3.246 \(\int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=118 \[ \frac {\left (a^2 A-2 a b B-A b^2\right ) \cot (c+d x)}{d}+x \left (a^2 A-2 a b B-A b^2\right )-\frac {a^2 A \cot ^3(c+d x)}{3 d}+\frac {\left (b^2 B-a (a B+2 A b)\right ) \log (\sin (c+d x))}{d}-\frac {a (a B+2 A b) \cot ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3604, 3628, 3529, 3531, 3475} \[ \frac {\left (a^2 A-2 a b B-A b^2\right ) \cot (c+d x)}{d}+x \left (a^2 A-2 a b B-A b^2\right )-\frac {a^2 A \cot ^3(c+d x)}{3 d}+\frac {\left (b^2 B-a (a B+2 A b)\right ) \log (\sin (c+d x))}{d}-\frac {a (a B+2 A b) \cot ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*A - A*b^2 - 2*a*b*B)*x + ((a^2*A - A*b^2 - 2*a*b*B)*Cot[c + d*x])/d - (a*(2*A*b + a*B)*Cot[c + d*x]^2)/(2
*d) - (a^2*A*Cot[c + d*x]^3)/(3*d) + ((b^2*B - a*(2*A*b + a*B))*Log[Sin[c + d*x]])/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 3529

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

Rule 3531

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

Rule 3604

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((B*c - A*d)*(b*c - a*d)^2*(c + d*Tan[e + f*x])^(n + 1))/(f*d^2*(n +
1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[B*(b*c - a*d)^2 + A*d*(a^2
*c - b^2*c + 2*a*b*d) + d*(B*(a^2*c - b^2*c + 2*a*b*d) + A*(2*a*b*c - a^2*d + b^2*d))*Tan[e + f*x] + b^2*B*(c^
2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^
2, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1]

Rule 3628

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

Rubi steps

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

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Mathematica [C]  time = 1.40, size = 152, normalized size = 1.29 \[ \frac {6 \left (a^2 A-2 a b B-A b^2\right ) \cot (c+d x)-6 \left (a^2 B+2 a A b-b^2 B\right ) \log (\tan (c+d x))-2 a^2 A \cot ^3(c+d x)-3 a (a B+2 A b) \cot ^2(c+d x)+3 (a+i b)^2 (B-i A) \log (-\tan (c+d x)+i)+3 (a-i b)^2 (B+i A) \log (\tan (c+d x)+i)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 157, normalized size = 1.33 \[ -\frac {3 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} + 3 \, {\left (B a^{2} + 2 \, A a b - 2 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} d x\right )} \tan \left (d x + c\right )^{3} + 2 \, A a^{2} - 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{6 \, d \tan \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(B*a^2 + 2*A*a*b - B*b^2)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^3 + 3*(B*a^2 + 2*A*a*b
- 2*(A*a^2 - 2*B*a*b - A*b^2)*d*x)*tan(d*x + c)^3 + 2*A*a^2 - 6*(A*a^2 - 2*B*a*b - A*b^2)*tan(d*x + c)^2 + 3*
(B*a^2 + 2*A*a*b)*tan(d*x + c))/(d*tan(d*x + c)^3)

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giac [B]  time = 2.49, size = 334, normalized size = 2.83 \[ \frac {A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )} + 24 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 24 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {44 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 88 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(A*a^2*tan(1/2*d*x + 1/2*c)^3 - 3*B*a^2*tan(1/2*d*x + 1/2*c)^2 - 6*A*a*b*tan(1/2*d*x + 1/2*c)^2 - 15*A*a^
2*tan(1/2*d*x + 1/2*c) + 24*B*a*b*tan(1/2*d*x + 1/2*c) + 12*A*b^2*tan(1/2*d*x + 1/2*c) + 24*(A*a^2 - 2*B*a*b -
A*b^2)*(d*x + c) + 24*(B*a^2 + 2*A*a*b - B*b^2)*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 24*(B*a^2 + 2*A*a*b - B*b^2
)*log(abs(tan(1/2*d*x + 1/2*c))) + (44*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 88*A*a*b*tan(1/2*d*x + 1/2*c)^3 - 44*B*b
^2*tan(1/2*d*x + 1/2*c)^3 + 15*A*a^2*tan(1/2*d*x + 1/2*c)^2 - 24*B*a*b*tan(1/2*d*x + 1/2*c)^2 - 12*A*b^2*tan(1
/2*d*x + 1/2*c)^2 - 3*B*a^2*tan(1/2*d*x + 1/2*c) - 6*A*a*b*tan(1/2*d*x + 1/2*c) - A*a^2)/tan(1/2*d*x + 1/2*c)^
3)/d

