3.456 \(\int \frac {\tan ^6(c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=154 \[ -\frac {b \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac {a x}{a^2+b^2}-\frac {a \left (a^2-b^2\right ) \tan (c+d x)}{b^4 d}+\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 b^3 d}+\frac {a^6 \log (a+b \tan (c+d x))}{b^5 d \left (a^2+b^2\right )}-\frac {a \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^4(c+d x)}{4 b d} \]

[Out]

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

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Rubi [A]  time = 0.58, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3566, 3647, 3648, 3627, 3617, 31, 3475} \[ \frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 b^3 d}-\frac {a \left (a^2-b^2\right ) \tan (c+d x)}{b^4 d}+\frac {a^6 \log (a+b \tan (c+d x))}{b^5 d \left (a^2+b^2\right )}-\frac {b \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac {a x}{a^2+b^2}-\frac {a \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^4(c+d x)}{4 b d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + b*Tan[c + d*x]),x]

[Out]

-((a*x)/(a^2 + b^2)) - (b*Log[Cos[c + d*x]])/((a^2 + b^2)*d) + (a^6*Log[a + b*Tan[c + d*x]])/(b^5*(a^2 + b^2)*
d) - (a*(a^2 - b^2)*Tan[c + d*x])/(b^4*d) + ((a^2 - b^2)*Tan[c + d*x]^2)/(2*b^3*d) - (a*Tan[c + d*x]^3)/(3*b^2
*d) + Tan[c + d*x]^4/(4*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 3566

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 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 3627

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

Rule 3647

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 3648

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 ^6(c+d x)}{a+b \tan (c+d x)} \, dx &=\frac {\tan ^4(c+d x)}{4 b d}+\frac {\int \frac {\tan ^3(c+d x) \left (-4 a-4 b \tan (c+d x)-4 a \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{4 b}\\ &=-\frac {a \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^4(c+d x)}{4 b d}+\frac {\int \frac {\tan ^2(c+d x) \left (12 a^2+12 \left (a^2-b^2\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{12 b^2}\\ &=\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 b^3 d}-\frac {a \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^4(c+d x)}{4 b d}+\frac {\int \frac {\tan (c+d x) \left (-24 a \left (a^2-b^2\right )+24 b^3 \tan (c+d x)-24 a \left (a^2-b^2\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{24 b^3}\\ &=-\frac {a \left (a^2-b^2\right ) \tan (c+d x)}{b^4 d}+\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 b^3 d}-\frac {a \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^4(c+d x)}{4 b d}+\frac {\int \frac {24 a^2 \left (a^2-b^2\right )+24 \left (a^4-a^2 b^2+b^4\right ) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{24 b^4}\\ &=-\frac {a x}{a^2+b^2}-\frac {a \left (a^2-b^2\right ) \tan (c+d x)}{b^4 d}+\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 b^3 d}-\frac {a \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^4(c+d x)}{4 b d}+\frac {a^6 \int \frac {1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^4 \left (a^2+b^2\right )}+\frac {b \int \tan (c+d x) \, dx}{a^2+b^2}\\ &=-\frac {a x}{a^2+b^2}-\frac {b \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}-\frac {a \left (a^2-b^2\right ) \tan (c+d x)}{b^4 d}+\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 b^3 d}-\frac {a \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^4(c+d x)}{4 b d}+\frac {a^6 \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^5 \left (a^2+b^2\right ) d}\\ &=-\frac {a x}{a^2+b^2}-\frac {b \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}+\frac {a^6 \log (a+b \tan (c+d x))}{b^5 \left (a^2+b^2\right ) d}-\frac {a \left (a^2-b^2\right ) \tan (c+d x)}{b^4 d}+\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 b^3 d}-\frac {a \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^4(c+d x)}{4 b d}\\ \end {align*}

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Mathematica [C]  time = 1.73, size = 167, normalized size = 1.08 \[ \frac {6 \left (2 a^6 \log (a+b \tan (c+d x))+b^5 (b+i a) \log (-\tan (c+d x)+i)+b^5 (b-i a) \log (\tan (c+d x)+i)\right )-12 a b \left (a^4-b^4\right ) \tan (c+d x)+6 b^2 \left (a^4-b^4\right ) \tan ^2(c+d x)+3 b^4 \left (a^2+b^2\right ) \tan ^4(c+d x)-4 a b^3 \left (a^2+b^2\right ) \tan ^3(c+d x)}{12 b^5 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^6/(a + b*Tan[c + d*x]),x]

[Out]

