3.5.57 \(\int \frac {\tan ^5(c+d x)}{a+b \tan (c+d x)} \, dx\) [457]

Optimal. Leaf size=125 \[ \frac {b x}{a^2+b^2}-\frac {a \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}-\frac {a^5 \log (a+b \tan (c+d x))}{b^4 \left (a^2+b^2\right ) d}+\frac {\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac {a \tan ^2(c+d x)}{2 b^2 d}+\frac {\tan ^3(c+d x)}{3 b d} \]

[Out]

b*x/(a^2+b^2)-a*ln(cos(d*x+c))/(a^2+b^2)/d-a^5*ln(a+b*tan(d*x+c))/b^4/(a^2+b^2)/d+(a^2-b^2)*tan(d*x+c)/b^3/d-1
/2*a*tan(d*x+c)^2/b^2/d+1/3*tan(d*x+c)^3/b/d

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Rubi [A]
time = 0.26, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3647, 3728, 3729, 3707, 3698, 31, 3556} \begin {gather*} -\frac {a \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac {b x}{a^2+b^2}+\frac {\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac {a^5 \log (a+b \tan (c+d x))}{b^4 d \left (a^2+b^2\right )}-\frac {a \tan ^2(c+d x)}{2 b^2 d}+\frac {\tan ^3(c+d x)}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + b*Tan[c + d*x]),x]

[Out]

(b*x)/(a^2 + b^2) - (a*Log[Cos[c + d*x]])/((a^2 + b^2)*d) - (a^5*Log[a + b*Tan[c + d*x]])/(b^4*(a^2 + b^2)*d)
+ ((a^2 - b^2)*Tan[c + d*x])/(b^3*d) - (a*Tan[c + d*x]^2)/(2*b^2*d) + Tan[c + d*x]^3/(3*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 3556

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

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3698

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 3707

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 3728

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] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d
*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3729

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

Rubi steps

\begin {align*} \int \frac {\tan ^5(c+d x)}{a+b \tan (c+d x)} \, dx &=\frac {\tan ^3(c+d x)}{3 b d}+\frac {\int \frac {\tan ^2(c+d x) \left (-3 a-3 b \tan (c+d x)-3 a \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{3 b}\\ &=-\frac {a \tan ^2(c+d x)}{2 b^2 d}+\frac {\tan ^3(c+d x)}{3 b d}+\frac {\int \frac {\tan (c+d x) \left (6 a^2+6 \left (a^2-b^2\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{6 b^2}\\ &=\frac {\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac {a \tan ^2(c+d x)}{2 b^2 d}+\frac {\tan ^3(c+d x)}{3 b d}+\frac {\int \frac {-6 a \left (a^2-b^2\right )+6 b^3 \tan (c+d x)-6 a \left (a^2-b^2\right ) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{6 b^3}\\ &=\frac {b x}{a^2+b^2}+\frac {\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac {a \tan ^2(c+d x)}{2 b^2 d}+\frac {\tan ^3(c+d x)}{3 b d}+\frac {a \int \tan (c+d x) \, dx}{a^2+b^2}-\frac {a^5 \int \frac {1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^3 \left (a^2+b^2\right )}\\ &=\frac {b x}{a^2+b^2}-\frac {a \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}+\frac {\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac {a \tan ^2(c+d x)}{2 b^2 d}+\frac {\tan ^3(c+d x)}{3 b d}-\frac {a^5 \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^4 \left (a^2+b^2\right ) d}\\ &=\frac {b x}{a^2+b^2}-\frac {a \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}-\frac {a^5 \log (a+b \tan (c+d x))}{b^4 \left (a^2+b^2\right ) d}+\frac {\left (a^2-b^2\right ) \tan (c+d x)}{b^3 d}-\frac {a \tan ^2(c+d x)}{2 b^2 d}+\frac {\tan ^3(c+d x)}{3 b d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.62, size = 155, normalized size = 1.24 \begin {gather*} \frac {\log (i-\tan (c+d x))}{2 (a+i b) d}+\frac {\log (i+\tan (c+d x))}{2 (a-i b) d}-\frac {a^5 \log (a+b \tan (c+d x))}{b^4 \left (a^2+b^2\right ) d}+\frac {a^2 \tan (c+d x)}{b^3 d}-\frac {\tan (c+d x)}{b d}-\frac {a \tan ^2(c+d x)}{2 b^2 d}+\frac {\tan ^3(c+d x)}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + b*Tan[c + d*x]),x]

