∫ tan2(c + dx)(a + b tan(c + dx))3(A + B tan(c + dx)) dx

3.248 \(\int \tan ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=201 \[ -\frac {b \left (a^2 B+2 a A b-b^2 B\right ) \tan (c+d x)}{d}+\frac {\left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \log (\cos (c+d x))}{d}-x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )+\frac {(5 A b-a B) (a+b \tan (c+d x))^4}{20 b^2 d}-\frac {(a B+A b) (a+b \tan (c+d x))^2}{2 d}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\frac {B (a+b \tan (c+d x))^3}{3 d} \]

[Out]

-(A*a^3-3*A*a*b^2-3*B*a^2*b+B*b^3)*x+(3*A*a^2*b-A*b^3+B*a^3-3*B*a*b^2)*ln(cos(d*x+c))/d-b*(2*A*a*b+B*a^2-B*b^2

)*tan(d*x+c)/d-1/2*(A*b+B*a)*(a+b*tan(d*x+c))^2/d-1/3*B*(a+b*tan(d*x+c))^3/d+1/20*(5*A*b-B*a)*(a+b*tan(d*x+c))

^4/b^2/d+1/5*B*tan(d*x+c)*(a+b*tan(d*x+c))^4/b/d

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Rubi [A] time = 0.37, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3607, 3630, 3528, 3525, 3475} \[ -\frac {b \left (a^2 B+2 a A b-b^2 B\right ) \tan (c+d x)}{d}+\frac {\left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \log (\cos (c+d x))}{d}-x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )+\frac {(5 A b-a B) (a+b \tan (c+d x))^4}{20 b^2 d}-\frac {(a B+A b) (a+b \tan (c+d x))^2}{2 d}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\frac {B (a+b \tan (c+d x))^3}{3 d} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Tan[c + d*x]^2*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

-((a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*x) + ((3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Log[Cos[c + d*x]])/d -

(b*(2*a*A*b + a^2*B - b^2*B)*Tan[c + d*x])/d - ((A*b + a*B)*(a + b*Tan[c + d*x])^2)/(2*d) - (B*(a + b*Tan[c +

d*x])^3)/(3*d) + ((5*A*b - a*B)*(a + b*Tan[c + d*x])^4)/(20*b^2*d) + (B*Tan[c + d*x]*(a + b*Tan[c + d*x])^4)/

(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 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 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 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (

f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])

^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]

&& !LeQ[m, -1]

Rubi steps

\begin {align*} \int \tan ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\frac {B \tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}+\frac {\int (a+b \tan (c+d x))^3 \left (-a B-5 b B \tan (c+d x)+(5 A b-a B) \tan ^2(c+d x)\right ) \, dx}{5 b}\\ &=\frac {(5 A b-a B) (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}+\frac {\int (a+b \tan (c+d x))^3 (-5 A b-5 b B \tan (c+d x)) \, dx}{5 b}\\ &=-\frac {B (a+b \tan (c+d x))^3}{3 d}+\frac {(5 A b-a B) (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}+\frac {\int (a+b \tan (c+d x))^2 (-5 b (a A-b B)-5 b (A b+a B) \tan (c+d x)) \, dx}{5 b}\\ &=-\frac {(A b+a B) (a+b \tan (c+d x))^2}{2 d}-\frac {B (a+b \tan (c+d x))^3}{3 d}+\frac {(5 A b-a B) (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}+\frac {\int (a+b \tan (c+d x)) \left (-5 b \left (a^2 A-A b^2-2 a b B\right )-5 b \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)\right ) \, dx}{5 b}\\ &=-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x-\frac {b \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)}{d}-\frac {(A b+a B) (a+b \tan (c+d x))^2}{2 d}-\frac {B (a+b \tan (c+d x))^3}{3 d}+\frac {(5 A b-a B) (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}+\left (-3 a^2 A b+A b^3-a^3 B+3 a b^2 B\right ) \int \tan (c+d x) \, dx\\ &=-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x+\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \log (\cos (c+d x))}{d}-\frac {b \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)}{d}-\frac {(A b+a B) (a+b \tan (c+d x))^2}{2 d}-\frac {B (a+b \tan (c+d x))^3}{3 d}+\frac {(5 A b-a B) (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {B \tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}\\ \end {align*}

