3.701 \(\int \frac{\sec ^2(x) (a+b \tan (x))}{c+d \tan (x)} \, dx\)

Optimal. Leaf size=28 \[ \frac{b \tan (x)}{d}-\frac{(b c-a d) \log (c+d \tan (x))}{d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0873822, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {4342, 43} \[ \frac{b \tan (x)}{d}-\frac{(b c-a d) \log (c+d \tan (x))}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4342

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^2, x_Symbol] :> With[{d = FreeFactors[Tan[c*(a + b*x)], x]}, Dist[d/
(b*c), Subst[Int[SubstFor[1, Tan[c*(a + b*x)]/d, u, x], x], x, Tan[c*(a + b*x)]/d], x] /; FunctionOfQ[Tan[c*(a
+ b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && NonsumQ[u] && (EqQ[F, Sec] || EqQ[F, sec])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^2(x) (a+b \tan (x))}{c+d \tan (x)} \, dx &=\operatorname{Subst}\left (\int \frac{a+b x}{c+d x} \, dx,x,\tan (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{b}{d}+\frac{-b c+a d}{d (c+d x)}\right ) \, dx,x,\tan (x)\right )\\ &=-\frac{(b c-a d) \log (c+d \tan (x))}{d^2}+\frac{b \tan (x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.353691, size = 54, normalized size = 1.93 \[ \frac{\cos (x) (a+b \tan (x)) ((b c-a d) (\log (\cos (x))-\log (c \cos (x)+d \sin (x)))+b d \tan (x))}{d^2 (a \cos (x)+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.042, size = 35, normalized size = 1.3 \begin{align*}{\frac{b\tan \left ( x \right ) }{d}}+{\frac{\ln \left ( c+d\tan \left ( x \right ) \right ) a}{d}}-{\frac{\ln \left ( c+d\tan \left ( x \right ) \right ) cb}{{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.958968, size = 38, normalized size = 1.36 \begin{align*} \frac{b \tan \left (x\right )}{d} - \frac{{\left (b c - a d\right )} \log \left (d \tan \left (x\right ) + c\right )}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.33689, size = 194, normalized size = 6.93 \begin{align*} -\frac{{\left (b c - a d\right )} \cos \left (x\right ) \log \left (2 \, c d \cos \left (x\right ) \sin \left (x\right ) +{\left (c^{2} - d^{2}\right )} \cos \left (x\right )^{2} + d^{2}\right ) -{\left (b c - a d\right )} \cos \left (x\right ) \log \left (\cos \left (x\right )^{2}\right ) - 2 \, b d \sin \left (x\right )}{2 \, d^{2} \cos \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 3.57323, size = 29, normalized size = 1.04 \begin{align*} \frac{b \tan{\left (x \right )}}{d} + \frac{\left (a d - b c\right ) \left (\begin{cases} \frac{\tan{\left (x \right )}}{c} & \text{for}\: d = 0 \\\frac{\log{\left (c + d \tan{\left (x \right )} \right )}}{d} & \text{otherwise} \end{cases}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*tan(x)/d + (a*d - b*c)*Piecewise((tan(x)/c, Eq(d, 0)), (log(c + d*tan(x))/d, True))/d

________________________________________________________________________________________

Giac [A]  time = 1.11106, size = 39, normalized size = 1.39 \begin{align*} \frac{b \tan \left (x\right )}{d} - \frac{{\left (b c - a d\right )} \log \left ({\left | d \tan \left (x\right ) + c \right |}\right )}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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