3.1.30 \(\int \frac {\cot ^3(c+d x) (B \tan (c+d x)+C \tan ^2(c+d x))}{a+b \tan (c+d x)} \, dx\) [30]

Optimal. Leaf size=103 \[ -\frac {(a B+b C) x}{a^2+b^2}-\frac {B \cot (c+d x)}{a d}-\frac {(b B-a C) \log (\sin (c+d x))}{a^2 d}+\frac {b^2 (b B-a C) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 \left (a^2+b^2\right ) d} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3713, 3690, 3732, 3611, 3556} \begin {gather*} \frac {b^2 (b B-a C) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 d \left (a^2+b^2\right )}-\frac {x (a B+b C)}{a^2+b^2}-\frac {(b B-a C) \log (\sin (c+d x))}{a^2 d}-\frac {B \cot (c+d x)}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^3*(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*Cot[c + d*x])/(a*d) - ((b*B - a*C)*Log[Sin[c + d*x]])/(a^2*d) + (b^2*(b*B
- a*C)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/(a^2*(a^2 + b^2)*d)

Rule 3556

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

Rule 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3690

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

Rule 3713

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 3732

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

Rubi steps

\begin {align*} \int \frac {\cot ^3(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 {\cot ^2(c+d x) (B+C \tan (c+d x))}{a+b \tan (c+d x)} \, dx\\ &=-\frac {B \cot (c+d x)}{a d}-\frac {\int \frac {\cot (c+d x) \left (b B-a C+a B \tan (c+d x)+b B \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{a}\\ &=-\frac {(a B+b C) x}{a^2+b^2}-\frac {B \cot (c+d x)}{a d}-\frac {(b B-a C) \int \cot (c+d x) \, dx}{a^2}+\frac {\left (b^2 (b B-a C)\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^2 \left (a^2+b^2\right )}\\ &=-\frac {(a B+b C) x}{a^2+b^2}-\frac {B \cot (c+d x)}{a d}-\frac {(b B-a C) \log (\sin (c+d x))}{a^2 d}+\frac {b^2 (b B-a C) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 \left (a^2+b^2\right ) d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.61, size = 138, normalized size = 1.34 \begin {gather*} \frac {-\frac {2 B \cot (c+d x)}{a}+\frac {i (B+i C) \log (i-\tan (c+d x))}{a+i b}+\frac {2 (-b B+a C) \log (\tan (c+d x))}{a^2}-\frac {(i B+C) \log (i+\tan (c+d x))}{a-i b}+\frac {2 b^2 (b B-a C) \log (a+b \tan (c+d x))}{a^2 \left (a^2+b^2\right )}}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^3*(B*Tan[c + d*x] + C*Tan[c + d*x]^2))/(a + b*Tan[c + d*x]),x]

[Out]

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

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Maple [A]
time = 0.48, size = 123, normalized size = 1.19

method	result	size
derivativedivides	\(\frac {-\frac {B}{a \tan \left (d x +c \right )}+\frac {\left (-B b +C a \right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{2}}+\frac {\left (B b -C a \right ) b^{2} \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2} \left (a^{2}+b^{2}\right )}+\frac {\frac {\left (B b -C a \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-a B -C b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\)	\(123\)
default	\(\frac {-\frac {B}{a \tan \left (d x +c \right )}+\frac {\left (-B b +C a \right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{2}}+\frac {\left (B b -C a \right ) b^{2} \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2} \left (a^{2}+b^{2}\right )}+\frac {\frac {\left (B b -C a \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-a B -C b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\)	\(123\)
norman	\(\frac {-\frac {B \tan \left (d x +c \right )}{a d}-\frac {\left (a B +C b \right ) x \left (\tan ^{2}\left (d x +c \right )\right )}{a^{2}+b^{2}}}{\tan \left (d x +c \right )^{2}}+\frac {b^{2} \left (B b -C a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2} d \left (a^{2}+b^{2}\right )}-\frac {\left (B b -C a \right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}+\frac {\left (B b -C a \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{2}+b^{2}\right )}\)	\(148\)
risch	\(\frac {x B}{i b -a}-\frac {i x C}{i b -a}+\frac {2 i B b x}{a^{2}}+\frac {2 i B b c}{a^{2} d}-\frac {2 i C x}{a}-\frac {2 i C c}{d a}-\frac {2 i b^{3} B x}{a^{2} \left (a^{2}+b^{2}\right )}-\frac {2 i b^{3} B c}{a^{2} d \left (a^{2}+b^{2}\right )}+\frac {2 i b^{2} C x}{a \left (a^{2}+b^{2}\right )}+\frac {2 i b^{2} C c}{a d \left (a^{2}+b^{2}\right )}-\frac {2 i B}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C}{d a}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B}{a^{2} d \left (a^{2}+b^{2}\right )}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) C}{a d \left (a^{2}+b^{2}\right )}\)	\(320\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 131, normalized size = 1.27 \begin {gather*} -\frac {\frac {2 \, {\left (B a + C b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} + \frac {2 \, {\left (C a b^{2} - B b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 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 - B b\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{2}} + \frac {2 \, B}{a \tan \left (d x + c\right )}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 1.19, size = 177, normalized size = 1.72 \begin {gather*} -\frac {2 \, B a^{3} + 2 \, B a b^{2} + 2 \, {\left (B a^{3} + C a^{2} b\right )} d x \tan \left (d x + c\right ) - {\left (C a^{3} - B a^{2} b + C a b^{2} - B b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) + {\left (C a b^{2} - B b^{3}\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 ) \tan \left (d x + c\right )}{2 \, {\left (a^{4} + a^{2} b^{2}\right )} d \tan \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [C] Result contains complex when optimal does not.
time = 5.02, size = 2064, normalized size = 20.04 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(B*tan(d*x+c)+C*tan(d*x+c)**2)/(a+b*tan(d*x+c)),x)

