3.381 \(\int (a+b \tan ^4(c+d x))^4 \, dx\)

Optimal. Leaf size=216 \[ \frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tan ^7(c+d x)}{7 d}-\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tan ^3(c+d x)}{3 d}-\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b^3 (4 a+b) \tan ^{11}(c+d x)}{11 d}-\frac {b^3 (4 a+b) \tan ^9(c+d x)}{9 d}+x (a+b)^4+\frac {b^4 \tan ^{15}(c+d x)}{15 d}-\frac {b^4 \tan ^{13}(c+d x)}{13 d} \]

[Out]

(a+b)^4*x-b*(2*a+b)*(2*a^2+2*a*b+b^2)*tan(d*x+c)/d+1/3*b*(2*a+b)*(2*a^2+2*a*b+b^2)*tan(d*x+c)^3/d-1/5*b^2*(6*a
^2+4*a*b+b^2)*tan(d*x+c)^5/d+1/7*b^2*(6*a^2+4*a*b+b^2)*tan(d*x+c)^7/d-1/9*b^3*(4*a+b)*tan(d*x+c)^9/d+1/11*b^3*
(4*a+b)*tan(d*x+c)^11/d-1/13*b^4*tan(d*x+c)^13/d+1/15*b^4*tan(d*x+c)^15/d

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Rubi [A]  time = 0.13, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3661, 1154, 203} \[ \frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tan ^7(c+d x)}{7 d}-\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tan ^3(c+d x)}{3 d}-\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b^3 (4 a+b) \tan ^{11}(c+d x)}{11 d}-\frac {b^3 (4 a+b) \tan ^9(c+d x)}{9 d}+x (a+b)^4+\frac {b^4 \tan ^{15}(c+d x)}{15 d}-\frac {b^4 \tan ^{13}(c+d x)}{13 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^4)^4,x]

[Out]

(a + b)^4*x - (b*(2*a + b)*(2*a^2 + 2*a*b + b^2)*Tan[c + d*x])/d + (b*(2*a + b)*(2*a^2 + 2*a*b + b^2)*Tan[c +
d*x]^3)/(3*d) - (b^2*(6*a^2 + 4*a*b + b^2)*Tan[c + d*x]^5)/(5*d) + (b^2*(6*a^2 + 4*a*b + b^2)*Tan[c + d*x]^7)/
(7*d) - (b^3*(4*a + b)*Tan[c + d*x]^9)/(9*d) + (b^3*(4*a + b)*Tan[c + d*x]^11)/(11*d) - (b^4*Tan[c + d*x]^13)/
(13*d) + (b^4*Tan[c + d*x]^15)/(15*d)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 1154

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a
+ c*x^4)^p, x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \left (a+b \tan ^4(c+d x)\right )^4 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^4\right )^4}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-b (2 a+b) \left (2 a^2+2 a b+b^2\right )+b (2 a+b) \left (2 a^2+2 a b+b^2\right ) x^2-b^2 \left (6 a^2+4 a b+b^2\right ) x^4+b^2 \left (6 a^2+4 a b+b^2\right ) x^6-b^3 (4 a+b) x^8+b^3 (4 a+b) x^{10}-b^4 x^{12}+b^4 x^{14}+\frac {(a+b)^4}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tan ^3(c+d x)}{3 d}-\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tan ^7(c+d x)}{7 d}-\frac {b^3 (4 a+b) \tan ^9(c+d x)}{9 d}+\frac {b^3 (4 a+b) \tan ^{11}(c+d x)}{11 d}-\frac {b^4 \tan ^{13}(c+d x)}{13 d}+\frac {b^4 \tan ^{15}(c+d x)}{15 d}+\frac {(a+b)^4 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=(a+b)^4 x-\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tan ^3(c+d x)}{3 d}-\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tan ^7(c+d x)}{7 d}-\frac {b^3 (4 a+b) \tan ^9(c+d x)}{9 d}+\frac {b^3 (4 a+b) \tan ^{11}(c+d x)}{11 d}-\frac {b^4 \tan ^{13}(c+d x)}{13 d}+\frac {b^4 \tan ^{15}(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 5.38, size = 196, normalized size = 0.91 \[ \frac {b \tan (c+d x) \left (6435 b \left (6 a^2+4 a b+b^2\right ) \tan ^6(c+d x)-9009 b \left (6 a^2+4 a b+b^2\right ) \tan ^4(c+d x)+15015 \left (4 a^3+6 a^2 b+4 a b^2+b^3\right ) \tan ^2(c+d x)-45045 \left (4 a^3+6 a^2 b+4 a b^2+b^3\right )+4095 b^2 (4 a+b) \tan ^{10}(c+d x)-5005 b^2 (4 a+b) \tan ^8(c+d x)+3003 b^3 \tan ^{14}(c+d x)-3465 b^3 \tan ^{12}(c+d x)\right )}{45045 d}+\frac {(a+b)^4 \tan ^{-1}(\tan (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x]^4)^4,x]