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maple [A]  time = 0.38, size = 188, normalized size = 1.59 \[ -\frac {a^{2} A \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} A \cot \left (d x +c \right )}{d}+a^{2} A x +\frac {A \,a^{2} c}{d}-\frac {a^{2} B \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{2} B \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {A a b \left (\cot ^{2}\left (d x +c \right )\right )}{d}-\frac {2 A a b \ln \left (\sin \left (d x +c \right )\right )}{d}-2 B x a b -\frac {2 B \cot \left (d x +c \right ) a b}{d}-\frac {2 B a b c}{d}-A x \,b^{2}-\frac {A \cot \left (d x +c \right ) b^{2}}{d}-\frac {A \,b^{2} c}{d}+\frac {b^{2} B \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.98, size = 149, normalized size = 1.26 \[ \frac {6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )} + 3 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {2 \, A a^{2} - 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(A*a^2 - 2*B*a*b - A*b^2)*(d*x + c) + 3*(B*a^2 + 2*A*a*b - B*b^2)*log(tan(d*x + c)^2 + 1) - 6*(B*a^2 +
2*A*a*b - B*b^2)*log(tan(d*x + c)) - (2*A*a^2 - 6*(A*a^2 - 2*B*a*b - A*b^2)*tan(d*x + c)^2 + 3*(B*a^2 + 2*A*a*
b)*tan(d*x + c))/tan(d*x + c)^3)/d

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mupad [B]  time = 6.37, size = 156, normalized size = 1.32 \[ -\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (\frac {A\,a^2}{3}+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-A\,a^2+2\,B\,a\,b+A\,b^2\right )+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^2+2\,A\,a\,b-B\,b^2\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c + d*x) - 1i)*(A*1i - B)*(a*1i - b)^2)/(2*d) - (log(tan(c + d*x))*(B*a^2 - B*b^2 + 2*A*a*b))/d - (co
t(c + d*x)^3*((A*a^2)/3 + tan(c + d*x)^2*(A*b^2 - A*a^2 + 2*B*a*b) + tan(c + d*x)*((B*a^2)/2 + A*a*b)))/d - (l
og(tan(c + d*x) + 1i)*(A*1i + B)*(a*1i + b)^2)/(2*d)

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sympy [A]  time = 2.50, size = 260, normalized size = 2.20 \[ \begin {cases} \tilde {\infty } A a^{2} x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (A + B \tan {\relax (c )}\right ) \left (a + b \tan {\relax (c )}\right )^{2} \cot ^{4}{\relax (c )} & \text {for}\: d = 0 \\A a^{2} x + \frac {A a^{2}}{d \tan {\left (c + d x \right )}} - \frac {A a^{2}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {A a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - \frac {2 A a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {A a b}{d \tan ^{2}{\left (c + d x \right )}} - A b^{2} x - \frac {A b^{2}}{d \tan {\left (c + d x \right )}} + \frac {B a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B a^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - 2 B a b x - \frac {2 B a b}{d \tan {\left (c + d x \right )}} - \frac {B b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*A*a**2*x, (Eq(c, 0) | Eq(c, -d*x)) & (Eq(d, 0) | Eq(c, -d*x))), (x*(A + B*tan(c))*(a + b*tan(c)
)**2*cot(c)**4, Eq(d, 0)), (A*a**2*x + A*a**2/(d*tan(c + d*x)) - A*a**2/(3*d*tan(c + d*x)**3) + A*a*b*log(tan(
c + d*x)**2 + 1)/d - 2*A*a*b*log(tan(c + d*x))/d - A*a*b/(d*tan(c + d*x)**2) - A*b**2*x - A*b**2/(d*tan(c + d*
x)) + B*a**2*log(tan(c + d*x)**2 + 1)/(2*d) - B*a**2*log(tan(c + d*x))/d - B*a**2/(2*d*tan(c + d*x)**2) - 2*B*
a*b*x - 2*B*a*b/(d*tan(c + d*x)) - B*b**2*log(tan(c + d*x)**2 + 1)/(2*d) + B*b**2*log(tan(c + d*x))/d, True))

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