a^2 - b^2)*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2916

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4482

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\csc (c+d x) \sec (c+d x)}{b+a \cos (c+d x)} \, dx \\ & = -\frac {a \text {Subst}\left (\int \frac {a}{x (b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{d} \\ & = -\frac {a^2 \text {Subst}\left (\int \frac {1}{x (b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{d} \\ & = -\frac {a^2 \text {Subst}\left (\int \left (\frac {1}{2 a^2 (a+b) (a-x)}+\frac {1}{a^2 b x}+\frac {1}{2 a^2 (a-b) (a+x)}+\frac {1}{b (-a+b) (a+b) (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{d} \\ & = \frac {\log (1-\cos (c+d x))}{2 (a+b) d}-\frac {\log (\cos (c+d x))}{b d}-\frac {\log (1+\cos (c+d x))}{2 (a-b) d}+\frac {a^2 \log (b+a \cos (c+d x))}{b \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=2 \left (\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 (-a+b) d}-\frac {\log (\cos (c+d x))}{2 b d}-\frac {a^2 \log (b+a \cos (c+d x))}{2 b \left (-a^2+b^2\right ) d}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 (a+b) d}\right ) \]

[In]

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

[Out]

2*(Log[Cos[(c + d*x)/2]]/(2*(-a + b)*d) - Log[Cos[c + d*x]]/(2*b*d) - (a^2*Log[b + a*Cos[c + d*x]])/(2*b*(-a^2
+ b^2)*d) + Log[Sin[(c + d*x)/2]]/(2*(a + b)*d))

Maple [A] (verified)

Time = 1.91 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\cos \left (d x +c \right )\right )}{b}+\frac {a^{2} \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right ) b}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\)	\(87\)
default	\(\frac {-\frac {\ln \left (\cos \left (d x +c \right )\right )}{b}+\frac {a^{2} \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right ) b}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\)	\(87\)
risch	\(\frac {i x}{a -b}+\frac {i c}{d \left (a -b \right )}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {2 i a^{2} x}{b \left (a^{2}-b^{2}\right )}-\frac {2 i a^{2} c}{b d \left (a^{2}-b^{2}\right )}+\frac {2 i x}{b}+\frac {2 i c}{b d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b d \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b d}\)	\(223\)

[In]

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

[Out]

1/d*(-1/b*ln(cos(d*x+c))+a^2/(a+b)/(a-b)/b*ln(b+cos(d*x+c)*a)+1/(2*a+2*b)*ln(cos(d*x+c)-1)-1/(2*a-2*b)*ln(cos(
d*x+c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {2 \, a^{2} \log \left (a \cos \left (d x + c\right ) + b\right ) - 2 \, {\left (a^{2} - b^{2}\right )} \log \left (-\cos \left (d x + c\right )\right ) - {\left (a b + b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a b - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d} \]

[In]

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

[Out]

1/2*(2*a^2*log(a*cos(d*x + c) + b) - 2*(a^2 - b^2)*log(-cos(d*x + c)) - (a*b + b^2)*log(1/2*cos(d*x + c) + 1/2
) + (a*b - b^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^2*b - b^3)*d)

Sympy [F]

\[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {a^{2} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} b - b^{3}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b} + \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b}}{d} \]

[In]

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

[Out]

(a^2*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^2*b - b^3) - log(sin(d*x + c)/(cos(d*x + c) +
1) + 1)/b - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/b + log(sin(d*x + c)/(cos(d*x + c) + 1))/(a + b))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (90) = 180\).

Time = 0.36 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.69 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {b \log \left ({\left | a + b + \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \right |}\right )}{a^{2} - b^{2}} + \frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {{\left | -2 \, a - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \, {\left | b \right |} \right |}}{{\left | -2 \, a - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + 2 \, {\left | b \right |} \right |}}\right )}{{\left (a^{2} - b^{2}\right )} {\left | b \right |}} + \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b}}{2 \, d} \]

[In]

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

[Out]

1/2*(b*log(abs(a + b + 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2
- b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2))/(a^2 - b^2) + (2*a^2 - b^2)*log(abs(-2*a - 2*a*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*abs(b))/abs(-2*a - 2*a*(cos(d*x + c) -
1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*abs(b)))/((a^2 - b^2)*abs(b)) + log(abs
(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b))/d

Mupad [B] (verification not implemented)

Time = 22.52 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}{b\,d}+\frac {a^2\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d\,\left (a^2\,b-b^3\right )} \]

[In]

int(1/(cos(c + d*x)^2*(a*sin(c + d*x) + b*tan(c + d*x))),x)

[Out]

log(tan(c/2 + (d*x)/2))/(d*(a + b)) - log(tan(c/2 + (d*x)/2)^2 - 1)/(b*d) + (a^2*log(a + b - a*tan(c/2 + (d*x)
/2)^2 + b*tan(c/2 + (d*x)/2)^2))/(d*(a^2*b - b^3))