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Mathematica [C] time = 2.21, size = 241, normalized size = 1.20 \[ \frac {10 B \left (6 b^2 \left (b^2-6 a^2\right ) \tan (c+d x)-12 a b^3 \tan ^2(c+d x)-3 i (a-i b)^4 \log (\tan (c+d x)+i)+3 i (a+i b)^4 \log (-\tan (c+d x)+i)-2 b^4 \tan ^3(c+d x)\right )-30 (A b-a B) \left (6 a b^2 \tan (c+d x)+(-b+i a)^3 \log (-\tan (c+d x)+i)-(b+i a)^3 \log (\tan (c+d x)+i)+b^3 \tan ^2(c+d x)\right )+\frac {3 (5 A b-a B) (a+b \tan (c+d x))^4}{b}+12 B \tan (c+d x) (a+b \tan (c+d x))^4}{60 b d} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Tan[c + d*x]^2*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

((3*(5*A*b - a*B)*(a + b*Tan[c + d*x])^4)/b + 12*B*Tan[c + d*x]*(a + b*Tan[c + d*x])^4 - 30*(A*b - a*B)*((I*a

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

+ 10*B*((3*I)*(a + I*b)^4*Log[I - Tan[c + d*x]] - (3*I)*(a - I*b)^4*Log[I + Tan[c + d*x]] + 6*b^2*(-6*a^2 + b^

2)*Tan[c + d*x] - 12*a*b^3*Tan[c + d*x]^2 - 2*b^4*Tan[c + d*x]^3))/(60*b*d)

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fricas [A] time = 0.67, size = 213, normalized size = 1.06 \[ \frac {12 \, B b^{3} \tan \left (d x + c\right )^{5} + 15 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \tan \left (d x + c\right )^{4} + 20 \, {\left (3 \, B a^{2} b + 3 \, A a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )^{3} - 60 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} d x + 30 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{2} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 60 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \tan \left (d x + c\right )}{60 \, d} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/60*(12*B*b^3*tan(d*x + c)^5 + 15*(3*B*a*b^2 + A*b^3)*tan(d*x + c)^4 + 20*(3*B*a^2*b + 3*A*a*b^2 - B*b^3)*tan

(d*x + c)^3 - 60*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*d*x + 30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*tan(

d*x + c)^2 + 30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*log(1/(tan(d*x + c)^2 + 1)) + 60*(A*a^3 - 3*B*a^2*b -

3*A*a*b^2 + B*b^3)*tan(d*x + c))/d

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giac [B] time = 8.12, size = 3997, normalized size = 19.89 \[ \text {result too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(60*A*a^3*d*x*tan(d*x)^5*tan(c)^5 - 180*B*a^2*b*d*x*tan(d*x)^5*tan(c)^5 - 180*A*a*b^2*d*x*tan(d*x)^5*tan

(c)^5 + 60*B*b^3*d*x*tan(d*x)^5*tan(c)^5 - 30*B*a^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 - 90*A*a^2*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 + 90*B*a*b^2*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 - 300*A*a^3*d*x*tan(d*x)^4*tan(c)^4 + 900*B*a^2*b*d*x*tan(d*x)^4*tan(c)^4 + 900*A*a*b^2*d*x*tan(d*x)^

4*tan(c)^4 - 300*B*b^3*d*x*tan(d*x)^4*tan(c)^4 - 30*B*a^3*tan(d*x)^5*tan(c)^5 - 90*A*a^2*b*tan(d*x)^5*tan(c)^5

+ 135*B*a*b^2*tan(d*x)^5*tan(c)^5 + 45*A*b^3*tan(d*x)^5*tan(c)^5 + 150*B*a^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*t

an(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 + 450*A*a^2*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 - 450*B*a*b^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*ta

n(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 + 60*A*a^3*tan(d*x)^5*tan(c)^4 - 180*B*a^2*b*tan(d*x)^5*tan(c)^4 - 180*A

