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

Optimal. Leaf size=127 \[ \frac {(a B+b C) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac {x (b B-a C)}{a^2+b^2}-\frac {a^3 (b B-a C) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )}+\frac {(b B-a C) \tan (c+d x)}{b^2 d}+\frac {C \tan ^2(c+d x)}{2 b d} \]

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

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Rubi [A]  time = 0.47, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {3632, 3607, 3647, 3626, 3617, 31, 3475} \[ -\frac {a^3 (b B-a C) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )}+\frac {(a B+b C) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac {x (b B-a C)}{a^2+b^2}+\frac {(b B-a C) \tan (c+d x)}{b^2 d}+\frac {C \tan ^2(c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^2*(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^3*(b*B - a*C)*Log[a + b*
Tan[c + d*x]])/(b^3*(a^2 + b^2)*d) + ((b*B - a*C)*Tan[c + d*x])/(b^2*d) + (C*Tan[c + d*x]^2)/(2*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 3607

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*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*
f*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[a^2*A*d*(m +
n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m
- 1) - b*(A*b + a*B)*d*(m + n))*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] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&
!(IGtQ[n, 1] && ( !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 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]

Rule 3632

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 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])))

Rubi steps

\begin {align*} \int \frac {\tan ^2(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 {\tan ^3(c+d x) (B+C \tan (c+d x))}{a+b \tan (c+d x)} \, dx\\ &=\frac {C \tan ^2(c+d x)}{2 b d}+\frac {\int \frac {\tan (c+d x) \left (-2 a C-2 b C \tan (c+d x)+2 (b B-a C) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 b}\\ &=\frac {(b B-a C) \tan (c+d x)}{b^2 d}+\frac {C \tan ^2(c+d x)}{2 b d}+\frac {\int \frac {-2 a (b B-a C)-2 b^2 B \tan (c+d x)-2 \left (a b B-a^2 C+b^2 C\right ) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{2 b^2}\\ &=-\frac {(b B-a C) x}{a^2+b^2}+\frac {(b B-a C) \tan (c+d x)}{b^2 d}+\frac {C \tan ^2(c+d x)}{2 b d}-\frac {\left (a^3 (b B-a C)\right ) \int \frac {1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^2 \left (a^2+b^2\right )}-\frac {(a B+b C) \int \tan (c+d x) \, dx}{a^2+b^2}\\ &=-\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 {(b B-a C) \tan (c+d x)}{b^2 d}+\frac {C \tan ^2(c+d x)}{2 b d}-\frac {\left (a^3 (b B-a C)\right ) \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^3 \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^3 (b B-a C) \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right ) d}+\frac {(b B-a C) \tan (c+d x)}{b^2 d}+\frac {C \tan ^2(c+d x)}{2 b d}\\ \end {align*}

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Mathematica [C]  time = 1.45, size = 138, normalized size = 1.09 \[ \frac {\frac {2 a^3 (a C-b B) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )}+\frac {2 (b B-a C) \tan (c+d x)}{b}-\frac {b (B+i C) \log (-\tan (c+d x)+i)}{a+i b}-\frac {b (B-i C) \log (\tan (c+d x)+i)}{a-i b}+C \tan ^2(c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.87, size = 190, normalized size = 1.50 \[ \frac {2 \, {\left (C a b^{3} - B b^{4}\right )} d x + {\left (C a^{2} b^{2} + C b^{4}\right )} \tan \left (d x + c\right )^{2} + {\left (C a^{4} - B a^{3} b\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - {\left (C a^{4} - B a^{3} b - B a b^{3} - C b^{4}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (C a^{3} b - B a^{2} b^{2} + C a b^{3} - B b^{4}\right )} \tan \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} + b^{5}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 2.11, size = 135, normalized size = 1.06 \[ \frac {\frac {2 \, {\left (C a - B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {{\left (B a + C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {2 \, {\left (C a^{4} - B a^{3} b\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{3} + b^{5}} + \frac {C b \tan \left (d x + c\right )^{2} - 2 \, C a \tan \left (d x + c\right ) + 2 \, B b \tan \left (d x + c\right )}{b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(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^4 - B*a^3*
b)*log(abs(b*tan(d*x + c) + a))/(a^2*b^3 + b^5) + (C*b*tan(d*x + c)^2 - 2*C*a*tan(d*x + c) + 2*B*b*tan(d*x + c
))/b^2)/d

