3.324 \(\int \cot ^4(c+d x) \sqrt{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=279 \[ \frac{\left (8 a^2 A b+16 a^3 B+2 a b^2 B-A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{8 a^{5/2} d}+\frac{\left (8 a^2 A-2 a b B+A b^2\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 a^2 d}-\frac{\sqrt{a-i b} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{\sqrt{a+i b} (-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}-\frac{(6 a B+A b) \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 a d}-\frac{A \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d} \]

[Out]

((8*a^2*A*b - A*b^3 + 16*a^3*B + 2*a*b^2*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(8*a^(5/2)*d) - (Sqrt[a
- I*b]*(I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + (Sqrt[a + I*b]*(I*A - B)*ArcTanh[Sqrt[a
+ b*Tan[c + d*x]]/Sqrt[a + I*b]])/d + ((8*a^2*A + A*b^2 - 2*a*b*B)*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(8*
a^2*d) - ((A*b + 6*a*B)*Cot[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]])/(12*a*d) - (A*Cot[c + d*x]^3*Sqrt[a + b*Tan[c
+ d*x]])/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 1.16793, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {3608, 3649, 3653, 3539, 3537, 63, 208, 3634} \[ \frac{\left (8 a^2 A b+16 a^3 B+2 a b^2 B-A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{8 a^{5/2} d}+\frac{\left (8 a^2 A-2 a b B+A b^2\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 a^2 d}-\frac{\sqrt{a-i b} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{\sqrt{a+i b} (-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}-\frac{(6 a B+A b) \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 a d}-\frac{A \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2*A*b - A*b^3 + 16*a^3*B + 2*a*b^2*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(8*a^(5/2)*d) - (Sqrt[a
- I*b]*(I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + (Sqrt[a + I*b]*(I*A - B)*ArcTanh[Sqrt[a
+ b*Tan[c + d*x]]/Sqrt[a + I*b]])/d + ((8*a^2*A + A*b^2 - 2*a*b*B)*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(8*
a^2*d) - ((A*b + 6*a*B)*Cot[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]])/(12*a*d) - (A*Cot[c + d*x]^3*Sqrt[a + b*Tan[c
+ d*x]])/(3*d)

Rule 3608

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

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3653

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

Rule 3539

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

Rule 3537

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rubi steps

\begin{align*} \int \cot ^4(c+d x) \sqrt{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=-\frac{A \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}-\frac{1}{3} \int \frac{\cot ^3(c+d x) \left (\frac{1}{2} (-A b-6 a B)+3 (a A-b B) \tan (c+d x)+\frac{5}{2} A b \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{(A b+6 a B) \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 a d}-\frac{A \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}+\frac{\int \frac{\cot ^2(c+d x) \left (-\frac{3}{4} \left (8 a^2 A+A b^2-2 a b B\right )-6 a (A b+a B) \tan (c+d x)-\frac{3}{4} b (A b+6 a B) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx}{6 a}\\ &=\frac{\left (8 a^2 A+A b^2-2 a b B\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 a^2 d}-\frac{(A b+6 a B) \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 a d}-\frac{A \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}-\frac{\int \frac{\cot (c+d x) \left (\frac{3}{8} \left (8 a^2 A b-A b^3+16 a^3 B+2 a b^2 B\right )-6 a^2 (a A-b B) \tan (c+d x)-\frac{3}{8} b \left (8 a^2 A+A b^2-2 a b B\right ) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx}{6 a^2}\\ &=\frac{\left (8 a^2 A+A b^2-2 a b B\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 a^2 d}-\frac{(A b+6 a B) \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 a d}-\frac{A \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}-\frac{\int \frac{-6 a^2 (a A-b B)-6 a^2 (A b+a B) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{6 a^2}-\frac{\left (8 a^2 A b-A b^3+16 a^3 B+2 a b^2 B\right ) \int \frac{\cot (c+d x) \left (1+\tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx}{16 a^2}\\ &=\frac{\left (8 a^2 A+A b^2-2 a b B\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 a^2 d}-\frac{(A b+6 a B) \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 a d}-\frac{A \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}+\frac{1}{2} ((a-i b) (A-i B)) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx+\frac{1}{2} ((a+i b) (A+i B)) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx-\frac{\left (8 a^2 A b-A b^3+16 a^3 B+2 a b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{16 a^2 d}\\ &=\frac{\left (8 a^2 A+A b^2-2 a b B\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 a^2 d}-\frac{(A b+6 a B) \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 a d}-\frac{A \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}+\frac{(i (a-i b) (A-i B)) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac{((i a-b) (A+i B)) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}-\frac{\left (8 a^2 A b-A b^3+16 a^3 B+2 a b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{8 a^2 b d}\\ &=\frac{\left (8 a^2 A b-A b^3+16 a^3 B+2 a b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{8 a^{5/2} d}+\frac{\left (8 a^2 A+A b^2-2 a b B\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 a^2 d}-\frac{(A b+6 a B) \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 a d}-\frac{A \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}-\frac{((a-i b) (A-i B)) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i a}{b}+\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}-\frac{((a+i b) (A+i B)) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i a}{b}-\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=\frac{\left (8 a^2 A b-A b^3+16 a^3 B+2 a b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{8 a^{5/2} d}-\frac{\sqrt{a-i b} (i A+B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{\sqrt{a+i b} (i A-B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}+\frac{\left (8 a^2 A+A b^2-2 a b B\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 a^2 d}-\frac{(A b+6 a B) \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 a d}-\frac{A \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}\\ \end{align*}

