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

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

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3606, 3626, 3617, 31, 3475} \[ \frac {a^2 (A b-a B) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )}-\frac {(A b-a B) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac {x (a A+b B)}{a^2+b^2}+\frac {B \tan (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.59, size = 118, normalized size = 1.17 \[ \frac {\frac {2 a^2 (A b-a B) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )}+\frac {i (A+i B) \log (-\tan (c+d x)+i)}{a+i b}-\frac {(B+i A) \log (\tan (c+d x)+i)}{a-i b}+\frac {2 B \tan (c+d x)}{b}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.84, size = 149, normalized size = 1.48 \[ -\frac {2 \, {\left (A a b^{2} + B b^{3}\right )} d x + {\left (B a^{3} - A a^{2} b\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - {\left (B a^{3} - A a^{2} b + B a b^{2} - A b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (B a^{2} b + B b^{3}\right )} \tan \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} + b^{4}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.59, size = 110, normalized size = 1.09 \[ -\frac {\frac {2 \, {\left (A a + B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} + \frac {{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{2} + b^{4}} - \frac {2 \, B \tan \left (d x + c\right )}{b}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.22, size = 179, normalized size = 1.77 \[ \frac {B \tan \left (d x +c \right )}{b d}+\frac {a^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) A}{d b \left (a^{2}+b^{2}\right )}-\frac {a^{3} B \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{2} \left (a^{2}+b^{2}\right ) d}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A b}{2 d \left (a^{2}+b^{2}\right )}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a B}{2 d \left (a^{2}+b^{2}\right )}-\frac {A \arctan \left (\tan \left (d x +c \right )\right ) a}{d \left (a^{2}+b^{2}\right )}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b}{d \left (a^{2}+b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.99, size = 109, normalized size = 1.08 \[ -\frac {\frac {2 \, {\left (A a + B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} + \frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{2} + b^{4}} + \frac {{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, B \tan \left (d x + c\right )}{b}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.41, size = 117, normalized size = 1.16 \[ \frac {B\,\mathrm {tan}\left (c+d\,x\right )}{b\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^3-A\,a^2\,b\right )}{d\,\left (a^2\,b^2+b^4\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.53, size = 1015, normalized size = 10.05 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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