[Out]

Log[I - Tan[c + d*x]]/(2*(a + I*b)*d) + Log[I + Tan[c + d*x]]/(2*(a - I*b)*d) - (a^5*Log[a + b*Tan[c + d*x]])/
(b^4*(a^2 + b^2)*d) + (a^2*Tan[c + d*x])/(b^3*d) - Tan[c + d*x]/(b*d) - (a*Tan[c + d*x]^2)/(2*b^2*d) + Tan[c +
d*x]^3/(3*b*d)

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Maple [A]
time = 0.18, size = 119, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {\frac {\frac {b^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {a b \left (\tan ^{2}\left (d x +c \right )\right )}{2}+a^{2} \tan \left (d x +c \right )-b^{2} \tan \left (d x +c \right )}{b^{3}}+\frac {\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+b \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {a^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{4} \left (a^{2}+b^{2}\right )}}{d}\)	\(119\)
default	\(\frac {\frac {\frac {b^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {a b \left (\tan ^{2}\left (d x +c \right )\right )}{2}+a^{2} \tan \left (d x +c \right )-b^{2} \tan \left (d x +c \right )}{b^{3}}+\frac {\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+b \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {a^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{4} \left (a^{2}+b^{2}\right )}}{d}\)	\(119\)
norman	\(\frac {b x}{a^{2}+b^{2}}+\frac {\left (a^{2}-b^{2}\right ) \tan \left (d x +c \right )}{b^{3} d}+\frac {\tan ^{3}\left (d x +c \right )}{3 b d}-\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 b^{2} d}+\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{2}+b^{2}\right )}-\frac {a^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{4} \left (a^{2}+b^{2}\right ) d}\)	\(126\)
risch	\(\frac {i x}{i b -a}-\frac {2 i a^{3} x}{b^{4}}-\frac {2 i a^{3} c}{b^{4} d}+\frac {2 i a x}{b^{2}}+\frac {2 i a c}{b^{2} d}+\frac {2 i a^{5} x}{b^{4} \left (a^{2}+b^{2}\right )}+\frac {2 i a^{5} c}{b^{4} d \left (a^{2}+b^{2}\right )}+\frac {2 i \left (3 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 a^{2}-4 b^{2}\right )}{3 b^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{4} d}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{2} d}-\frac {a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{4} d \left (a^{2}+b^{2}\right )}\)	\(308\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/b^3*(1/3*b^2*tan(d*x+c)^3-1/2*a*b*tan(d*x+c)^2+a^2*tan(d*x+c)-b^2*tan(d*x+c))+1/(a^2+b^2)*(1/2*a*ln(1+t
an(d*x+c)^2)+b*arctan(tan(d*x+c)))-1/b^4*a^5/(a^2+b^2)*ln(a+b*tan(d*x+c)))

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Maxima [A]
time = 0.53, size = 123, normalized size = 0.98 \begin {gather*} -\frac {\frac {6 \, a^{5} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{4} + b^{6}} - \frac {6 \, {\left (d x + c\right )} b}{a^{2} + b^{2}} - \frac {3 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, b^{2} \tan \left (d x + c\right )^{3} - 3 \, a b \tan \left (d x + c\right )^{2} + 6 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )}{b^{3}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/6*(6*a^5*log(b*tan(d*x + c) + a)/(a^2*b^4 + b^6) - 6*(d*x + c)*b/(a^2 + b^2) - 3*a*log(tan(d*x + c)^2 + 1)/
(a^2 + b^2) - (2*b^2*tan(d*x + c)^3 - 3*a*b*tan(d*x + c)^2 + 6*(a^2 - b^2)*tan(d*x + c))/b^3)/d

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Fricas [A]
time = 1.72, size = 159, normalized size = 1.27 \begin {gather*} \frac {6 \, b^{5} d x - 3 \, a^{5} \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 ) + 2 \, {\left (a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )^{3} - 3 \, {\left (a^{3} b^{2} + a b^{4}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{5} - a b^{4}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 6 \, {\left (a^{4} b - b^{5}\right )} \tan \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} + b^{6}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(6*b^5*d*x - 3*a^5*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) + 2*(a^2*b^3
+ b^5)*tan(d*x + c)^3 - 3*(a^3*b^2 + a*b^4)*tan(d*x + c)^2 + 3*(a^5 - a*b^4)*log(1/(tan(d*x + c)^2 + 1)) + 6*(
a^4*b - b^5)*tan(d*x + c))/((a^2*b^4 + b^6)*d)