[Out]

((a + b)^4*ArcTan[Tan[c + d*x]])/d + (b*Tan[c + d*x]*(-45045*(4*a^3 + 6*a^2*b + 4*a*b^2 + b^3) + 15015*(4*a^3
+ 6*a^2*b + 4*a*b^2 + b^3)*Tan[c + d*x]^2 - 9009*b*(6*a^2 + 4*a*b + b^2)*Tan[c + d*x]^4 + 6435*b*(6*a^2 + 4*a*
b + b^2)*Tan[c + d*x]^6 - 5005*b^2*(4*a + b)*Tan[c + d*x]^8 + 4095*b^2*(4*a + b)*Tan[c + d*x]^10 - 3465*b^3*Ta
n[c + d*x]^12 + 3003*b^3*Tan[c + d*x]^14))/(45045*d)

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fricas [A]  time = 0.68, size = 225, normalized size = 1.04 \[ \frac {3003 \, b^{4} \tan \left (d x + c\right )^{15} - 3465 \, b^{4} \tan \left (d x + c\right )^{13} + 4095 \, {\left (4 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{11} - 5005 \, {\left (4 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{9} + 6435 \, {\left (6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{7} - 9009 \, {\left (6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{5} + 15015 \, {\left (4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{3} + 45045 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} d x - 45045 \, {\left (4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )}{45045 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+tan(d*x+c)^4*b)^4,x, algorithm="fricas")

[Out]

1/45045*(3003*b^4*tan(d*x + c)^15 - 3465*b^4*tan(d*x + c)^13 + 4095*(4*a*b^3 + b^4)*tan(d*x + c)^11 - 5005*(4*
a*b^3 + b^4)*tan(d*x + c)^9 + 6435*(6*a^2*b^2 + 4*a*b^3 + b^4)*tan(d*x + c)^7 - 9009*(6*a^2*b^2 + 4*a*b^3 + b^
4)*tan(d*x + c)^5 + 15015*(4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*tan(d*x + c)^3 + 45045*(a^4 + 4*a^3*b + 6*a^2*
b^2 + 4*a*b^3 + b^4)*d*x - 45045*(4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*tan(d*x + c))/d

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+tan(d*x+c)^4*b)^4,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.03, size = 412, normalized size = 1.91 \[ -\frac {b^{4} \left (\tan ^{9}\left (d x +c \right )\right )}{9 d}+\frac {4 \left (\tan ^{11}\left (d x +c \right )\right ) a \,b^{3}}{11 d}-\frac {6 \left (\tan ^{5}\left (d x +c \right )\right ) a^{2} b^{2}}{5 d}+\frac {2 \left (\tan ^{3}\left (d x +c \right )\right ) a^{2} b^{2}}{d}+\frac {4 \left (\tan ^{7}\left (d x +c \right )\right ) a \,b^{3}}{7 d}+\frac {6 \left (\tan ^{7}\left (d x +c \right )\right ) a^{2} b^{2}}{7 d}-\frac {4 \left (\tan ^{9}\left (d x +c \right )\right ) a \,b^{3}}{9 d}+\frac {b^{4} \left (\tan ^{15}\left (d x +c \right )\right )}{15 d}-\frac {\left (\tan ^{5}\left (d x +c \right )\right ) b^{4}}{5 d}+\frac {4 \left (\tan ^{3}\left (d x +c \right )\right ) a \,b^{3}}{3 d}+\frac {4 \left (\tan ^{3}\left (d x +c \right )\right ) a^{3} b}{3 d}-\frac {4 \left (\tan ^{5}\left (d x +c \right )\right ) a \,b^{3}}{5 d}+\frac {4 \arctan \left (\tan \left (d x +c \right )\right ) a^{3} b}{d}+\frac {4 \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{3}}{d}+\frac {6 \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b^{2}}{d}-\frac {4 a^{3} b \tan \left (d x +c \right )}{d}+\frac {b^{4} \left (\tan ^{11}\left (d x +c \right )\right )}{11 d}+\frac {b^{4} \left (\tan ^{7}\left (d x +c \right )\right )}{7 d}-\frac {b^{4} \tan \left (d x +c \right )}{d}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{4}}{d}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) b^{4}}{d}-\frac {b^{4} \left (\tan ^{13}\left (d x +c \right )\right )}{13 d}+\frac {b^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {6 a^{2} b^{2} \tan \left (d x +c \right )}{d}-\frac {4 a \,b^{3} \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c)^4)^4,x)

