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

Optimal. Leaf size=128 \[ -\frac {b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}+\frac {4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}-x \left (a^4-6 a^2 b^2+b^4\right )+\frac {(a+b \tan (c+d x))^5}{5 b d}-\frac {b (a+b \tan (c+d x))^3}{3 d}-\frac {a b (a+b \tan (c+d x))^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3543, 3482, 3528, 3525, 3475} \[ -\frac {b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}+\frac {4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}-x \left (-6 a^2 b^2+a^4+b^4\right )+\frac {(a+b \tan (c+d x))^5}{5 b d}-\frac {b (a+b \tan (c+d x))^3}{3 d}-\frac {a b (a+b \tan (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2*(a + b*Tan[c + d*x])^4,x]

[Out]

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

Rule 3475

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

Rule 3482

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps

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

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Mathematica [C]  time = 0.75, size = 122, normalized size = 0.95 \[ \frac {30 b^2 \left (b^2-6 a^2\right ) \tan (c+d x)-60 a b^3 \tan ^2(c+d x)+\frac {6 (a+b \tan (c+d x))^5}{b}+15 i (a+i b)^4 \log (-\tan (c+d x)+i)-15 i (a-i b)^4 \log (\tan (c+d x)+i)-10 b^4 \tan ^3(c+d x)}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2*(a + b*Tan[c + d*x])^4,x]

[Out]

((15*I)*(a + I*b)^4*Log[I - Tan[c + d*x]] - (15*I)*(a - I*b)^4*Log[I + Tan[c + d*x]] + 30*b^2*(-6*a^2 + b^2)*T
an[c + d*x] - 60*a*b^3*Tan[c + d*x]^2 - 10*b^4*Tan[c + d*x]^3 + (6*(a + b*Tan[c + d*x])^5)/b)/(30*d)

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fricas [A]  time = 0.43, size = 148, normalized size = 1.16 \[ \frac {3 \, b^{4} \tan \left (d x + c\right )^{5} + 15 \, a b^{3} \tan \left (d x + c\right )^{4} + 5 \, {\left (6 \, a^{2} b^{2} - b^{4}\right )} \tan \left (d x + c\right )^{3} - 15 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} d x + 30 \, {\left (a^{3} b - a b^{3}\right )} \tan \left (d x + c\right )^{2} + 30 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 15 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^4*tan(d*x + c)^5 + 15*a*b^3*tan(d*x + c)^4 + 5*(6*a^2*b^2 - b^4)*tan(d*x + c)^3 - 15*(a^4 - 6*a^2*b^
2 + b^4)*d*x + 30*(a^3*b - a*b^3)*tan(d*x + c)^2 + 30*(a^3*b - a*b^3)*log(1/(tan(d*x + c)^2 + 1)) + 15*(a^4 -
6*a^2*b^2 + b^4)*tan(d*x + c))/d