*a*b^2*tan(d*x)^5*tan(c)^4 + 60*B*b^3*tan(d*x)^5*tan(c)^4 + 60*A*a^3*tan(d*x)^4*tan(c)^5 - 180*B*a^2*b*tan(d*x

)^4*tan(c)^5 - 180*A*a*b^2*tan(d*x)^4*tan(c)^5 + 60*B*b^3*tan(d*x)^4*tan(c)^5 + 600*A*a^3*d*x*tan(d*x)^3*tan(c

)^3 - 1800*B*a^2*b*d*x*tan(d*x)^3*tan(c)^3 - 1800*A*a*b^2*d*x*tan(d*x)^3*tan(c)^3 + 600*B*b^3*d*x*tan(d*x)^3*t

an(c)^3 - 30*B*a^3*tan(d*x)^5*tan(c)^3 - 90*A*a^2*b*tan(d*x)^5*tan(c)^3 + 90*B*a*b^2*tan(d*x)^5*tan(c)^3 + 30*

A*b^3*tan(d*x)^5*tan(c)^3 + 90*B*a^3*tan(d*x)^4*tan(c)^4 + 270*A*a^2*b*tan(d*x)^4*tan(c)^4 - 495*B*a*b^2*tan(d

*x)^4*tan(c)^4 - 165*A*b^3*tan(d*x)^4*tan(c)^4 - 30*B*a^3*tan(d*x)^3*tan(c)^5 - 90*A*a^2*b*tan(d*x)^3*tan(c)^5

+ 90*B*a*b^2*tan(d*x)^3*tan(c)^5 + 30*A*b^3*tan(d*x)^3*tan(c)^5 + 60*B*a^2*b*tan(d*x)^5*tan(c)^2 + 60*A*a*b^2

*tan(d*x)^5*tan(c)^2 - 20*B*b^3*tan(d*x)^5*tan(c)^2 - 300*B*a^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 - 900*A*a^2

*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 + 900*B*a*b^2*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 - 240*A*a^3*tan(d*x)^4*tan(c)^3 + 900*B*a^2*b*tan(d*x)^4*tan(c)^3 + 900*A*a*b^2*tan(d*

x)^4*tan(c)^3 - 300*B*b^3*tan(d*x)^4*tan(c)^3 - 240*A*a^3*tan(d*x)^3*tan(c)^4 + 900*B*a^2*b*tan(d*x)^3*tan(c)^

4 + 900*A*a*b^2*tan(d*x)^3*tan(c)^4 - 300*B*b^3*tan(d*x)^3*tan(c)^4 + 60*B*a^2*b*tan(d*x)^2*tan(c)^5 + 60*A*a*

b^2*tan(d*x)^2*tan(c)^5 - 20*B*b^3*tan(d*x)^2*tan(c)^5 - 45*B*a*b^2*tan(d*x)^5*tan(c) - 15*A*b^3*tan(d*x)^5*ta

n(c) - 600*A*a^3*d*x*tan(d*x)^2*tan(c)^2 + 1800*B*a^2*b*d*x*tan(d*x)^2*tan(c)^2 + 1800*A*a*b^2*d*x*tan(d*x)^2*

tan(c)^2 - 600*B*b^3*d*x*tan(d*x)^2*tan(c)^2 + 90*B*a^3*tan(d*x)^4*tan(c)^2 + 270*A*a^2*b*tan(d*x)^4*tan(c)^2

- 450*B*a*b^2*tan(d*x)^4*tan(c)^2 - 150*A*b^3*tan(d*x)^4*tan(c)^2 - 120*B*a^3*tan(d*x)^3*tan(c)^3 - 360*A*a^2*

b*tan(d*x)^3*tan(c)^3 + 540*B*a*b^2*tan(d*x)^3*tan(c)^3 + 180*A*b^3*tan(d*x)^3*tan(c)^3 + 90*B*a^3*tan(d*x)^2*

tan(c)^4 + 270*A*a^2*b*tan(d*x)^2*tan(c)^4 - 450*B*a*b^2*tan(d*x)^2*tan(c)^4 - 150*A*b^3*tan(d*x)^2*tan(c)^4 -

