3.356 \(\int \frac{\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=285 \[ \frac{b \left (7 a^2 A b-4 a^3 B-12 a b^2 B+15 A b^3\right )}{4 a^3 d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{\left (8 a^2 A+12 a b B-15 A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{7/2} d}+\frac{(5 A b-4 a B) \cot (c+d x)}{4 a^2 d \sqrt{a+b \tan (c+d x)}}-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{3/2}}-\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{3/2}}-\frac{A \cot ^2(c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}} \]

[Out]

((8*a^2*A - 15*A*b^2 + 12*a*b*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(4*a^(7/2)*d) - ((A - I*B)*ArcTanh
[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((a - I*b)^(3/2)*d) - ((A + I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sq
rt[a + I*b]])/((a + I*b)^(3/2)*d) + (b*(7*a^2*A*b + 15*A*b^3 - 4*a^3*B - 12*a*b^2*B))/(4*a^3*(a^2 + b^2)*d*Sqr
t[a + b*Tan[c + d*x]]) + ((5*A*b - 4*a*B)*Cot[c + d*x])/(4*a^2*d*Sqrt[a + b*Tan[c + d*x]]) - (A*Cot[c + d*x]^2
)/(2*a*d*Sqrt[a + b*Tan[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.20969, antiderivative size = 285, 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 = {3609, 3649, 3653, 3539, 3537, 63, 208, 3634} \[ \frac{b \left (7 a^2 A b-4 a^3 B-12 a b^2 B+15 A b^3\right )}{4 a^3 d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{\left (8 a^2 A+12 a b B-15 A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{7/2} d}+\frac{(5 A b-4 a B) \cot (c+d x)}{4 a^2 d \sqrt{a+b \tan (c+d x)}}-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{3/2}}-\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{3/2}}-\frac{A \cot ^2(c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2*A - 15*A*b^2 + 12*a*b*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(4*a^(7/2)*d) - ((A - I*B)*ArcTanh
[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((a - I*b)^(3/2)*d) - ((A + I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sq
rt[a + I*b]])/((a + I*b)^(3/2)*d) + (b*(7*a^2*A*b + 15*A*b^3 - 4*a^3*B - 12*a*b^2*B))/(4*a^3*(a^2 + b^2)*d*Sqr
t[a + b*Tan[c + d*x]]) + ((5*A*b - 4*a*B)*Cot[c + d*x])/(4*a^2*d*Sqrt[a + b*Tan[c + d*x]]) - (A*Cot[c + d*x]^2
)/(2*a*d*Sqrt[a + b*Tan[c + d*x]])

Rule 3609

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*(A*b - a*B)*(a + b*Tan[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[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*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]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ
[a, 0])))

