3.250 \(\int (a+b \tan ^2(c+d x))^4 \, dx\)

Optimal. Leaf size=115 \[ \frac {b^2 \left (6 a^2-4 a b+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {b (2 a-b) \left (2 a^2-2 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b^3 (4 a-b) \tan ^5(c+d x)}{5 d}+x (a-b)^4+\frac {b^4 \tan ^7(c+d x)}{7 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 115, 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, 390, 203} \[ \frac {b^2 \left (6 a^2-4 a b+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {b (2 a-b) \left (2 a^2-2 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b^3 (4 a-b) \tan ^5(c+d x)}{5 d}+x (a-b)^4+\frac {b^4 \tan ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^2)^4,x]

[Out]

(a - b)^4*x + ((2*a - b)*b*(2*a^2 - 2*a*b + b^2)*Tan[c + d*x])/d + (b^2*(6*a^2 - 4*a*b + b^2)*Tan[c + d*x]^3)/
(3*d) + ((4*a - b)*b^3*Tan[c + d*x]^5)/(5*d) + (b^4*Tan[c + d*x]^7)/(7*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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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 ^2(c+d x)\right )^4 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^4}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left ((2 a-b) b \left (2 a^2-2 a b+b^2\right )+b^2 \left (6 a^2-4 a b+b^2\right ) x^2+(4 a-b) b^3 x^4+b^4 x^6+\frac {(a-b)^4}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {(2 a-b) b \left (2 a^2-2 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b^2 \left (6 a^2-4 a b+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {(4 a-b) b^3 \tan ^5(c+d x)}{5 d}+\frac {b^4 \tan ^7(c+d x)}{7 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 {(2 a-b) b \left (2 a^2-2 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b^2 \left (6 a^2-4 a b+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {(4 a-b) b^3 \tan ^5(c+d x)}{5 d}+\frac {b^4 \tan ^7(c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 1.91, size = 137, normalized size = 1.19 \[ \frac {\tan (c+d x) \left (b \left (35 b \left (6 a^2-4 a b+b^2\right ) \tan ^2(c+d x)+105 \left (4 a^3-6 a^2 b+4 a b^2-b^3\right )+21 b^2 (4 a-b) \tan ^4(c+d x)+15 b^3 \tan ^6(c+d x)\right )+\frac {105 (a-b)^4 \tanh ^{-1}\left (\sqrt {-\tan ^2(c+d x)}\right )}{\sqrt {-\tan ^2(c+d x)}}\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x]^2)^4,x]

[Out]

(Tan[c + d*x]*((105*(a - b)^4*ArcTanh[Sqrt[-Tan[c + d*x]^2]])/Sqrt[-Tan[c + d*x]^2] + b*(105*(4*a^3 - 6*a^2*b
+ 4*a*b^2 - b^3) + 35*b*(6*a^2 - 4*a*b + b^2)*Tan[c + d*x]^2 + 21*(4*a - b)*b^2*Tan[c + d*x]^4 + 15*b^3*Tan[c
+ d*x]^6)))/(105*d)

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fricas [A]  time = 0.41, size = 134, normalized size = 1.17 \[ \frac {15 \, b^{4} \tan \left (d x + c\right )^{7} + 21 \, {\left (4 \, a b^{3} - b^{4}\right )} \tan \left (d x + c\right )^{5} + 35 \, {\left (6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{3} + 105 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} d x + 105 \, {\left (4 \, a^{3} b - 6 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} \tan \left (d x + c\right )}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c)^2)^4,x, algorithm="fricas")

[Out]

1/105*(15*b^4*tan(d*x + c)^7 + 21*(4*a*b^3 - b^4)*tan(d*x + c)^5 + 35*(6*a^2*b^2 - 4*a*b^3 + b^4)*tan(d*x + c)
^3 + 105*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d*x + 105*(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 [B]  time = 35.01, size = 2209, normalized size = 19.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c)^2)^4,x, algorithm="giac")

[Out]

