3.526 \(\int \cot ^4(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=237 \[ \frac{5 b \left (8 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{8 \sqrt{a} d}+\frac{\left (8 a^2-11 b^2\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 d}-\frac{a^2 \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}-\frac{i (a-i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{i (a+i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}-\frac{13 a b \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 d} \]

[Out]

(5*b*(8*a^2 - b^2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(8*Sqrt[a]*d) - (I*(a - I*b)^(5/2)*ArcTanh[Sqrt[
a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + (I*(a + I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/d
+ ((8*a^2 - 11*b^2)*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(8*d) - (13*a*b*Cot[c + d*x]^2*Sqrt[a + b*Tan[c + d
*x]])/(12*d) - (a^2*Cot[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]])/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 1.0665, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {3565, 3649, 3653, 3539, 3537, 63, 208, 3634} \[ \frac{5 b \left (8 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{8 \sqrt{a} d}+\frac{\left (8 a^2-11 b^2\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 d}-\frac{a^2 \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}-\frac{i (a-i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{i (a+i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}-\frac{13 a b \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^(5/2),x]

[Out]

(5*b*(8*a^2 - b^2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(8*Sqrt[a]*d) - (I*(a - I*b)^(5/2)*ArcTanh[Sqrt[
a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + (I*(a + I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/d
+ ((8*a^2 - 11*b^2)*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(8*d) - (13*a*b*Cot[c + d*x]^2*Sqrt[a + b*Tan[c + d
*x]])/(12*d) - (a^2*Cot[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]])/(3*d)

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

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) (a+b \tan (c+d x))^{5/2} \, dx &=-\frac{a^2 \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{13 a^2 b}{2}-3 a \left (a^2-3 b^2\right ) \tan (c+d x)-\frac{1}{2} b \left (5 a^2-6 b^2\right ) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{13 a b \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 d}-\frac{a^2 \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} a^2 \left (8 a^2-11 b^2\right )+6 a b \left (3 a^2-b^2\right ) \tan (c+d x)+\frac{39}{4} a^2 b^2 \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx}{6 a}\\ &=\frac{\left (8 a^2-11 b^2\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 d}-\frac{13 a b \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 d}-\frac{a^2 \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}+\frac{\int \frac{\cot (c+d x) \left (-\frac{15}{8} a^2 b \left (8 a^2-b^2\right )+6 a^3 \left (a^2-3 b^2\right ) \tan (c+d x)+\frac{3}{8} a^2 b \left (8 a^2-11 b^2\right ) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx}{6 a^2}\\ &=\frac{\left (8 a^2-11 b^2\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 d}-\frac{13 a b \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 d}-\frac{a^2 \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}+\frac{\int \frac{6 a^3 \left (a^2-3 b^2\right )+6 a^2 b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{6 a^2}-\frac{1}{16} \left (5 b \left (8 a^2-b^2\right )\right ) \int \frac{\cot (c+d x) \left (1+\tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{\left (8 a^2-11 b^2\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 d}-\frac{13 a b \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 d}-\frac{a^2 \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}+\frac{1}{2} (a-i b)^3 \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx+\frac{1}{2} (a+i b)^3 \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx-\frac{\left (5 b \left (8 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{16 d}\\ &=\frac{\left (8 a^2-11 b^2\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 d}-\frac{13 a b \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 d}-\frac{a^2 \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}+\frac{(i a-b)^3 \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)^3 \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 (5 \left (8 a^2-b^2\right )\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 d}\\ &=\frac{5 b \left (8 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{8 \sqrt{a} d}+\frac{\left (8 a^2-11 b^2\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 d}-\frac{13 a b \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 d}-\frac{a^2 \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}-\frac{(a-i b)^3 \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)^3 \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{5 b \left (8 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{8 \sqrt{a} d}-\frac{i (a-i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{i (a+i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}+\frac{\left (8 a^2-11 b^2\right ) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{8 d}-\frac{13 a b \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}}{12 d}-\frac{a^2 \cot ^3(c+d x) \sqrt{a+b \tan (c+d x)}}{3 d}\\ \end{align*}

Mathematica [A]  time = 3.52187, size = 185, normalized size = 0.78 \[ -\frac{\frac{15 b \left (b^2-8 a^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a}}+\cot (c+d x) \sqrt{a+b \tan (c+d x)} \left (8 a^2 \cot ^2(c+d x)-24 a^2+26 a b \cot (c+d x)+33 b^2\right )+24 i (a-i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )-24 i (a+i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^(5/2),x]

[Out]

-((15*b*(-8*a^2 + b^2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/Sqrt[a] + (24*I)*(a - I*b)^(5/2)*ArcTanh[Sqr
t[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] - (24*I)*(a + I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]]
+ Cot[c + d*x]*(-24*a^2 + 33*b^2 + 26*a*b*Cot[c + d*x] + 8*a^2*Cot[c + d*x]^2)*Sqrt[a + b*Tan[c + d*x]])/(24*d
)

________________________________________________________________________________________

Maple [C]  time = 1.677, size = 88284, normalized size = 372.5 \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))^(5/2),x)

[Out]

result too large to display

________________________________________________________________________________________

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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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))^(5/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [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))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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))^(5/2),x, algorithm="giac")

[Out]

Timed out