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 \frac{\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx &=-\frac{A \cot ^2(c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}}-\frac{\int \frac{\cot ^2(c+d x) \left (\frac{1}{2} (5 A b-4 a B)+2 a A \tan (c+d x)+\frac{5}{2} A b \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^{3/2}} \, dx}{2 a}\\ &=\frac{(5 A b-4 a B) \cot (c+d x)}{4 a^2 d \sqrt{a+b \tan (c+d x)}}-\frac{A \cot ^2(c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}}+\frac{\int \frac{\cot (c+d x) \left (\frac{1}{4} \left (-8 a^2 A+15 A b^2-12 a b B\right )-2 a^2 B \tan (c+d x)+\frac{3}{4} b (5 A b-4 a B) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^{3/2}} \, dx}{2 a^2}\\ &=\frac{b \left (7 a^2 A b+15 A b^3-4 a^3 B-12 a b^2 B\right )}{4 a^3 \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{(5 A b-4 a B) \cot (c+d x)}{4 a^2 d \sqrt{a+b \tan (c+d x)}}-\frac{A \cot ^2(c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}}+\frac{\int \frac{\cot (c+d x) \left (-\frac{1}{8} \left (a^2+b^2\right ) \left (8 a^2 A-15 A b^2+12 a b B\right )+a^3 (A b-a B) \tan (c+d x)+\frac{1}{8} b \left (7 a^2 A b+15 A b^3-4 a^3 B-12 a b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx}{a^3 \left (a^2+b^2\right )}\\ &=\frac{b \left (7 a^2 A b+15 A b^3-4 a^3 B-12 a b^2 B\right )}{4 a^3 \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{(5 A b-4 a B) \cot (c+d x)}{4 a^2 d \sqrt{a+b \tan (c+d x)}}-\frac{A \cot ^2(c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}}+\frac{\int \frac{a^3 (A b-a B)+a^3 (a A+b B) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{a^3 \left (a^2+b^2\right )}-\frac{\left (8 a^2 A-15 A b^2+12 a b B\right ) \int \frac{\cot (c+d x) \left (1+\tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac{b \left (7 a^2 A b+15 A b^3-4 a^3 B-12 a b^2 B\right )}{4 a^3 \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{(5 A b-4 a B) \cot (c+d x)}{4 a^2 d \sqrt{a+b \tan (c+d x)}}-\frac{A \cot ^2(c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}}+\frac{(A-i B) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (i a+b)}+\frac{((i a+b) (A+i B)) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 \left (a^2+b^2\right )}-\frac{\left (8 a^2 A-15 A b^2+12 a b B\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{8 a^3 d}\\ &=\frac{b \left (7 a^2 A b+15 A b^3-4 a^3 B-12 a b^2 B\right )}{4 a^3 \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{(5 A b-4 a B) \cot (c+d x)}{4 a^2 d \sqrt{a+b \tan (c+d x)}}-\frac{A \cot ^2(c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}}+\frac{(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 (a-i b) d}+\frac{(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 (a+i b) d}-\frac{\left (8 a^2 A-15 A b^2+12 a b 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 )}{4 a^3 b d}\\ &=\frac{\left (8 a^2 A-15 A b^2+12 a b B\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{7/2} d}+\frac{b \left (7 a^2 A b+15 A b^3-4 a^3 B-12 a b^2 B\right )}{4 a^3 \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{(5 A b-4 a B) \cot (c+d x)}{4 a^2 d \sqrt{a+b \tan (c+d x)}}-\frac{A \cot ^2(c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}}-\frac{(i (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 )}{(a+i b) b d}+\frac{(i A+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 )}{(a-i b) b d}\\ &=\frac{\left (8 a^2 A-15 A b^2+12 a b B\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{4 a^{7/2} d}-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{(a-i b)^{3/2} d}-\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{(a+i b)^{3/2} d}+\frac{b \left (7 a^2 A b+15 A b^3-4 a^3 B-12 a b^2 B\right )}{4 a^3 \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{(5 A b-4 a B) \cot (c+d x)}{4 a^2 d \sqrt{a+b \tan (c+d x)}}-\frac{A \cot ^2(c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 6.21701, size = 409, normalized size = 1.44 \[ -\frac{A \cot ^2(c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}}-\frac{-\frac{(5 A b-4 a B) \cot (c+d x)}{2 a d \sqrt{a+b \tan (c+d x)}}-\frac{\frac{2 \left (\frac{1}{4} b^2 \left (-8 a^2 A-12 a b B+15 A b^2\right )-a \left (-2 a^2 b B-\frac{3}{4} a b (5 A b-4 a B)\right )\right )}{a d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (\frac{\left (a^2+b^2\right ) \left (8 a^2 A+12 a b B-15 A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{4 \sqrt{a} d}+\frac{i \sqrt{a-i b} \left (a^3 (A b-a B)-i a^3 (a A+b B)\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (-a+i b)}-\frac{i \sqrt{a+i b} \left (a^3 (A b-a B)+i a^3 (a A+b B)\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (-a-i b)}\right )}{a \left (a^2+b^2\right )}}{a}}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Cot[c + d*x]^2)/(2*a*d*Sqrt[a + b*Tan[c + d*x]]) - (-((5*A*b - 4*a*B)*Cot[c + d*x])/(2*a*d*Sqrt[a + b*Tan[
c + d*x]]) - ((2*(((a^2 + b^2)*(8*a^2*A - 15*A*b^2 + 12*a*b*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(4*S
qrt[a]*d) + (I*Sqrt[a - I*b]*(a^3*(A*b - a*B) - I*a^3*(a*A + b*B))*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I
*b]])/((-a + I*b)*d) - (I*Sqrt[a + I*b]*(a^3*(A*b - a*B) + I*a^3*(a*A + b*B))*ArcTanh[Sqrt[a + b*Tan[c + d*x]]
/Sqrt[a + I*b]])/((-a - I*b)*d)))/(a*(a^2 + b^2)) + (2*((b^2*(-8*a^2*A + 15*A*b^2 - 12*a*b*B))/4 - a*(-2*a^2*b
*B - (3*a*b*(5*A*b - 4*a*B))/4)))/(a*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]))/a)/(2*a)

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Maple [C]  time = 3.467, size = 174418, normalized size = 612. \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)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),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)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),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)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{3}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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