45*B*a*b^2*tan(d*x)*tan(c)^5 - 15*A*b^3*tan(d*x)*tan(c)^5 + 12*B*b^3*tan(d*x)^5 - 120*B*a^2*b*tan(d*x)^4*tan(

c) - 120*A*a*b^2*tan(d*x)^4*tan(c) + 100*B*b^3*tan(d*x)^4*tan(c) + 300*B*a^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*ta

n(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 + 900*A*a^2*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 - 900*B*a*b^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan

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 + 360*A*a^3*tan(d*x)^3*tan(c)^2 - 1440*B*a^2*b*tan(d*x)^3*tan(c)^2 - 1440

*A*a*b^2*tan(d*x)^3*tan(c)^2 + 600*B*b^3*tan(d*x)^3*tan(c)^2 + 360*A*a^3*tan(d*x)^2*tan(c)^3 - 1440*B*a^2*b*ta

n(d*x)^2*tan(c)^3 - 1440*A*a*b^2*tan(d*x)^2*tan(c)^3 + 600*B*b^3*tan(d*x)^2*tan(c)^3 - 120*B*a^2*b*tan(d*x)*ta

n(c)^4 - 120*A*a*b^2*tan(d*x)*tan(c)^4 + 100*B*b^3*tan(d*x)*tan(c)^4 + 12*B*b^3*tan(c)^5 + 45*B*a*b^2*tan(d*x)

^4 + 15*A*b^3*tan(d*x)^4 + 300*A*a^3*d*x*tan(d*x)*tan(c) - 900*B*a^2*b*d*x*tan(d*x)*tan(c) - 900*A*a*b^2*d*x*t

an(d*x)*tan(c) + 300*B*b^3*d*x*tan(d*x)*tan(c) - 90*B*a^3*tan(d*x)^3*tan(c) - 270*A*a^2*b*tan(d*x)^3*tan(c) +

450*B*a*b^2*tan(d*x)^3*tan(c) + 150*A*b^3*tan(d*x)^3*tan(c) + 120*B*a^3*tan(d*x)^2*tan(c)^2 + 360*A*a^2*b*tan(

d*x)^2*tan(c)^2 - 540*B*a*b^2*tan(d*x)^2*tan(c)^2 - 180*A*b^3*tan(d*x)^2*tan(c)^2 - 90*B*a^3*tan(d*x)*tan(c)^3

- 270*A*a^2*b*tan(d*x)*tan(c)^3 + 450*B*a*b^2*tan(d*x)*tan(c)^3 + 150*A*b^3*tan(d*x)*tan(c)^3 + 45*B*a*b^2*ta

n(c)^4 + 15*A*b^3*tan(c)^4 + 60*B*a^2*b*tan(d*x)^3 + 60*A*a*b^2*tan(d*x)^3 - 20*B*b^3*tan(d*x)^3 - 150*B*a^3*l

og(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)/(t

an(c)^2 + 1))*tan(d*x)*tan(c) - 450*A*a^2*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) + 450*B*a*b^2*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) - 240*A*a^3*tan(d*x)^2*tan(c) + 900*B*a^2*b*tan(d*x)^2*ta

n(c) + 900*A*a*b^2*tan(d*x)^2*tan(c) - 300*B*b^3*tan(d*x)^2*tan(c) - 240*A*a^3*tan(d*x)*tan(c)^2 + 900*B*a^2*b

*tan(d*x)*tan(c)^2 + 900*A*a*b^2*tan(d*x)*tan(c)^2 - 300*B*b^3*tan(d*x)*tan(c)^2 + 60*B*a^2*b*tan(c)^3 + 60*A*

a*b^2*tan(c)^3 - 20*B*b^3*tan(c)^3 - 60*A*a^3*d*x + 180*B*a^2*b*d*x + 180*A*a*b^2*d*x - 60*B*b^3*d*x + 30*B*a^

3*tan(d*x)^2 + 90*A*a^2*b*tan(d*x)^2 - 90*B*a*b^2*tan(d*x)^2 - 30*A*b^3*tan(d*x)^2 - 90*B*a^3*tan(d*x)*tan(c)