1/105*(105*a^4*d*x*tan(d*x)^7*tan(c)^7 - 420*a^3*b*d*x*tan(d*x)^7*tan(c)^7 + 630*a^2*b^2*d*x*tan(d*x)^7*tan(c)
^7 - 420*a*b^3*d*x*tan(d*x)^7*tan(c)^7 + 105*b^4*d*x*tan(d*x)^7*tan(c)^7 - 735*a^4*d*x*tan(d*x)^6*tan(c)^6 + 2
940*a^3*b*d*x*tan(d*x)^6*tan(c)^6 - 4410*a^2*b^2*d*x*tan(d*x)^6*tan(c)^6 + 2940*a*b^3*d*x*tan(d*x)^6*tan(c)^6
- 735*b^4*d*x*tan(d*x)^6*tan(c)^6 - 420*a^3*b*tan(d*x)^7*tan(c)^6 + 630*a^2*b^2*tan(d*x)^7*tan(c)^6 - 420*a*b^
3*tan(d*x)^7*tan(c)^6 + 105*b^4*tan(d*x)^7*tan(c)^6 - 420*a^3*b*tan(d*x)^6*tan(c)^7 + 630*a^2*b^2*tan(d*x)^6*t
an(c)^7 - 420*a*b^3*tan(d*x)^6*tan(c)^7 + 105*b^4*tan(d*x)^6*tan(c)^7 + 2205*a^4*d*x*tan(d*x)^5*tan(c)^5 - 882
0*a^3*b*d*x*tan(d*x)^5*tan(c)^5 + 13230*a^2*b^2*d*x*tan(d*x)^5*tan(c)^5 - 8820*a*b^3*d*x*tan(d*x)^5*tan(c)^5 +
2205*b^4*d*x*tan(d*x)^5*tan(c)^5 - 210*a^2*b^2*tan(d*x)^7*tan(c)^4 + 140*a*b^3*tan(d*x)^7*tan(c)^4 - 35*b^4*t
an(d*x)^7*tan(c)^4 + 2520*a^3*b*tan(d*x)^6*tan(c)^5 - 4410*a^2*b^2*tan(d*x)^6*tan(c)^5 + 2940*a*b^3*tan(d*x)^6
*tan(c)^5 - 735*b^4*tan(d*x)^6*tan(c)^5 + 2520*a^3*b*tan(d*x)^5*tan(c)^6 - 4410*a^2*b^2*tan(d*x)^5*tan(c)^6 +
2940*a*b^3*tan(d*x)^5*tan(c)^6 - 735*b^4*tan(d*x)^5*tan(c)^6 - 210*a^2*b^2*tan(d*x)^4*tan(c)^7 + 140*a*b^3*tan
(d*x)^4*tan(c)^7 - 35*b^4*tan(d*x)^4*tan(c)^7 - 3675*a^4*d*x*tan(d*x)^4*tan(c)^4 + 14700*a^3*b*d*x*tan(d*x)^4*
tan(c)^4 - 22050*a^2*b^2*d*x*tan(d*x)^4*tan(c)^4 + 14700*a*b^3*d*x*tan(d*x)^4*tan(c)^4 - 3675*b^4*d*x*tan(d*x)
^4*tan(c)^4 - 84*a*b^3*tan(d*x)^7*tan(c)^2 + 21*b^4*tan(d*x)^7*tan(c)^2 + 840*a^2*b^2*tan(d*x)^6*tan(c)^3 - 98
0*a*b^3*tan(d*x)^6*tan(c)^3 + 245*b^4*tan(d*x)^6*tan(c)^3 - 6300*a^3*b*tan(d*x)^5*tan(c)^4 + 11970*a^2*b^2*tan
(d*x)^5*tan(c)^4 - 8820*a*b^3*tan(d*x)^5*tan(c)^4 + 2205*b^4*tan(d*x)^5*tan(c)^4 - 6300*a^3*b*tan(d*x)^4*tan(c
)^5 + 11970*a^2*b^2*tan(d*x)^4*tan(c)^5 - 8820*a*b^3*tan(d*x)^4*tan(c)^5 + 2205*b^4*tan(d*x)^4*tan(c)^5 + 840*
a^2*b^2*tan(d*x)^3*tan(c)^6 - 980*a*b^3*tan(d*x)^3*tan(c)^6 + 245*b^4*tan(d*x)^3*tan(c)^6 - 84*a*b^3*tan(d*x)^
2*tan(c)^7 + 21*b^4*tan(d*x)^2*tan(c)^7 + 3675*a^4*d*x*tan(d*x)^3*tan(c)^3 - 14700*a^3*b*d*x*tan(d*x)^3*tan(c)
^3 + 22050*a^2*b^2*d*x*tan(d*x)^3*tan(c)^3 - 14700*a*b^3*d*x*tan(d*x)^3*tan(c)^3 + 3675*b^4*d*x*tan(d*x)^3*tan