(6*(b^5*(I*a + b)*Log[I - Tan[c + d*x]] + b^5*((-I)*a + b)*Log[I + Tan[c + d*x]] + 2*a^6*Log[a + b*Tan[c + d*x
]]) - 12*a*b*(a^4 - b^4)*Tan[c + d*x] + 6*b^2*(a^4 - b^4)*Tan[c + d*x]^2 - 4*a*b^3*(a^2 + b^2)*Tan[c + d*x]^3
+ 3*b^4*(a^2 + b^2)*Tan[c + d*x]^4)/(12*b^5*(a^2 + b^2)*d)

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fricas [A]  time = 0.51, size = 181, normalized size = 1.18 \[ -\frac {12 \, a b^{5} d x - 6 \, a^{6} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 3 \, {\left (a^{2} b^{4} + b^{6}\right )} \tan \left (d x + c\right )^{4} + 4 \, {\left (a^{3} b^{3} + a b^{5}\right )} \tan \left (d x + c\right )^{3} - 6 \, {\left (a^{4} b^{2} - b^{6}\right )} \tan \left (d x + c\right )^{2} + 6 \, {\left (a^{6} + b^{6}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 12 \, {\left (a^{5} b - a b^{5}\right )} \tan \left (d x + c\right )}{12 \, {\left (a^{2} b^{5} + b^{7}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(12*a*b^5*d*x - 6*a^6*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) - 3*(a^2
*b^4 + b^6)*tan(d*x + c)^4 + 4*(a^3*b^3 + a*b^5)*tan(d*x + c)^3 - 6*(a^4*b^2 - b^6)*tan(d*x + c)^2 + 6*(a^6 +
b^6)*log(1/(tan(d*x + c)^2 + 1)) + 12*(a^5*b - a*b^5)*tan(d*x + c))/((a^2*b^5 + b^7)*d)

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giac [A]  time = 17.07, size = 158, normalized size = 1.03 \[ \frac {\frac {12 \, a^{6} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{5} + b^{7}} - \frac {12 \, {\left (d x + c\right )} a}{a^{2} + b^{2}} + \frac {6 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {3 \, b^{3} \tan \left (d x + c\right )^{4} - 4 \, a b^{2} \tan \left (d x + c\right )^{3} + 6 \, a^{2} b \tan \left (d x + c\right )^{2} - 6 \, b^{3} \tan \left (d x + c\right )^{2} - 12 \, a^{3} \tan \left (d x + c\right ) + 12 \, a b^{2} \tan \left (d x + c\right )}{b^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

1/12*(12*a^6*log(abs(b*tan(d*x + c) + a))/(a^2*b^5 + b^7) - 12*(d*x + c)*a/(a^2 + b^2) + 6*b*log(tan(d*x + c)^
2 + 1)/(a^2 + b^2) + (3*b^3*tan(d*x + c)^4 - 4*a*b^2*tan(d*x + c)^3 + 6*a^2*b*tan(d*x + c)^2 - 6*b^3*tan(d*x +
c)^2 - 12*a^3*tan(d*x + c) + 12*a*b^2*tan(d*x + c))/b^4)/d

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maple [A]  time = 0.18, size = 179, normalized size = 1.16 \[ \frac {\tan ^{4}\left (d x +c \right )}{4 b d}-\frac {a \left (\tan ^{3}\left (d x +c \right )\right )}{3 b^{2} d}+\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d \,b^{3}}-\frac {\tan ^{2}\left (d x +c \right )}{2 b d}-\frac {a^{3} \tan \left (d x +c \right )}{d \,b^{4}}+\frac {a \tan \left (d x +c \right )}{b^{2} d}+\frac {a^{6} \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{5} \left (a^{2}+b^{2}\right ) d}+\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{2}+b^{2}\right )}-\frac {a \arctan \left (\tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^6/(a+b*tan(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.81, size = 146, normalized size = 0.95 \[ \frac {\frac {12 \, a^{6} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{5} + b^{7}} - \frac {12 \, {\left (d x + c\right )} a}{a^{2} + b^{2}} + \frac {6 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {3 \, b^{3} \tan \left (d x + c\right )^{4} - 4 \, a b^{2} \tan \left (d x + c\right )^{3} + 6 \, {\left (a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{2} - 12 \, {\left (a^{3} - a b^{2}\right )} \tan \left (d x + c\right )}{b^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*a^6*log(b*tan(d*x + c) + a)/(a^2*b^5 + b^7) - 12*(d*x + c)*a/(a^2 + b^2) + 6*b*log(tan(d*x + c)^2 + 1
)/(a^2 + b^2) + (3*b^3*tan(d*x + c)^4 - 4*a*b^2*tan(d*x + c)^3 + 6*(a^2*b - b^3)*tan(d*x + c)^2 - 12*(a^3 - a*
b^2)*tan(d*x + c))/b^4)/d

