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

Optimal. Leaf size=85 \[ -\frac {a (b B-a C) \log (a+b \tan (c+d x))}{b d \left (a^2+b^2\right )}-\frac {(a B+b C) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac {x (b B-a C)}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1629, 635, 203, 260} \[ -\frac {a (b B-a C) \log (a+b \tan (c+d x))}{b d \left (a^2+b^2\right )}-\frac {(a B+b C) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac {x (b B-a C)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

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

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [C]  time = 0.20, size = 98, normalized size = 1.15 \[ \frac {b (a-i b) (B+i C) \log (-\tan (c+d x)+i)+b (a+i b) (B-i C) \log (\tan (c+d x)+i)+2 a (a C-b B) \log (a+b \tan (c+d x))}{2 b d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 2.31, size = 110, normalized size = 1.29 \[ -\frac {2 \, {\left (C a b - B b^{2}\right )} d x - {\left (C a^{2} - B a b\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + {\left (C a^{2} + C b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, {\left (a^{2} b + b^{3}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.52, size = 95, normalized size = 1.12 \[ -\frac {\frac {2 \, {\left (C a - B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {{\left (B a + C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, {\left (C a^{2} - B a b\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.28, size = 159, normalized size = 1.87 \[ -\frac {a \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )}+\frac {a^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) C}{d \left (a^{2}+b^{2}\right ) b}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a B}{2 d \left (a^{2}+b^{2}\right )}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) C b}{2 d \left (a^{2}+b^{2}\right )}+\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b}{d \left (a^{2}+b^{2}\right )}-\frac {C \arctan \left (\tan \left (d x +c \right )\right ) a}{d \left (a^{2}+b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.77, size = 94, normalized size = 1.11 \[ -\frac {\frac {2 \, {\left (C a - B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {2 \, {\left (C a^{2} - B a b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b + b^{3}} - \frac {{\left (B a + C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 9.07, size = 100, normalized size = 1.18 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}-\frac {a\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,b-C\,a\right )}{b\,d\,\left (a^2+b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.12, size = 724, normalized size = 8.52 \[ \begin {cases} \frac {\tilde {\infty } x \left (B \tan {\relax (c )} + C \tan ^{2}{\relax (c )}\right )}{\tan {\relax (c )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\- \frac {B d x \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i B d x}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {B}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {i C d x \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {C d x}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {C \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i C \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i C}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} & \text {for}\: a = - i b \\- \frac {B d x \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i B d x}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {B}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {i C d x \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {C d x}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {C \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i C \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i C}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} & \text {for}\: a = i b \\\frac {\frac {B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - C x + \frac {C \tan {\left (c + d x \right )}}{d}}{a} & \text {for}\: b = 0 \\\frac {x \left (B \tan {\relax (c )} + C \tan ^{2}{\relax (c )}\right )}{a + b \tan {\relax (c )}} & \text {for}\: d = 0 \\- \frac {2 B a b \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} b d + 2 b^{3} d} + \frac {B a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} b d + 2 b^{3} d} + \frac {2 B b^{2} d x}{2 a^{2} b d + 2 b^{3} d} + \frac {2 C a^{2} \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} b d + 2 b^{3} d} - \frac {2 C a b d x}{2 a^{2} b d + 2 b^{3} d} + \frac {C b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} b d + 2 b^{3} d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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