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giac [B]  time = 34.21, size = 2281, normalized size = 17.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*a^4*d*x*tan(d*x)^5*tan(c)^5 - 90*a^2*b^2*d*x*tan(d*x)^5*tan(c)^5 + 15*b^4*d*x*tan(d*x)^5*tan(c)^5 -
30*a^3*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(
c) + 1)/(tan(c)^2 + 1))*tan(d*x)^5*tan(c)^5 + 30*a*b^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(
d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^5*tan(c)^5 - 75*a^4*d*x*tan(d*x
)^4*tan(c)^4 + 450*a^2*b^2*d*x*tan(d*x)^4*tan(c)^4 - 75*b^4*d*x*tan(d*x)^4*tan(c)^4 - 30*a^3*b*tan(d*x)^5*tan(
c)^5 + 45*a*b^3*tan(d*x)^5*tan(c)^5 + 150*a^3*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*
tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^4*tan(c)^4 - 150*a*b^3*log(4*(tan(d*x)
^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*
tan(d*x)^4*tan(c)^4 + 15*a^4*tan(d*x)^5*tan(c)^4 - 90*a^2*b^2*tan(d*x)^5*tan(c)^4 + 15*b^4*tan(d*x)^5*tan(c)^4
+ 15*a^4*tan(d*x)^4*tan(c)^5 - 90*a^2*b^2*tan(d*x)^4*tan(c)^5 + 15*b^4*tan(d*x)^4*tan(c)^5 + 150*a^4*d*x*tan(
d*x)^3*tan(c)^3 - 900*a^2*b^2*d*x*tan(d*x)^3*tan(c)^3 + 150*b^4*d*x*tan(d*x)^3*tan(c)^3 - 30*a^3*b*tan(d*x)^5*
tan(c)^3 + 30*a*b^3*tan(d*x)^5*tan(c)^3 + 90*a^3*b*tan(d*x)^4*tan(c)^4 - 165*a*b^3*tan(d*x)^4*tan(c)^4 - 30*a^
3*b*tan(d*x)^3*tan(c)^5 + 30*a*b^3*tan(d*x)^3*tan(c)^5 + 30*a^2*b^2*tan(d*x)^5*tan(c)^2 - 5*b^4*tan(d*x)^5*tan
(c)^2 - 300*a^3*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(
d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 + 300*a*b^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan
(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 - 60*a^4*t
an(d*x)^4*tan(c)^3 + 450*a^2*b^2*tan(d*x)^4*tan(c)^3 - 75*b^4*tan(d*x)^4*tan(c)^3 - 60*a^4*tan(d*x)^3*tan(c)^4
+ 450*a^2*b^2*tan(d*x)^3*tan(c)^4 - 75*b^4*tan(d*x)^3*tan(c)^4 + 30*a^2*b^2*tan(d*x)^2*tan(c)^5 - 5*b^4*tan(d
*x)^2*tan(c)^5 - 15*a*b^3*tan(d*x)^5*tan(c) - 150*a^4*d*x*tan(d*x)^2*tan(c)^2 + 900*a^2*b^2*d*x*tan(d*x)^2*tan
(c)^2 - 150*b^4*d*x*tan(d*x)^2*tan(c)^2 + 90*a^3*b*tan(d*x)^4*tan(c)^2 - 150*a*b^3*tan(d*x)^4*tan(c)^2 - 120*a
^3*b*tan(d*x)^3*tan(c)^3 + 180*a*b^3*tan(d*x)^3*tan(c)^3 + 90*a^3*b*tan(d*x)^2*tan(c)^4 - 150*a*b^3*tan(d*x)^2
*tan(c)^4 - 15*a*b^3*tan(d*x)*tan(c)^5 + 3*b^4*tan(d*x)^5 - 60*a^2*b^2*tan(d*x)^4*tan(c) + 25*b^4*tan(d*x)^4*t
an(c) + 300*a^3*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(
d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 - 300*a*b^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan
(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + 90*a^4*t
an(d*x)^3*tan(c)^2 - 720*a^2*b^2*tan(d*x)^3*tan(c)^2 + 150*b^4*tan(d*x)^3*tan(c)^2 + 90*a^4*tan(d*x)^2*tan(c)^
3 - 720*a^2*b^2*tan(d*x)^2*tan(c)^3 + 150*b^4*tan(d*x)^2*tan(c)^3 - 60*a^2*b^2*tan(d*x)*tan(c)^4 + 25*b^4*tan(
d*x)*tan(c)^4 + 3*b^4*tan(c)^5 + 15*a*b^3*tan(d*x)^4 + 75*a^4*d*x*tan(d*x)*tan(c) - 450*a^2*b^2*d*x*tan(d*x)*t
an(c) + 75*b^4*d*x*tan(d*x)*tan(c) - 90*a^3*b*tan(d*x)^3*tan(c) + 150*a*b^3*tan(d*x)^3*tan(c) + 120*a^3*b*tan(
d*x)^2*tan(c)^2 - 180*a*b^3*tan(d*x)^2*tan(c)^2 - 90*a^3*b*tan(d*x)*tan(c)^3 + 150*a*b^3*tan(d*x)*tan(c)^3 + 1
5*a*b^3*tan(c)^4 + 30*a^2*b^2*tan(d*x)^3 - 5*b^4*tan(d*x)^3 - 150*a^3*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x
)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)*tan(c) + 150*a
*b^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) +
1)/(tan(c)^2 + 1))*tan(d*x)*tan(c) - 60*a^4*tan(d*x)^2*tan(c) + 450*a^2*b^2*tan(d*x)^2*tan(c) - 75*b^4*tan(d*
x)^2*tan(c) - 60*a^4*tan(d*x)*tan(c)^2 + 450*a^2*b^2*tan(d*x)*tan(c)^2 - 75*b^4*tan(d*x)*tan(c)^2 + 30*a^2*b^2
*tan(c)^3 - 5*b^4*tan(c)^3 - 15*a^4*d*x + 90*a^2*b^2*d*x - 15*b^4*d*x + 30*a^3*b*tan(d*x)^2 - 30*a*b^3*tan(d*x
)^2 - 90*a^3*b*tan(d*x)*tan(c) + 165*a*b^3*tan(d*x)*tan(c) + 30*a^3*b*tan(c)^2 - 30*a*b^3*tan(c)^2 + 30*a^3*b*
log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(
tan(c)^2 + 1)) - 30*a*b^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2
- 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1)) + 15*a^4*tan(d*x) - 90*a^2*b^2*tan(d*x) + 15*b^4*tan(d*x) + 15*a^4*ta
n(c) - 90*a^2*b^2*tan(c) + 15*b^4*tan(c) + 30*a^3*b - 45*a*b^3)/(d*tan(d*x)^5*tan(c)^5 - 5*d*tan(d*x)^4*tan(c)
^4 + 10*d*tan(d*x)^3*tan(c)^3 - 10*d*tan(d*x)^2*tan(c)^2 + 5*d*tan(d*x)*tan(c) - d)