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mupad [B]  time = 4.26, size = 165, normalized size = 1.07 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {1}{2\,b}-\frac {a^2}{2\,b^3}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{4\,b\,d}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,b^2\,d}+\frac {a^6\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d\,\left (a^2\,b^5+b^7\right )}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )}{b\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^6/(a + b*tan(c + d*x)),x)

[Out]

(log(tan(c + d*x) - 1i)*1i)/(2*d*(a + b*1i)) + log(tan(c + d*x) + 1i)/(2*d*(a*1i + b)) - (tan(c + d*x)^2*(1/(2
*b) - a^2/(2*b^3)))/d + tan(c + d*x)^4/(4*b*d) - (a*tan(c + d*x)^3)/(3*b^2*d) + (a^6*log(a + b*tan(c + d*x)))/
(d*(b^7 + a^2*b^5)) + (a*tan(c + d*x)*(1/b - a^2/b^3))/(b*d)

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sympy [A]  time = 3.54, size = 944, normalized size = 6.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**6/(a+b*tan(d*x+c)),x)

[Out]

Piecewise((zoo*x*tan(c)**5, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-30*d*x*tan(c + d*x)/(12*I*b*d*tan(c + d*x) + 12
*b*d) + 30*I*d*x/(12*I*b*d*tan(c + d*x) + 12*b*d) + 18*I*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(12*I*b*d*tan(c
+ d*x) + 12*b*d) + 18*log(tan(c + d*x)**2 + 1)/(12*I*b*d*tan(c + d*x) + 12*b*d) + 3*I*tan(c + d*x)**5/(12*I*b
*d*tan(c + d*x) + 12*b*d) - tan(c + d*x)**4/(12*I*b*d*tan(c + d*x) + 12*b*d) - 8*I*tan(c + d*x)**3/(12*I*b*d*t
an(c + d*x) + 12*b*d) + 12*tan(c + d*x)**2/(12*I*b*d*tan(c + d*x) + 12*b*d) + 30/(12*I*b*d*tan(c + d*x) + 12*b
*d), Eq(a, -I*b)), (-30*d*x*tan(c + d*x)/(-12*I*b*d*tan(c + d*x) + 12*b*d) - 30*I*d*x/(-12*I*b*d*tan(c + d*x)
+ 12*b*d) - 18*I*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(-12*I*b*d*tan(c + d*x) + 12*b*d) + 18*log(tan(c + d*x)
**2 + 1)/(-12*I*b*d*tan(c + d*x) + 12*b*d) - 3*I*tan(c + d*x)**5/(-12*I*b*d*tan(c + d*x) + 12*b*d) - tan(c + d
*x)**4/(-12*I*b*d*tan(c + d*x) + 12*b*d) + 8*I*tan(c + d*x)**3/(-12*I*b*d*tan(c + d*x) + 12*b*d) + 12*tan(c +
d*x)**2/(-12*I*b*d*tan(c + d*x) + 12*b*d) + 30/(-12*I*b*d*tan(c + d*x) + 12*b*d), Eq(a, I*b)), ((-x + tan(c +
d*x)**5/(5*d) - tan(c + d*x)**3/(3*d) + tan(c + d*x)/d)/a, Eq(b, 0)), (x*tan(c)**6/(a + b*tan(c)), Eq(d, 0)),
(12*a**6*log(a/b + tan(c + d*x))/(12*a**2*b**5*d + 12*b**7*d) - 12*a**5*b*tan(c + d*x)/(12*a**2*b**5*d + 12*b*
*7*d) + 6*a**4*b**2*tan(c + d*x)**2/(12*a**2*b**5*d + 12*b**7*d) - 4*a**3*b**3*tan(c + d*x)**3/(12*a**2*b**5*d
+ 12*b**7*d) + 3*a**2*b**4*tan(c + d*x)**4/(12*a**2*b**5*d + 12*b**7*d) - 12*a*b**5*d*x/(12*a**2*b**5*d + 12*
b**7*d) - 4*a*b**5*tan(c + d*x)**3/(12*a**2*b**5*d + 12*b**7*d) + 12*a*b**5*tan(c + d*x)/(12*a**2*b**5*d + 12*
b**7*d) + 6*b**6*log(tan(c + d*x)**2 + 1)/(12*a**2*b**5*d + 12*b**7*d) + 3*b**6*tan(c + d*x)**4/(12*a**2*b**5*
d + 12*b**7*d) - 6*b**6*tan(c + d*x)**2/(12*a**2*b**5*d + 12*b**7*d), True))

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