(c)^3 - 15*b^4*tan(d*x)^7 + 168*a*b^3*tan(d*x)^6*tan(c) - 147*b^4*tan(d*x)^6*tan(c) - 1260*a^2*b^2*tan(d*x)^5*
tan(c)^2 + 1680*a*b^3*tan(d*x)^5*tan(c)^2 - 735*b^4*tan(d*x)^5*tan(c)^2 + 8400*a^3*b*tan(d*x)^4*tan(c)^3 - 163
80*a^2*b^2*tan(d*x)^4*tan(c)^3 + 12600*a*b^3*tan(d*x)^4*tan(c)^3 - 3675*b^4*tan(d*x)^4*tan(c)^3 + 8400*a^3*b*t
an(d*x)^3*tan(c)^4 - 16380*a^2*b^2*tan(d*x)^3*tan(c)^4 + 12600*a*b^3*tan(d*x)^3*tan(c)^4 - 3675*b^4*tan(d*x)^3
*tan(c)^4 - 1260*a^2*b^2*tan(d*x)^2*tan(c)^5 + 1680*a*b^3*tan(d*x)^2*tan(c)^5 - 735*b^4*tan(d*x)^2*tan(c)^5 +
168*a*b^3*tan(d*x)*tan(c)^6 - 147*b^4*tan(d*x)*tan(c)^6 - 15*b^4*tan(c)^7 - 2205*a^4*d*x*tan(d*x)^2*tan(c)^2 +
8820*a^3*b*d*x*tan(d*x)^2*tan(c)^2 - 13230*a^2*b^2*d*x*tan(d*x)^2*tan(c)^2 + 8820*a*b^3*d*x*tan(d*x)^2*tan(c)
^2 - 2205*b^4*d*x*tan(d*x)^2*tan(c)^2 - 84*a*b^3*tan(d*x)^5 + 21*b^4*tan(d*x)^5 + 840*a^2*b^2*tan(d*x)^4*tan(c
) - 980*a*b^3*tan(d*x)^4*tan(c) + 245*b^4*tan(d*x)^4*tan(c) - 6300*a^3*b*tan(d*x)^3*tan(c)^2 + 11970*a^2*b^2*t
an(d*x)^3*tan(c)^2 - 8820*a*b^3*tan(d*x)^3*tan(c)^2 + 2205*b^4*tan(d*x)^3*tan(c)^2 - 6300*a^3*b*tan(d*x)^2*tan
(c)^3 + 11970*a^2*b^2*tan(d*x)^2*tan(c)^3 - 8820*a*b^3*tan(d*x)^2*tan(c)^3 + 2205*b^4*tan(d*x)^2*tan(c)^3 + 84
0*a^2*b^2*tan(d*x)*tan(c)^4 - 980*a*b^3*tan(d*x)*tan(c)^4 + 245*b^4*tan(d*x)*tan(c)^4 - 84*a*b^3*tan(c)^5 + 21
*b^4*tan(c)^5 + 735*a^4*d*x*tan(d*x)*tan(c) - 2940*a^3*b*d*x*tan(d*x)*tan(c) + 4410*a^2*b^2*d*x*tan(d*x)*tan(c
) - 2940*a*b^3*d*x*tan(d*x)*tan(c) + 735*b^4*d*x*tan(d*x)*tan(c) - 210*a^2*b^2*tan(d*x)^3 + 140*a*b^3*tan(d*x)
^3 - 35*b^4*tan(d*x)^3 + 2520*a^3*b*tan(d*x)^2*tan(c) - 4410*a^2*b^2*tan(d*x)^2*tan(c) + 2940*a*b^3*tan(d*x)^2
*tan(c) - 735*b^4*tan(d*x)^2*tan(c) + 2520*a^3*b*tan(d*x)*tan(c)^2 - 4410*a^2*b^2*tan(d*x)*tan(c)^2 + 2940*a*b
^3*tan(d*x)*tan(c)^2 - 735*b^4*tan(d*x)*tan(c)^2 - 210*a^2*b^2*tan(c)^3 + 140*a*b^3*tan(c)^3 - 35*b^4*tan(c)^3
- 105*a^4*d*x + 420*a^3*b*d*x - 630*a^2*b^2*d*x + 420*a*b^3*d*x - 105*b^4*d*x - 420*a^3*b*tan(d*x) + 630*a^2*
b^2*tan(d*x) - 420*a*b^3*tan(d*x) + 105*b^4*tan(d*x) - 420*a^3*b*tan(c) + 630*a^2*b^2*tan(c) - 420*a*b^3*tan(c
) + 105*b^4*tan(c))/(d*tan(d*x)^7*tan(c)^7 - 7*d*tan(d*x)^6*tan(c)^6 + 21*d*tan(d*x)^5*tan(c)^5 - 35*d*tan(d*x
)^4*tan(c)^4 + 35*d*tan(d*x)^3*tan(c)^3 - 21*d*tan(d*x)^2*tan(c)^2 + 7*d*tan(d*x)*tan(c) - d)

