3.26 \(\int \frac{\tan (c+d x) (B \tan (c+d x)+C \tan ^2(c+d x))}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=101 \[ \frac{a^2 (b B-a C) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )}-\frac{(b B-a C) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac{x (a B+b C)}{a^2+b^2}+\frac{C \tan (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.243285, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3632, 3606, 3626, 3617, 31, 3475} \[ \frac{a^2 (b B-a C) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )}-\frac{(b B-a C) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac{x (a B+b C)}{a^2+b^2}+\frac{C \tan (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3632

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Tan[e + f*x])
^(m + 1)*(c + d*Tan[e + f*x])^n*(b*B - a*C + b*C*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3606

Int[(((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b^2*B*Tan[e + f*x])/(d*f), x] + Dist[1/d, Int[(a^2*A*d - b^2*B*c + (2*a*
A*b + B*(a^2 - b^2))*d*Tan[e + f*x] + (A*b^2*d - b*B*(b*c - 2*a*d))*Tan[e + f*x]^2)/(c + d*Tan[e + f*x]), x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3626

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[((a*A + b*B - a*C)*x)/(a^2 + b^2), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

Rule 3617

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3475

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

Rubi steps

\begin{align*} \int \frac{\tan (c+d x) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx &=\int \frac{\tan ^2(c+d x) (B+C \tan (c+d x))}{a+b \tan (c+d x)} \, dx\\ &=\frac{C \tan (c+d x)}{b d}+\frac{\int \frac{-a C-b C \tan (c+d x)+(b B-a C) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b}\\ &=-\frac{(a B+b C) x}{a^2+b^2}+\frac{C \tan (c+d x)}{b d}+\frac{(b B-a C) \int \tan (c+d x) \, dx}{a^2+b^2}+\frac{\left (a^2 (b B-a C)\right ) \int \frac{1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac{(a B+b C) x}{a^2+b^2}-\frac{(b B-a C) \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}+\frac{C \tan (c+d x)}{b d}+\frac{\left (a^2 (b B-a C)\right ) \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^2 \left (a^2+b^2\right ) d}\\ &=-\frac{(a B+b C) x}{a^2+b^2}-\frac{(b B-a C) \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}+\frac{a^2 (b B-a C) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right ) d}+\frac{C \tan (c+d x)}{b d}\\ \end{align*}

Mathematica [C]  time = 0.571322, size = 118, normalized size = 1.17 \[ \frac{\frac{2 a^2 (b B-a C) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )}+\frac{i (B+i C) \log (-\tan (c+d x)+i)}{a+i b}-\frac{(C+i B) \log (\tan (c+d x)+i)}{a-i b}+\frac{2 C \tan (c+d x)}{b}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.034, size = 179, normalized size = 1.8 \begin{align*}{\frac{C\tan \left ( dx+c \right ) }{bd}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Bb}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Ca}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{C\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{bd \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{{a}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) C}{{b}^{2}d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.77521, size = 147, normalized size = 1.46 \begin{align*} -\frac{\frac{2 \,{\left (B a + C b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} + \frac{2 \,{\left (C a^{3} - B a^{2} b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{2} + b^{4}} + \frac{{\left (C a - B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, C \tan \left (d x + c\right )}{b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(B*a + C*b)*(d*x + c)/(a^2 + b^2) + 2*(C*a^3 - B*a^2*b)*log(b*tan(d*x + c) + a)/(a^2*b^2 + b^4) + (C*a
- B*b)*log(tan(d*x + c)^2 + 1)/(a^2 + b^2) - 2*C*tan(d*x + c)/b)/d

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Fricas [A]  time = 1.20611, size = 333, normalized size = 3.3 \begin{align*} -\frac{2 \,{\left (B a b^{2} + C b^{3}\right )} d x +{\left (C a^{3} - B a^{2} b\right )} \log \left (\frac{b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) -{\left (C a^{3} - B a^{2} b + C a b^{2} - B b^{3}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \,{\left (C a^{2} b + C b^{3}\right )} \tan \left (d x + c\right )}{2 \,{\left (a^{2} b^{2} + b^{4}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(B*a*b^2 + C*b^3)*d*x + (C*a^3 - B*a^2*b)*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x
+ c)^2 + 1)) - (C*a^3 - B*a^2*b + C*a*b^2 - B*b^3)*log(1/(tan(d*x + c)^2 + 1)) - 2*(C*a^2*b + C*b^3)*tan(d*x
+ c))/((a^2*b^2 + b^4)*d)