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Sympy [C] Result contains complex when optimal does not.
time = 0.79, size = 821, normalized size = 6.57 \begin {gather*} \begin {cases} \tilde {\infty } x \tan ^{4}{\left (c \right )} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {\frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {\tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {\tan ^{2}{\left (c + d x \right )}}{2 d}}{a} & \text {for}\: b = 0 \\\frac {15 d x \tan {\left (c + d x \right )}}{6 b d \tan {\left (c + d x \right )} - 6 i b d} - \frac {15 i d x}{6 b d \tan {\left (c + d x \right )} - 6 i b d} - \frac {6 i \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{6 b d \tan {\left (c + d x \right )} - 6 i b d} - \frac {6 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{6 b d \tan {\left (c + d x \right )} - 6 i b d} + \frac {2 \tan ^{4}{\left (c + d x \right )}}{6 b d \tan {\left (c + d x \right )} - 6 i b d} + \frac {i \tan ^{3}{\left (c + d x \right )}}{6 b d \tan {\left (c + d x \right )} - 6 i b d} - \frac {9 \tan ^{2}{\left (c + d x \right )}}{6 b d \tan {\left (c + d x \right )} - 6 i b d} - \frac {15}{6 b d \tan {\left (c + d x \right )} - 6 i b d} & \text {for}\: a = - i b \\\frac {15 d x \tan {\left (c + d x \right )}}{6 b d \tan {\left (c + d x \right )} + 6 i b d} + \frac {15 i d x}{6 b d \tan {\left (c + d x \right )} + 6 i b d} + \frac {6 i \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{6 b d \tan {\left (c + d x \right )} + 6 i b d} - \frac {6 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{6 b d \tan {\left (c + d x \right )} + 6 i b d} + \frac {2 \tan ^{4}{\left (c + d x \right )}}{6 b d \tan {\left (c + d x \right )} + 6 i b d} - \frac {i \tan ^{3}{\left (c + d x \right )}}{6 b d \tan {\left (c + d x \right )} + 6 i b d} - \frac {9 \tan ^{2}{\left (c + d x \right )}}{6 b d \tan {\left (c + d x \right )} + 6 i b d} - \frac {15}{6 b d \tan {\left (c + d x \right )} + 6 i b d} & \text {for}\: a = i b \\\frac {x \tan ^{5}{\left (c \right )}}{a + b \tan {\left (c \right )}} & \text {for}\: d = 0 \\- \frac {6 a^{5} \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{6 a^{2} b^{4} d + 6 b^{6} d} + \frac {6 a^{4} b \tan {\left (c + d x \right )}}{6 a^{2} b^{4} d + 6 b^{6} d} - \frac {3 a^{3} b^{2} \tan ^{2}{\left (c + d x \right )}}{6 a^{2} b^{4} d + 6 b^{6} d} + \frac {2 a^{2} b^{3} \tan ^{3}{\left (c + d x \right )}}{6 a^{2} b^{4} d + 6 b^{6} d} + \frac {3 a b^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{6 a^{2} b^{4} d + 6 b^{6} d} - \frac {3 a b^{4} \tan ^{2}{\left (c + d x \right )}}{6 a^{2} b^{4} d + 6 b^{6} d} + \frac {6 b^{5} d x}{6 a^{2} b^{4} d + 6 b^{6} d} + \frac {2 b^{5} \tan ^{3}{\left (c + d x \right )}}{6 a^{2} b^{4} d + 6 b^{6} d} - \frac {6 b^{5} \tan {\left (c + d x \right )}}{6 a^{2} b^{4} d + 6 b^{6} d} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5/(a+b*tan(d*x+c)),x)

[Out]