- 270*A*a^2*b*tan(d*x)*tan(c) + 495*B*a*b^2*tan(d*x)*tan(c) + 165*A*b^3*tan(d*x)*tan(c) + 30*B*a^3*tan(c)^2 +

90*A*a^2*b*tan(c)^2 - 90*B*a*b^2*tan(c)^2 - 30*A*b^3*tan(c)^2 + 30*B*a^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)) + 90*A*a^2*b*log(4*(ta

n(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)) - 90*B*a*b^2*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)) + 60*A*a^3*tan(d*x) - 180*B*a^2*b*tan(d*x) - 180*A

*a*b^2*tan(d*x) + 60*B*b^3*tan(d*x) + 60*A*a^3*tan(c) - 180*B*a^2*b*tan(c) - 180*A*a*b^2*tan(c) + 60*B*b^3*tan

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 = 0.03, size = 383, normalized size = 1.91 \[ \frac {B \left (\tan ^{5}\left (d x +c \right )\right ) b^{3}}{5 d}+\frac {A \left (\tan ^{4}\left (d x +c \right )\right ) b^{3}}{4 d}+\frac {3 B \left (\tan ^{4}\left (d x +c \right )\right ) a \,b^{2}}{4 d}+\frac {A \left (\tan ^{3}\left (d x +c \right )\right ) a \,b^{2}}{d}+\frac {B \left (\tan ^{3}\left (d x +c \right )\right ) a^{2} b}{d}-\frac {B \left (\tan ^{3}\left (d x +c \right )\right ) b^{3}}{3 d}+\frac {3 A \left (\tan ^{2}\left (d x +c \right )\right ) a^{2} b}{2 d}-\frac {A \left (\tan ^{2}\left (d x +c \right )\right ) b^{3}}{2 d}+\frac {a^{3} B \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 B \left (\tan ^{2}\left (d x +c \right )\right ) a \,b^{2}}{2 d}+\frac {A \,a^{3} \tan \left (d x +c \right )}{d}-\frac {3 A \tan \left (d x +c \right ) a \,b^{2}}{d}-\frac {3 B \tan \left (d x +c \right ) a^{2} b}{d}+\frac {B \tan \left (d x +c \right ) b^{3}}{d}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A \,a^{2} b}{2 d}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A \,b^{3}}{2 d}-\frac {a^{3} B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B a \,b^{2}}{2 d}-\frac {a^{3} A \arctan \left (\tan \left (d x +c \right )\right )}{d}+\frac {3 A \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{d}+\frac {3 B \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b}{d}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b^{3}}{d} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

1/5/d*B*tan(d*x+c)^5*b^3+1/4/d*A*tan(d*x+c)^4*b^3+3/4/d*B*tan(d*x+c)^4*a*b^2+1/d*A*tan(d*x+c)^3*a*b^2+1/d*B*ta

n(d*x+c)^3*a^2*b-1/3/d*B*tan(d*x+c)^3*b^3+3/2/d*A*tan(d*x+c)^2*a^2*b-1/2/d*A*tan(d*x+c)^2*b^3+1/2/d*B*tan(d*x+

c)^2*a^3-3/2/d*B*tan(d*x+c)^2*a*b^2+1/d*A*a^3*tan(d*x+c)-3/d*A*tan(d*x+c)*a*b^2-3/d*B*tan(d*x+c)*a^2*b+1/d*B*t

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

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

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

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maxima [A] time = 0.77, size = 214, normalized size = 1.06 \[ \frac {12 \, B b^{3} \tan \left (d x + c\right )^{5} + 15 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \tan \left (d x + c\right )^{4} + 20 \, {\left (3 \, B a^{2} b + 3 \, A a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )^{3} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{2} - 60 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} {\left (d x + c\right )} - 30 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 60 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \tan \left (d x + c\right )}{60 \, d} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/60*(12*B*b^3*tan(d*x + c)^5 + 15*(3*B*a*b^2 + A*b^3)*tan(d*x + c)^4 + 20*(3*B*a^2*b + 3*A*a*b^2 - B*b^3)*tan

(d*x + c)^3 + 30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*tan(d*x + c)^2 - 60*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 +