[Out]

4/3/d*tan(d*x+c)^3*a*b^3+4/3/d*tan(d*x+c)^3*a^3*b-4/5/d*tan(d*x+c)^5*a*b^3+4/7/d*tan(d*x+c)^7*a*b^3+6/7/d*tan(
d*x+c)^7*a^2*b^2-4/9/d*tan(d*x+c)^9*a*b^3+4/11/d*tan(d*x+c)^11*a*b^3+4/d*arctan(tan(d*x+c))*a^3*b+4/d*arctan(t
an(d*x+c))*a*b^3-6/5/d*tan(d*x+c)^5*a^2*b^2+2/d*tan(d*x+c)^3*a^2*b^2+6/d*arctan(tan(d*x+c))*a^2*b^2-4/d*a^3*b*
tan(d*x+c)-1/5/d*tan(d*x+c)^5*b^4-1/d*b^4*tan(d*x+c)+1/d*arctan(tan(d*x+c))*a^4+1/d*arctan(tan(d*x+c))*b^4+1/3
*b^4*tan(d*x+c)^3/d-6*a^2*b^2*tan(d*x+c)/d-4*a*b^3*tan(d*x+c)/d-1/9*b^4*tan(d*x+c)^9/d+1/11*b^4*tan(d*x+c)^11/
d-1/13*b^4*tan(d*x+c)^13/d+1/15*b^4*tan(d*x+c)^15/d+1/7*b^4*tan(d*x+c)^7/d

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maxima [A]  time = 0.79, size = 265, normalized size = 1.23 \[ a^{4} x + \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} b}{3 \, d} + \frac {2 \, {\left (15 \, \tan \left (d x + c\right )^{7} - 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 105 \, d x + 105 \, c - 105 \, \tan \left (d x + c\right )\right )} a^{2} b^{2}}{35 \, d} + \frac {4 \, {\left (315 \, \tan \left (d x + c\right )^{11} - 385 \, \tan \left (d x + c\right )^{9} + 495 \, \tan \left (d x + c\right )^{7} - 693 \, \tan \left (d x + c\right )^{5} + 1155 \, \tan \left (d x + c\right )^{3} + 3465 \, d x + 3465 \, c - 3465 \, \tan \left (d x + c\right )\right )} a b^{3}}{3465 \, d} + \frac {{\left (3003 \, \tan \left (d x + c\right )^{15} - 3465 \, \tan \left (d x + c\right )^{13} + 4095 \, \tan \left (d x + c\right )^{11} - 5005 \, \tan \left (d x + c\right )^{9} + 6435 \, \tan \left (d x + c\right )^{7} - 9009 \, \tan \left (d x + c\right )^{5} + 15015 \, \tan \left (d x + c\right )^{3} + 45045 \, d x + 45045 \, c - 45045 \, \tan \left (d x + c\right )\right )} b^{4}}{45045 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+tan(d*x+c)^4*b)^4,x, algorithm="maxima")

[Out]

a^4*x + 4/3*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^3*b/d + 2/35*(15*tan(d*x + c)^7 - 21*tan(d*x + c
)^5 + 35*tan(d*x + c)^3 + 105*d*x + 105*c - 105*tan(d*x + c))*a^2*b^2/d + 4/3465*(315*tan(d*x + c)^11 - 385*ta
n(d*x + c)^9 + 495*tan(d*x + c)^7 - 693*tan(d*x + c)^5 + 1155*tan(d*x + c)^3 + 3465*d*x + 3465*c - 3465*tan(d*
x + c))*a*b^3/d + 1/45045*(3003*tan(d*x + c)^15 - 3465*tan(d*x + c)^13 + 4095*tan(d*x + c)^11 - 5005*tan(d*x +
c)^9 + 6435*tan(d*x + c)^7 - 9009*tan(d*x + c)^5 + 15015*tan(d*x + c)^3 + 45045*d*x + 45045*c - 45045*tan(d*x
+ c))*b^4/d

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mupad [B]  time = 11.59, size = 271, normalized size = 1.25 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {4\,a^3\,b}{3}+2\,a^2\,b^2+\frac {4\,a\,b^3}{3}+\frac {b^4}{3}\right )}{d}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,{\left (a+b\right )}^4}{a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4}\right )\,{\left (a+b\right )}^4}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4\right )}{d}-\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^{13}}{13\,d}+\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^{15}}{15\,d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {6\,a^2\,b^2}{5}+\frac {4\,a\,b^3}{5}+\frac {b^4}{5}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^7\,\left (\frac {6\,a^2\,b^2}{7}+\frac {4\,a\,b^3}{7}+\frac {b^4}{7}\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^9\,\left (\frac {b^4}{9}+\frac {4\,a\,b^3}{9}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^{11}\,\left (\frac {b^4}{11}+\frac {4\,a\,b^3}{11}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x)^4)^4,x)