Piecewise((zoo*x*tan(c)**4, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((log(tan(c + d*x)**2 + 1)/(2*d) + tan(c + d*x)**
4/(4*d) - tan(c + d*x)**2/(2*d))/a, Eq(b, 0)), (15*d*x*tan(c + d*x)/(6*b*d*tan(c + d*x) - 6*I*b*d) - 15*I*d*x/
(6*b*d*tan(c + d*x) - 6*I*b*d) - 6*I*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(6*b*d*tan(c + d*x) - 6*I*b*d) - 6*
log(tan(c + d*x)**2 + 1)/(6*b*d*tan(c + d*x) - 6*I*b*d) + 2*tan(c + d*x)**4/(6*b*d*tan(c + d*x) - 6*I*b*d) + I
*tan(c + d*x)**3/(6*b*d*tan(c + d*x) - 6*I*b*d) - 9*tan(c + d*x)**2/(6*b*d*tan(c + d*x) - 6*I*b*d) - 15/(6*b*d
*tan(c + d*x) - 6*I*b*d), Eq(a, -I*b)), (15*d*x*tan(c + d*x)/(6*b*d*tan(c + d*x) + 6*I*b*d) + 15*I*d*x/(6*b*d*
tan(c + d*x) + 6*I*b*d) + 6*I*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(6*b*d*tan(c + d*x) + 6*I*b*d) - 6*log(tan
(c + d*x)**2 + 1)/(6*b*d*tan(c + d*x) + 6*I*b*d) + 2*tan(c + d*x)**4/(6*b*d*tan(c + d*x) + 6*I*b*d) - I*tan(c
+ d*x)**3/(6*b*d*tan(c + d*x) + 6*I*b*d) - 9*tan(c + d*x)**2/(6*b*d*tan(c + d*x) + 6*I*b*d) - 15/(6*b*d*tan(c
+ d*x) + 6*I*b*d), Eq(a, I*b)), (x*tan(c)**5/(a + b*tan(c)), Eq(d, 0)), (-6*a**5*log(a/b + tan(c + d*x))/(6*a*
*2*b**4*d + 6*b**6*d) + 6*a**4*b*tan(c + d*x)/(6*a**2*b**4*d + 6*b**6*d) - 3*a**3*b**2*tan(c + d*x)**2/(6*a**2
*b**4*d + 6*b**6*d) + 2*a**2*b**3*tan(c + d*x)**3/(6*a**2*b**4*d + 6*b**6*d) + 3*a*b**4*log(tan(c + d*x)**2 +
1)/(6*a**2*b**4*d + 6*b**6*d) - 3*a*b**4*tan(c + d*x)**2/(6*a**2*b**4*d + 6*b**6*d) + 6*b**5*d*x/(6*a**2*b**4*
d + 6*b**6*d) + 2*b**5*tan(c + d*x)**3/(6*a**2*b**4*d + 6*b**6*d) - 6*b**5*tan(c + d*x)/(6*a**2*b**4*d + 6*b**
6*d), True))

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Giac [A]
time = 1.96, size = 129, normalized size = 1.03 \begin {gather*} -\frac {\frac {6 \, a^{5} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{4} + b^{6}} - \frac {6 \, {\left (d x + c\right )} b}{a^{2} + b^{2}} - \frac {3 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, b^{2} \tan \left (d x + c\right )^{3} - 3 \, a b \tan \left (d x + c\right )^{2} + 6 \, a^{2} \tan \left (d x + c\right ) - 6 \, b^{2} \tan \left (d x + c\right )}{b^{3}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(6*a^5*log(abs(b*tan(d*x + c) + a))/(a^2*b^4 + b^6) - 6*(d*x + c)*b/(a^2 + b^2) - 3*a*log(tan(d*x + c)^2
+ 1)/(a^2 + b^2) - (2*b^2*tan(d*x + c)^3 - 3*a*b*tan(d*x + c)^2 + 6*a^2*tan(d*x + c) - 6*b^2*tan(d*x + c))/b^3
)/d

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Mupad [B]
time = 4.08, size = 138, normalized size = 1.10 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,b\,d}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,b^2\,d}-\frac {a^5\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d\,\left (a^2\,b^4+b^6\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^5/(a + b*tan(c + d*x)),x)

[Out]

log(tan(c + d*x) + 1i)/(2*d*(a - b*1i)) - (tan(c + d*x)*(1/b - a^2/b^3))/d + (log(tan(c + d*x) - 1i)*1i)/(2*d*
(a*1i - b)) + tan(c + d*x)^3/(3*b*d) - (a*tan(c + d*x)^2)/(2*b^2*d) - (a^5*log(a + b*tan(c + d*x)))/(d*(b^6 +
a^2*b^4))

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