[Out]

Piecewise((nan, Eq(a, 0) & Eq(b, 0) & Eq(c, 0) & Eq(d, 0)), ((-B*x - B/(d*tan(c + d*x)) - C*log(tan(c + d*x)**
2 + 1)/(2*d) + C*log(tan(c + d*x))/d)/a, Eq(b, 0)), ((B*log(tan(c + d*x)**2 + 1)/(2*d) - B*log(tan(c + d*x))/d
- B/(2*d*tan(c + d*x)**2) - C*x - C/(d*tan(c + d*x)))/b, Eq(a, 0)), (-3*I*B*d*x*tan(c + d*x)**2/(2*b*d*tan(c
+ d*x)**2 - 2*I*b*d*tan(c + d*x)) - 3*B*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - B*lo
g(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) + I*B*log(tan(c + d*x)**
2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) + 2*B*log(tan(c + d*x))*tan(c + d*x)**2/(2*
b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - 2*I*B*log(tan(c + d*x))*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 - 2*
I*b*d*tan(c + d*x)) - 3*I*B*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - 2*B/(2*b*d*tan(c + d
*x)**2 - 2*I*b*d*tan(c + d*x)) + C*d*x*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - I*C*d*
x*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - I*C*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(
2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - C*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x)**2
- 2*I*b*d*tan(c + d*x)) + 2*I*C*log(tan(c + d*x))*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*
x)) + 2*C*log(tan(c + d*x))*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) + C*tan(c + d*x)/(2*b*
d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)), Eq(a, -I*b)), (3*I*B*d*x*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 + 2
*I*b*d*tan(c + d*x)) - 3*B*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) - B*log(tan(c + d*x
)**2 + 1)*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) - I*B*log(tan(c + d*x)**2 + 1)*tan(c
+ d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) + 2*B*log(tan(c + d*x))*tan(c + d*x)**2/(2*b*d*tan(c + d
*x)**2 + 2*I*b*d*tan(c + d*x)) + 2*I*B*log(tan(c + d*x))*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c +
d*x)) + 3*I*B*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) - 2*B/(2*b*d*tan(c + d*x)**2 + 2*I*
b*d*tan(c + d*x)) + C*d*x*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) + I*C*d*x*tan(c + d*x
)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) + I*C*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(2*b*d*tan(c +
d*x)**2 + 2*I*b*d*tan(c + d*x)) - C*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*ta
n(c + d*x)) - 2*I*C*log(tan(c + d*x))*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) + 2*C*log
(tan(c + d*x))*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) + C*tan(c + d*x)/(2*b*d*tan(c + d*x
)**2 + 2*I*b*d*tan(c + d*x)), Eq(a, I*b)), (nan, Eq(c, -d*x)), (x*(B*tan(c) + C*tan(c)**2)*cot(c)**3/(a + b*ta
n(c)), Eq(d, 0)), (-2*B*a**3*d*x*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) - 2*B*a**3/
(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) + B*a**2*b*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**4*
d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) - 2*B*a**2*b*log(tan(c + d*x))*tan(c + d*x)/(2*a**4*d*tan(c + d*x
) + 2*a**2*b**2*d*tan(c + d*x)) - 2*B*a*b**2/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) + 2*B*b**3*l
og(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) - 2*B*b**3*log(tan(c
+ d*x))*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) - C*a**3*log(tan(c + d*x)**2 + 1)*ta
n(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) + 2*C*a**3*log(tan(c + d*x))*tan(c + d*x)/(2*a
**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) - 2*C*a**2*b*d*x*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2
*b**2*d*tan(c + d*x)) - 2*C*a*b**2*log(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d
*tan(c + d*x)) + 2*C*a*b**2*log(tan(c + d*x))*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)
), True))

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Giac [A]
time = 1.14, size = 157, normalized size = 1.52 \begin {gather*} -\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 b^{3} - B b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + a^{2} b^{3}} - \frac {2 \, {\left (C a - B b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac {2 \, {\left (C a \tan \left (d x + c\right ) - B b \tan \left (d x + c\right ) + B a\right )}}{a^{2} \tan \left (d x + c\right )}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 10.34, size = 140, normalized size = 1.36 \begin {gather*} \frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,b^3-C\,a\,b^2\right )}{d\,\left (a^4+a^2\,b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,b-C\,a\right )}{a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}-\frac {B\,\mathrm {cot}\left (c+d\,x\right )}{a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cot(c + d*x)^3*(B*tan(c + d*x) + C*tan(c + d*x)^2))/(a + b*tan(c + d*x)),x)

[Out]

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

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