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Sympy [A]  time = 16.6724, size = 1020, normalized size = 10.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*(B*tan(c) + C*tan(c)**2), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((-B*x + B*tan(c + d*x)/d - C*log(
tan(c + d*x)**2 + 1)/(2*d) + C*tan(c + d*x)**2/(2*d))/a, Eq(b, 0)), (-I*B*d*x*tan(c + d*x)/(-2*b*d*tan(c + d*x
) + 2*I*b*d) - B*d*x/(-2*b*d*tan(c + d*x) + 2*I*b*d) - B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(-2*b*d*tan(c +
d*x) + 2*I*b*d) + I*B*log(tan(c + d*x)**2 + 1)/(-2*b*d*tan(c + d*x) + 2*I*b*d) + I*B/(-2*b*d*tan(c + d*x) + 2
*I*b*d) + 3*C*d*x*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d) - 3*I*C*d*x/(-2*b*d*tan(c + d*x) + 2*I*b*d) - I
*C*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d) - C*log(tan(c + d*x)**2 + 1)/(-2*b*d*
tan(c + d*x) + 2*I*b*d) - 2*C*tan(c + d*x)**2/(-2*b*d*tan(c + d*x) + 2*I*b*d) - 3*C/(-2*b*d*tan(c + d*x) + 2*I
*b*d), Eq(a, -I*b)), (-I*B*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + B*d*x/(2*b*d*tan(c + d*x) + 2*I*b
*d) + B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + I*B*log(tan(c + d*x)**2 + 1)/(2
*b*d*tan(c + d*x) + 2*I*b*d) + I*B/(2*b*d*tan(c + d*x) + 2*I*b*d) - 3*C*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) +
2*I*b*d) - 3*I*C*d*x/(2*b*d*tan(c + d*x) + 2*I*b*d) - I*C*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c
+ d*x) + 2*I*b*d) + C*log(tan(c + d*x)**2 + 1)/(2*b*d*tan(c + d*x) + 2*I*b*d) + 2*C*tan(c + d*x)**2/(2*b*d*tan
(c + d*x) + 2*I*b*d) + 3*C/(2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, I*b)), (x*(B*tan(c) + C*tan(c)**2)*tan(c)/(a
+ b*tan(c)), Eq(d, 0)), (2*B*a**2*b*log(a/b + tan(c + d*x))/(2*a**2*b**2*d + 2*b**4*d) - 2*B*a*b**2*d*x/(2*a**
2*b**2*d + 2*b**4*d) + B*b**3*log(tan(c + d*x)**2 + 1)/(2*a**2*b**2*d + 2*b**4*d) - 2*C*a**3*log(a/b + tan(c +
d*x))/(2*a**2*b**2*d + 2*b**4*d) + 2*C*a**2*b*tan(c + d*x)/(2*a**2*b**2*d + 2*b**4*d) - C*a*b**2*log(tan(c +
d*x)**2 + 1)/(2*a**2*b**2*d + 2*b**4*d) - 2*C*b**3*d*x/(2*a**2*b**2*d + 2*b**4*d) + 2*C*b**3*tan(c + d*x)/(2*a
**2*b**2*d + 2*b**4*d), True))

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Giac [A]  time = 1.5514, size = 149, normalized size = 1.48 \begin{align*} -\frac{\frac{2 \,{\left (B a + C b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} + \frac{{\left (C a - B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac{2 \,{\left (C a^{3} - B a^{2} b\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{2} + b^{4}} - \frac{2 \, C \tan \left (d x + c\right )}{b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(B*a + C*b)*(d*x + c)/(a^2 + b^2) + (C*a - B*b)*log(tan(d*x + c)^2 + 1)/(a^2 + b^2) + 2*(C*a^3 - B*a^2
*b)*log(abs(b*tan(d*x + c) + a))/(a^2*b^2 + b^4) - 2*C*tan(d*x + c)/b)/d