Mathematica [B]  time = 6.39186, size = 564, normalized size = 2.02 \[ \frac{2 b^4 \left (-\frac{(a A-b B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{2 a^{3/2} b^3}-\frac{3 (a B+A b) \left (\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{\cot (c+d x) \sqrt{a+b \tan (c+d x)}}{a b}\right )}{8 a b^2}+\frac{5 A \left (\frac{3 \left (\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{\cot (c+d x) \sqrt{a+b \tan (c+d x)}}{a b}\right )}{a}+\frac{2 \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{a b^2}\right )}{48 b}+\frac{(a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} b^4}+\frac{\left (a A b-a \sqrt{-b^2} B-A \sqrt{-b^2} b+b^2 (-B)\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-\sqrt{-b^2}}}\right )}{2 b^4 \sqrt{-b^2} \sqrt{a-\sqrt{-b^2}}}-\frac{\left (a A b+a \sqrt{-b^2} B+A \sqrt{-b^2} b+b^2 (-B)\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+\sqrt{-b^2}}}\right )}{2 \left (-b^2\right )^{5/2} \sqrt{a+\sqrt{-b^2}}}-\frac{(a B+A b) \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{4 a b^4}+\frac{(a A-b B) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{2 a b^4}-\frac{A \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{6 b^4}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^4*(((A*b + a*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(Sqrt[a]*b^4) - ((a*A - b*B)*ArcTanh[Sqrt[a +
b*Tan[c + d*x]]/Sqrt[a]])/(2*a^(3/2)*b^3) + ((a*A*b - A*b*Sqrt[-b^2] - b^2*B - a*Sqrt[-b^2]*B)*ArcTanh[Sqrt[a
+ b*Tan[c + d*x]]/Sqrt[a - Sqrt[-b^2]]])/(2*b^4*Sqrt[-b^2]*Sqrt[a - Sqrt[-b^2]]) - ((a*A*b + A*b*Sqrt[-b^2] -
b^2*B + a*Sqrt[-b^2]*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + Sqrt[-b^2]]])/(2*(-b^2)^(5/2)*Sqrt[a + Sqrt[
-b^2]]) + ((a*A - b*B)*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(2*a*b^4) - ((A*b + a*B)*Cot[c + d*x]^2*Sqrt[a +
b*Tan[c + d*x]])/(4*a*b^4) - (A*Cot[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]])/(6*b^4) - (3*(A*b + a*B)*(ArcTanh[Sq
rt[a + b*Tan[c + d*x]]/Sqrt[a]]/a^(3/2) - (Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(a*b)))/(8*a*b^2) + (5*A*((2
*Cot[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]])/(a*b^2) + (3*(ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]]/a^(3/2) - (C
ot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(a*b)))/a))/(48*b)))/d

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Maple [C]  time = 2.46, size = 118304, normalized size = 424. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x)

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \tan{\left (c + d x \right )}\right ) \sqrt{a + b \tan{\left (c + d x \right )}} \cot ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(a+b*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*sqrt(a + b*tan(c + d*x))*cot(c + d*x)**4, x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out