\(\int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) [606]

Optimal result

Integrand size = 36, antiderivative size = 156 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {B \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {2 B \cot ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {B \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {B \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d} \]

[Out]

-2/3*B*cot(d*x+c)^(3/2)/d+1/2*B*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)+1/2*B*arctan(1+2^(1/2)*cot(d*x+c
)^(1/2))/d*2^(1/2)+1/4*B*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)-1/4*B*ln(1+cot(d*x+c)+2^(1/2)*cot
(d*x+c)^(1/2))/d*2^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {21, 3554, 3557, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {B \arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}-\frac {2 B \cot ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {B \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {B \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d} \]

[In]

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

[Out]

-((B*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d)) + (B*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2
]*d) - (2*B*Cot[c + d*x]^(3/2))/(3*d) + (B*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d) -
(B*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = B \int \cot ^{\frac {5}{2}}(c+d x) \, dx \\ & = -\frac {2 B \cot ^{\frac {3}{2}}(c+d x)}{3 d}-B \int \sqrt {\cot (c+d x)} \, dx \\ & = -\frac {2 B \cot ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {B \text {Subst}\left (\int \frac {\sqrt {x}}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {2 B \cot ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {(2 B) \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d} \\ & = -\frac {2 B \cot ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {B \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}+\frac {B \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d} \\ & = -\frac {2 B \cot ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {B \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}+\frac {B \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}+\frac {B \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} d}+\frac {B \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} d} \\ & = -\frac {2 B \cot ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {B \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {B \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}+\frac {B \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {B \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d} \\ & = -\frac {B \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {B \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {2 B \cot ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {B \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {B \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.57 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \left (-3 \arctan \left (\sqrt [4]{-\cot ^2(c+d x)}\right ) \sqrt [4]{-\cot (c+d x)}+3 \text {arctanh}\left (\sqrt [4]{-\cot ^2(c+d x)}\right ) \sqrt [4]{-\cot (c+d x)}+2 \cot ^{\frac {7}{4}}(c+d x)\right )}{3 d \sqrt [4]{\cot (c+d x)}} \]

[In]

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

[Out]

-1/3*(B*(-3*ArcTan[(-Cot[c + d*x]^2)^(1/4)]*(-Cot[c + d*x])^(1/4) + 3*ArcTanh[(-Cot[c + d*x]^2)^(1/4)]*(-Cot[c
+ d*x])^(1/4) + 2*Cot[c + d*x]^(7/4)))/(d*Cot[c + d*x]^(1/4))

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {B \left (-\frac {2 \cot \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(102\)

[In]

int(cot(d*x+c)^(5/2)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

B/d*(-2/3*cot(d*x+c)^(3/2)+1/4*2^(1/2)*(ln((1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/(1+cot(d*x+c)+2^(1/2)*cot(d
*x+c)^(1/2)))+2*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 458, normalized size of antiderivative = 2.94 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {3 \, d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {B \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + {\left (d \cos \left (2 \, d x + 2 \, c\right ) + d\right )} \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}}}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) \sin \left (2 \, d x + 2 \, c\right ) - 3 \, d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {B \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) - {\left (d \cos \left (2 \, d x + 2 \, c\right ) + d\right )} \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}}}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) \sin \left (2 \, d x + 2 \, c\right ) - 3 i \, d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {B \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) - {\left (i \, d \cos \left (2 \, d x + 2 \, c\right ) + i \, d\right )} \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}}}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) \sin \left (2 \, d x + 2 \, c\right ) + 3 i \, d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {B \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) - {\left (-i \, d \cos \left (2 \, d x + 2 \, c\right ) - i \, d\right )} \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}}}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, {\left (B \cos \left (2 \, d x + 2 \, c\right ) + B\right )} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}}}{6 \, d \sin \left (2 \, d x + 2 \, c\right )} \]

[In]

integrate(cot(d*x+c)^(5/2)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(3*d*(-B^4/d^4)^(1/4)*log((B*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) + (d*cos(2*d*
x + 2*c) + d)*(-B^4/d^4)^(1/4))/(cos(2*d*x + 2*c) + 1))*sin(2*d*x + 2*c) - 3*d*(-B^4/d^4)^(1/4)*log((B*sqrt((c
os(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) - (d*cos(2*d*x + 2*c) + d)*(-B^4/d^4)^(1/4))/(cos(2*d*
x + 2*c) + 1))*sin(2*d*x + 2*c) - 3*I*d*(-B^4/d^4)^(1/4)*log((B*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))*
sin(2*d*x + 2*c) - (I*d*cos(2*d*x + 2*c) + I*d)*(-B^4/d^4)^(1/4))/(cos(2*d*x + 2*c) + 1))*sin(2*d*x + 2*c) + 3
*I*d*(-B^4/d^4)^(1/4)*log((B*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) - (-I*d*cos(2*d*x
+ 2*c) - I*d)*(-B^4/d^4)^(1/4))/(cos(2*d*x + 2*c) + 1))*sin(2*d*x + 2*c) + 4*(B*cos(2*d*x + 2*c) + B)*sqrt((co
s(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c)))/(d*sin(2*d*x + 2*c))

Sympy [F]

\[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=B \int \cot ^{\frac {5}{2}}{\left (c + d x \right )}\, dx \]

[In]

integrate(cot(d*x+c)**(5/2)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x)

[Out]

B*Integral(cot(c + d*x)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.81 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {3 \, {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} B - \frac {8 \, B}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \]

[In]

integrate(cot(d*x+c)^(5/2)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(
2) - 2/sqrt(tan(d*x + c)))) - sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(2)*log(-sqrt
(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*B - 8*B/tan(d*x + c)^(3/2))/d

Giac [F]

\[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\int { \frac {{\left (B b \tan \left (d x + c\right ) + B a\right )} \cot \left (d x + c\right )^{\frac {5}{2}}}{b \tan \left (d x + c\right ) + a} \,d x } \]

[In]

integrate(cot(d*x+c)^(5/2)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*b*tan(d*x + c) + B*a)*cot(d*x + c)^(5/2)/(b*tan(d*x + c) + a), x)

Mupad [B] (verification not implemented)

Time = 9.93 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.41 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\frac {1}{\mathrm {tan}\left (c+d\,x\right )}}\right )}{d}-\frac {2\,B\,{\left (\frac {1}{\mathrm {tan}\left (c+d\,x\right )}\right )}^{3/2}}{3\,d}-\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\frac {1}{\mathrm {tan}\left (c+d\,x\right )}}\right )}{d} \]

[In]

int((cot(c + d*x)^(5/2)*(B*a + B*b*tan(c + d*x)))/(a + b*tan(c + d*x)),x)

[Out]

((-1)^(1/4)*B*atan((-1)^(1/4)*(1/tan(c + d*x))^(1/2)))/d - (2*B*(1/tan(c + d*x))^(3/2))/(3*d) - ((-1)^(1/4)*B*
atanh((-1)^(1/4)*(1/tan(c + d*x))^(1/2)))/d