[Out]

(tan(c + d*x)^3*((4*a*b^3)/3 + (4*a^3*b)/3 + b^4/3 + 2*a^2*b^2))/d + (atan((tan(c + d*x)*(a + b)^4)/(4*a*b^3 +
4*a^3*b + a^4 + b^4 + 6*a^2*b^2))*(a + b)^4)/d - (tan(c + d*x)*(4*a*b^3 + 4*a^3*b + b^4 + 6*a^2*b^2))/d - (b^
4*tan(c + d*x)^13)/(13*d) + (b^4*tan(c + d*x)^15)/(15*d) - (tan(c + d*x)^5*((4*a*b^3)/5 + b^4/5 + (6*a^2*b^2)/
5))/d + (tan(c + d*x)^7*((4*a*b^3)/7 + b^4/7 + (6*a^2*b^2)/7))/d - (tan(c + d*x)^9*((4*a*b^3)/9 + b^4/9))/d +
(tan(c + d*x)^11*((4*a*b^3)/11 + b^4/11))/d

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sympy [A]  time = 7.07, size = 386, normalized size = 1.79 \[ \begin {cases} a^{4} x + 4 a^{3} b x + \frac {4 a^{3} b \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 a^{3} b \tan {\left (c + d x \right )}}{d} + 6 a^{2} b^{2} x + \frac {6 a^{2} b^{2} \tan ^{7}{\left (c + d x \right )}}{7 d} - \frac {6 a^{2} b^{2} \tan ^{5}{\left (c + d x \right )}}{5 d} + \frac {2 a^{2} b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac {6 a^{2} b^{2} \tan {\left (c + d x \right )}}{d} + 4 a b^{3} x + \frac {4 a b^{3} \tan ^{11}{\left (c + d x \right )}}{11 d} - \frac {4 a b^{3} \tan ^{9}{\left (c + d x \right )}}{9 d} + \frac {4 a b^{3} \tan ^{7}{\left (c + d x \right )}}{7 d} - \frac {4 a b^{3} \tan ^{5}{\left (c + d x \right )}}{5 d} + \frac {4 a b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 a b^{3} \tan {\left (c + d x \right )}}{d} + b^{4} x + \frac {b^{4} \tan ^{15}{\left (c + d x \right )}}{15 d} - \frac {b^{4} \tan ^{13}{\left (c + d x \right )}}{13 d} + \frac {b^{4} \tan ^{11}{\left (c + d x \right )}}{11 d} - \frac {b^{4} \tan ^{9}{\left (c + d x \right )}}{9 d} + \frac {b^{4} \tan ^{7}{\left (c + d x \right )}}{7 d} - \frac {b^{4} \tan ^{5}{\left (c + d x \right )}}{5 d} + \frac {b^{4} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{4} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan ^{4}{\relax (c )}\right )^{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+tan(d*x+c)**4*b)**4,x)

[Out]

Piecewise((a**4*x + 4*a**3*b*x + 4*a**3*b*tan(c + d*x)**3/(3*d) - 4*a**3*b*tan(c + d*x)/d + 6*a**2*b**2*x + 6*
a**2*b**2*tan(c + d*x)**7/(7*d) - 6*a**2*b**2*tan(c + d*x)**5/(5*d) + 2*a**2*b**2*tan(c + d*x)**3/d - 6*a**2*b
**2*tan(c + d*x)/d + 4*a*b**3*x + 4*a*b**3*tan(c + d*x)**11/(11*d) - 4*a*b**3*tan(c + d*x)**9/(9*d) + 4*a*b**3
*tan(c + d*x)**7/(7*d) - 4*a*b**3*tan(c + d*x)**5/(5*d) + 4*a*b**3*tan(c + d*x)**3/(3*d) - 4*a*b**3*tan(c + d*
x)/d + b**4*x + b**4*tan(c + d*x)**15/(15*d) - b**4*tan(c + d*x)**13/(13*d) + b**4*tan(c + d*x)**11/(11*d) - b
**4*tan(c + d*x)**9/(9*d) + b**4*tan(c + d*x)**7/(7*d) - b**4*tan(c + d*x)**5/(5*d) + b**4*tan(c + d*x)**3/(3*
d) - b**4*tan(c + d*x)/d, Ne(d, 0)), (x*(a + b*tan(c)**4)**4, True))

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