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maple [B]  time = 0.04, size = 242, normalized size = 2.10 \[ \frac {b^{4} \left (\tan ^{7}\left (d x +c \right )\right )}{7 d}+\frac {4 \left (\tan ^{5}\left (d x +c \right )\right ) a \,b^{3}}{5 d}-\frac {\left (\tan ^{5}\left (d x +c \right )\right ) b^{4}}{5 d}+\frac {2 \left (\tan ^{3}\left (d x +c \right )\right ) a^{2} b^{2}}{d}-\frac {4 \left (\tan ^{3}\left (d x +c \right )\right ) a \,b^{3}}{3 d}+\frac {b^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {4 a^{3} b \tan \left (d x +c \right )}{d}-\frac {6 a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {4 a \,b^{3} \tan \left (d x +c \right )}{d}-\frac {b^{4} \tan \left (d x +c \right )}{d}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{4}}{d}-\frac {4 \arctan \left (\tan \left (d x +c \right )\right ) a^{3} b}{d}+\frac {6 \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b^{2}}{d}-\frac {4 \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{3}}{d}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) b^{4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c)^2)^4,x)

[Out]

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

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maxima [A]  time = 0.42, size = 162, normalized size = 1.41 \[ a^{4} x - \frac {4 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} b}{d} + \frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} b^{2}}{d} + \frac {4 \, {\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} a b^{3}}{15 \, d} + \frac {{\left (15 \, \tan \left (d x + c\right )^{7} - 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 105 \, d x + 105 \, c - 105 \, \tan \left (d x + c\right )\right )} b^{4}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c)^2)^4,x, algorithm="maxima")

[Out]

a^4*x - 4*(d*x + c - tan(d*x + c))*a^3*b/d + 2*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^2*b^2/d + 4/1
5*(3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c))*a*b^3/d + 1/105*(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))*b^4/d

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mupad [B]  time = 11.38, size = 164, normalized size = 1.43 \[ \frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,{\left (a-b\right )}^4}{a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4}\right )\,{\left (a-b\right )}^4}{d}+\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^7}{7\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (2\,a^2\,b^2-\frac {4\,a\,b^3}{3}+\frac {b^4}{3}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {4\,a\,b^3}{5}-\frac {b^4}{5}\right )}{d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3\,b-6\,a^2\,b^2+4\,a\,b^3-b^4\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x)^2)^4,x)

[Out]

(atan((tan(c + d*x)*(a - b)^4)/(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))*(a - b)^4)/d + (b^4*tan(c + d*x)^7
)/(7*d) + (tan(c + d*x)^3*(b^4/3 - (4*a*b^3)/3 + 2*a^2*b^2))/d + (tan(c + d*x)^5*((4*a*b^3)/5 - b^4/5))/d + (t
an(c + d*x)*(4*a*b^3 + 4*a^3*b - b^4 - 6*a^2*b^2))/d

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sympy [A]  time = 1.27, size = 209, normalized size = 1.82 \[ \begin {cases} a^{4} x - 4 a^{3} b x + \frac {4 a^{3} b \tan {\left (c + d x \right )}}{d} + 6 a^{2} b^{2} x + \frac {2 a^{2} b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac {6 a^{2} b^{2} \tan {\left (c + d x \right )}}{d} - 4 a b^{3} x + \frac {4 a b^{3} \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac {4 a b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac {4 a b^{3} \tan {\left (c + d x \right )}}{d} + b^{4} x + \frac {b^{4} \tan ^{7}{\left (c + d x \right )}}{7 d} - \frac {b^{4} \tan ^{5}{\left (c + d x \right )}}{5 d} + \frac {b^{4} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{4} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan ^{2}{\relax (c )}\right )^{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c)**2)**4,x)

[Out]

Piecewise((a**4*x - 4*a**3*b*x + 4*a**3*b*tan(c + d*x)/d + 6*a**2*b**2*x + 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)**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)**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)**2)**4, True))

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