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maple [A]  time = 0.02, size = 234, normalized size = 1.83 \[ \frac {\left (\tan ^{5}\left (d x +c \right )\right ) b^{4}}{5 d}+\frac {a \,b^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{d}+\frac {2 \left (\tan ^{3}\left (d x +c \right )\right ) a^{2} b^{2}}{d}-\frac {b^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {2 a^{3} b \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {2 a \,b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {a^{4} \tan \left (d x +c \right )}{d}-\frac {6 a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {b^{4} \tan \left (d x +c \right )}{d}-\frac {2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3} b}{d}+\frac {2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a \,b^{3}}{d}-\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{4}}{d}+\frac {6 \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b^{2}}{d}-\frac {\arctan \left (\tan \left (d x +c \right )\right ) b^{4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.83, size = 149, normalized size = 1.16 \[ \frac {3 \, b^{4} \tan \left (d x + c\right )^{5} + 15 \, a b^{3} \tan \left (d x + c\right )^{4} + 5 \, {\left (6 \, a^{2} b^{2} - b^{4}\right )} \tan \left (d x + c\right )^{3} + 30 \, {\left (a^{3} b - a b^{3}\right )} \tan \left (d x + c\right )^{2} - 15 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )} - 30 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 15 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^4*tan(d*x + c)^5 + 15*a*b^3*tan(d*x + c)^4 + 5*(6*a^2*b^2 - b^4)*tan(d*x + c)^3 + 30*(a^3*b - a*b^3)
*tan(d*x + c)^2 - 15*(a^4 - 6*a^2*b^2 + b^4)*(d*x + c) - 30*(a^3*b - a*b^3)*log(tan(d*x + c)^2 + 1) + 15*(a^4
- 6*a^2*b^2 + b^4)*tan(d*x + c))/d

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mupad [B]  time = 3.98, size = 221, normalized size = 1.73 \[ \frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (2\,a\,b^3-2\,a^3\,b\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {b^4}{3}-2\,a^2\,b^2\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a\,b^3-2\,a^3\,b\right )}{d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a^4-6\,a^2\,b^2+b^4\right )}{d}+\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,d}-\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-a^2+2\,a\,b+b^2\right )\,\left (a^2+2\,a\,b-b^2\right )}{a^4-6\,a^2\,b^2+b^4}\right )\,\left (-a^2+2\,a\,b+b^2\right )\,\left (a^2+2\,a\,b-b^2\right )}{d}+\frac {a\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^4}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.81, size = 214, normalized size = 1.67 \[ \begin {cases} - a^{4} x + \frac {a^{4} \tan {\left (c + d x \right )}}{d} - \frac {2 a^{3} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {2 a^{3} b \tan ^{2}{\left (c + d x \right )}}{d} + 6 a^{2} b^{2} x + \frac {2 a^{2} b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac {6 a^{2} b^{2} \tan {\left (c + d x \right )}}{d} + \frac {2 a b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {a b^{3} \tan ^{4}{\left (c + d x \right )}}{d} - \frac {2 a b^{3} \tan ^{2}{\left (c + d x \right )}}{d} - b^{4} x + \frac {b^{4} \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac {b^{4} \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{4} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\relax (c )}\right )^{4} \tan ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**4*x + a**4*tan(c + d*x)/d - 2*a**3*b*log(tan(c + d*x)**2 + 1)/d + 2*a**3*b*tan(c + d*x)**2/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 + 2*a*b**3*log(tan(c + d*x)**2 + 1)
/d + a*b**3*tan(c + d*x)**4/d - 2*a*b**3*tan(c + d*x)**2/d - b**4*x + 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*tan(c)**2, True))

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