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maple [A]  time = 0.26, size = 211, normalized size = 1.66 \[ \frac {C \left (\tan ^{2}\left (d x +c \right )\right )}{2 b d}+\frac {B \tan \left (d x +c \right )}{b d}-\frac {C \tan \left (d x +c \right ) a}{d \,b^{2}}-\frac {a^{3} B \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{2} \left (a^{2}+b^{2}\right ) d}+\frac {a^{4} \ln \left (a +b \tan \left (d x +c \right )\right ) C}{d \,b^{3} \left (a^{2}+b^{2}\right )}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a B}{2 d \left (a^{2}+b^{2}\right )}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) C b}{2 d \left (a^{2}+b^{2}\right )}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b}{d \left (a^{2}+b^{2}\right )}+\frac {C \arctan \left (\tan \left (d x +c \right )\right ) a}{d \left (a^{2}+b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*C*tan(d*x+c)^2/b/d+B*tan(d*x+c)/b/d-1/d/b^2*C*tan(d*x+c)*a-a^3*B*ln(a+b*tan(d*x+c))/b^2/(a^2+b^2)/d+1/d/b^
3*a^4/(a^2+b^2)*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 = 0.72, size = 130, normalized size = 1.02 \[ \frac {\frac {2 \, {\left (C a - B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} + \frac {2 \, {\left (C a^{4} - B a^{3} b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{3} + b^{5}} - \frac {{\left (B a + C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {C b \tan \left (d x + c\right )^{2} - 2 \, {\left (C a - B b\right )} \tan \left (d x + c\right )}{b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 9.07, size = 144, normalized size = 1.13 \[ \frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B}{b}-\frac {C\,a}{b^2}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (C\,a^4-B\,a^3\,b\right )}{d\,\left (a^2\,b^3+b^5\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}+\frac {C\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.03, size = 1309, normalized size = 10.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(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)), (-3*I*B*d*x*tan(c + d*x)/(2
*I*b*d*tan(c + d*x) + 2*b*d) - 3*B*d*x/(2*I*b*d*tan(c + d*x) + 2*b*d) - B*log(tan(c + d*x)**2 + 1)*tan(c + d*x
)/(2*I*b*d*tan(c + d*x) + 2*b*d) + I*B*log(tan(c + d*x)**2 + 1)/(2*I*b*d*tan(c + d*x) + 2*b*d) + 2*I*B*tan(c +
d*x)**2/(2*I*b*d*tan(c + d*x) + 2*b*d) + 3*I*B/(2*I*b*d*tan(c + d*x) + 2*b*d) + 3*C*d*x*tan(c + d*x)/(2*I*b*d
*tan(c + d*x) + 2*b*d) - 3*I*C*d*x/(2*I*b*d*tan(c + d*x) + 2*b*d) - 2*I*C*log(tan(c + d*x)**2 + 1)*tan(c + d*x
)/(2*I*b*d*tan(c + d*x) + 2*b*d) - 2*C*log(tan(c + d*x)**2 + 1)/(2*I*b*d*tan(c + d*x) + 2*b*d) + I*C*tan(c + d
*x)**3/(2*I*b*d*tan(c + d*x) + 2*b*d) - C*tan(c + d*x)**2/(2*I*b*d*tan(c + d*x) + 2*b*d) - 3*C/(2*I*b*d*tan(c
+ d*x) + 2*b*d), Eq(a, -I*b)), (3*I*B*d*x*tan(c + d*x)/(-2*I*b*d*tan(c + d*x) + 2*b*d) - 3*B*d*x/(-2*I*b*d*tan
(c + d*x) + 2*b*d) - B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(-2*I*b*d*tan(c + d*x) + 2*b*d) - I*B*log(tan(c +
d*x)**2 + 1)/(-2*I*b*d*tan(c + d*x) + 2*b*d) - 2*I*B*tan(c + d*x)**2/(-2*I*b*d*tan(c + d*x) + 2*b*d) - 3*I*B/
(-2*I*b*d*tan(c + d*x) + 2*b*d) + 3*C*d*x*tan(c + d*x)/(-2*I*b*d*tan(c + d*x) + 2*b*d) + 3*I*C*d*x/(-2*I*b*d*t
an(c + d*x) + 2*b*d) + 2*I*C*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(-2*I*b*d*tan(c + d*x) + 2*b*d) - 2*C*log(t
an(c + d*x)**2 + 1)/(-2*I*b*d*tan(c + d*x) + 2*b*d) - I*C*tan(c + d*x)**3/(-2*I*b*d*tan(c + d*x) + 2*b*d) - C*
tan(c + d*x)**2/(-2*I*b*d*tan(c + d*x) + 2*b*d) - 3*C/(-2*I*b*d*tan(c + d*x) + 2*b*d), Eq(a, I*b)), ((-B*log(t
an(c + d*x)**2 + 1)/(2*d) + B*tan(c + d*x)**2/(2*d) + C*x + C*tan(c + d*x)**3/(3*d) - C*tan(c + d*x)/d)/a, Eq(
b, 0)), (x*(B*tan(c) + C*tan(c)**2)*tan(c)**2/(a + b*tan(c)), Eq(d, 0)), (-2*B*a**3*b*log(a/b + tan(c + d*x))/
(2*a**2*b**3*d + 2*b**5*d) + 2*B*a**2*b**2*tan(c + d*x)/(2*a**2*b**3*d + 2*b**5*d) - B*a*b**3*log(tan(c + d*x)
**2 + 1)/(2*a**2*b**3*d + 2*b**5*d) - 2*B*b**4*d*x/(2*a**2*b**3*d + 2*b**5*d) + 2*B*b**4*tan(c + d*x)/(2*a**2*
b**3*d + 2*b**5*d) + 2*C*a**4*log(a/b + tan(c + d*x))/(2*a**2*b**3*d + 2*b**5*d) - 2*C*a**3*b*tan(c + d*x)/(2*
a**2*b**3*d + 2*b**5*d) + C*a**2*b**2*tan(c + d*x)**2/(2*a**2*b**3*d + 2*b**5*d) + 2*C*a*b**3*d*x/(2*a**2*b**3
*d + 2*b**5*d) - 2*C*a*b**3*tan(c + d*x)/(2*a**2*b**3*d + 2*b**5*d) - C*b**4*log(tan(c + d*x)**2 + 1)/(2*a**2*
b**3*d + 2*b**5*d) + C*b**4*tan(c + d*x)**2/(2*a**2*b**3*d + 2*b**5*d), True))

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