B*b^3)*(d*x + c) - 30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*log(tan(d*x + c)^2 + 1) + 60*(A*a^3 - 3*B*a^2*b

- 3*A*a*b^2 + B*b^3)*tan(d*x + c))/d

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mupad [B] time = 6.39, size = 217, normalized size = 1.08 \[ \frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,a^3+B\,b^3-3\,a\,b\,\left (A\,b+B\,a\right )\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {B\,b^3}{3}-a\,b\,\left (A\,b+B\,a\right )\right )}{d}-x\,\left (A\,a^3-3\,B\,a^2\,b-3\,A\,a\,b^2+B\,b^3\right )+\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (-\frac {B\,a^3}{2}-\frac {3\,A\,a^2\,b}{2}+\frac {3\,B\,a\,b^2}{2}+\frac {A\,b^3}{2}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {A\,b^3}{4}+\frac {3\,B\,a\,b^2}{4}\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {B\,a^3}{2}-\frac {3\,A\,a^2\,b}{2}+\frac {3\,B\,a\,b^2}{2}+\frac {A\,b^3}{2}\right )}{d}+\frac {B\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,d} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3,x)

[Out]

(tan(c + d*x)*(A*a^3 + B*b^3 - 3*a*b*(A*b + B*a)))/d - (tan(c + d*x)^3*((B*b^3)/3 - a*b*(A*b + B*a)))/d - x*(A

*a^3 + B*b^3 - 3*A*a*b^2 - 3*B*a^2*b) + (log(tan(c + d*x)^2 + 1)*((A*b^3)/2 - (B*a^3)/2 - (3*A*a^2*b)/2 + (3*B

*a*b^2)/2))/d + (tan(c + d*x)^4*((A*b^3)/4 + (3*B*a*b^2)/4))/d - (tan(c + d*x)^2*((A*b^3)/2 - (B*a^3)/2 - (3*A

*a^2*b)/2 + (3*B*a*b^2)/2))/d + (B*b^3*tan(c + d*x)^5)/(5*d)

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sympy [A] time = 0.96, size = 384, normalized size = 1.91 \[ \begin {cases} - A a^{3} x + \frac {A a^{3} \tan {\left (c + d x \right )}}{d} - \frac {3 A a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 A a^{2} b \tan ^{2}{\left (c + d x \right )}}{2 d} + 3 A a b^{2} x + \frac {A a b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac {3 A a b^{2} \tan {\left (c + d x \right )}}{d} + \frac {A b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {A b^{3} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {A b^{3} \tan ^{2}{\left (c + d x \right )}}{2 d} - \frac {B a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B a^{3} \tan ^{2}{\left (c + d x \right )}}{2 d} + 3 B a^{2} b x + \frac {B a^{2} b \tan ^{3}{\left (c + d x \right )}}{d} - \frac {3 B a^{2} b \tan {\left (c + d x \right )}}{d} + \frac {3 B a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 B a b^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {3 B a b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} - B b^{3} x + \frac {B b^{3} \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac {B b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac {B b^{3} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \tan {\relax (c )}\right ) \left (a + b \tan {\relax (c )}\right )^{3} \tan ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(a+b*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

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

d*x)**2/(2*d) + 3*A*a*b**2*x + A*a*b**2*tan(c + d*x)**3/d - 3*A*a*b**2*tan(c + d*x)/d + A*b**3*log(tan(c + d*x

)**2 + 1)/(2*d) + A*b**3*tan(c + d*x)**4/(4*d) - A*b**3*tan(c + d*x)**2/(2*d) - B*a**3*log(tan(c + d*x)**2 + 1

)/(2*d) + B*a**3*tan(c + d*x)**2/(2*d) + 3*B*a**2*b*x + B*a**2*b*tan(c + d*x)**3/d - 3*B*a**2*b*tan(c + d*x)/d

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

2*d) - B*b**3*x + B*b**3*tan(c + d*x)**5/(5*d) - B*b**3*tan(c + d*x)**3/(3*d) + B*b**3*tan(c + d*x)/d, Ne(d, 0

)), (x*(A + B*tan(c))*(a + b*tan(c))**3*tan(c)**2, True))

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