\(\int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx\) [858]

Optimal result

Integrand size = 25, antiderivative size = 248 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\frac {i (i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {i (i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \]

[Out]

I*(I*a-b)^(5/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2
)/d+5*a*b^(3/2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+I
*(I*a+b)^(5/2)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2
)/d+b^2*(a+b*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 1.71 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4326, 3647, 3736, 6857, 65, 223, 212, 95, 211, 214} \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\frac {i (-b+i a)^{5/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {5 a b^{3/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i (b+i a)^{5/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \]

[In]

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

[Out]

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

Rule 65

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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 211

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

Rule 212

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

Rule 214

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3736

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] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^2)/(1 + ff^2*x^2)), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 4326

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+b \tan (c+d x))^{5/2}}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{2} a \left (2 a^2-b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)+\frac {5}{2} a b^2 \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx \\ & = \frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {\frac {1}{2} a \left (2 a^2-b^2\right )+b \left (3 a^2-b^2\right ) x+\frac {5}{2} a b^2 x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \left (\frac {5 a b^2}{2 \sqrt {x} \sqrt {a+b x}}+\frac {a^3-3 a b^2+b \left (3 a^2-b^2\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {a^3-3 a b^2+b \left (3 a^2-b^2\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (5 a b^2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \left (\frac {-b \left (3 a^2-b^2\right )+i \left (a^3-3 a b^2\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {b \left (3 a^2-b^2\right )+i \left (a^3-3 a b^2\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (5 a b^2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {\left ((i a-b)^3 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (5 a b^2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\left ((i a+b)^3 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {\left ((i a-b)^3 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\left ((i a+b)^3 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \\ & = \frac {i (i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {i (i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.97 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left ((-1)^{3/4} (-a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(-1)^{3/4} (a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {5 a^{3/2} b^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]

[In]

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

[Out]

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-1)^(3/4)*(-a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan
[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - (-1)^(3/4)*(a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c
+ d*x]])/Sqrt[a + b*Tan[c + d*x]]] + b^2*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + (5*a^(3/2)*b^(3/2)*Arc
Sinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/Sqrt[a + b*Tan[c + d*x]]))/d

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2949\) vs. \(2(202)=404\).

Time = 36.74 (sec) , antiderivative size = 2950, normalized size of antiderivative = 11.90

method	result	size
default	\(\text {Expression too large to display}\)	\(2950\)

[In]

int(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/4/d*(a+b*tan(d*x+c))^(1/2)*cot(d*x+c)^(1/2)/(cos(d*x+c)+1)/((sin(d*x+c)*cos(d*x+c)*a-cos(d*x+c)^2*b+b)/(cos(
d*x+c)+1)^2)^(1/2)*(10*sin(d*x+c)*2^(1/2)*b^(3/2)*arctanh(((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+
1)^2)^(1/2)*(csc(d*x+c)+cot(d*x+c))/b^(1/2))*(-b+(a^2+b^2)^(1/2))^(1/2)*a^2+2*sin(d*x+c)*2^(1/2)*((sin(d*x+c)*
cos(d*x+c)*a-cos(d*x+c)^2*b+b)/(cos(d*x+c)+1)^2)^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*a*b^2-sin(d*x+c)*(a^2+b^2)^(
1/2)*ln((cos(d*x+c)*cot(d*x+c)*a-2*a*cot(d*x+c)+2*sin(d*x+c)*(csc(d*x+c)*(cot(d*x+c)^2*a-2*cot(d*x+c)*csc(d*x+
c)*a+csc(d*x+c)^2*a-2*b*csc(d*x+c)+2*cot(d*x+c)*b-a)*(cos(d*x+c)-1))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*(a^2+b^
2)^(1/2)*cos(d*x+c)+2*b*cos(d*x+c)-sin(d*x+c)*a+csc(d*x+c)*a-2*(a^2+b^2)^(1/2)-2*b)/(cos(d*x+c)-1))*(b+(a^2+b^
2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*a^2+sin(d*x+c)*(a^2+b^2)^(1/2)*ln((cos(d*x+c)*cot(d*x+c)*a-2*a*cot(
d*x+c)+2*sin(d*x+c)*(csc(d*x+c)*(cot(d*x+c)^2*a-2*cot(d*x+c)*csc(d*x+c)*a+csc(d*x+c)^2*a-2*b*csc(d*x+c)+2*cot(
d*x+c)*b-a)*(cos(d*x+c)-1))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*(a^2+b^2)^(1/2)*cos(d*x+c)+2*b*cos(d*x+c)-sin(d*
x+c)*a+csc(d*x+c)*a-2*(a^2+b^2)^(1/2)-2*b)/(cos(d*x+c)-1))*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2
)*b^2+sin(d*x+c)*(a^2+b^2)^(1/2)*ln(-(-cos(d*x+c)*cot(d*x+c)*a+2*sin(d*x+c)*(csc(d*x+c)*(cot(d*x+c)^2*a-2*cot(
d*x+c)*csc(d*x+c)*a+csc(d*x+c)^2*a-2*b*csc(d*x+c)+2*cot(d*x+c)*b-a)*(cos(d*x+c)-1))^(1/2)*(b+(a^2+b^2)^(1/2))^
(1/2)+2*a*cot(d*x+c)+sin(d*x+c)*a-2*(a^2+b^2)^(1/2)*cos(d*x+c)-2*b*cos(d*x+c)-csc(d*x+c)*a+2*(a^2+b^2)^(1/2)+2
*b)/(cos(d*x+c)-1))*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*a^2-sin(d*x+c)*(a^2+b^2)^(1/2)*ln(-(-
cos(d*x+c)*cot(d*x+c)*a+2*sin(d*x+c)*(csc(d*x+c)*(cot(d*x+c)^2*a-2*cot(d*x+c)*csc(d*x+c)*a+csc(d*x+c)^2*a-2*b*
csc(d*x+c)+2*cot(d*x+c)*b-a)*(cos(d*x+c)-1))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*a*cot(d*x+c)+sin(d*x+c)*a-2*(a^
2+b^2)^(1/2)*cos(d*x+c)-2*b*cos(d*x+c)-csc(d*x+c)*a+2*(a^2+b^2)^(1/2)+2*b)/(cos(d*x+c)-1))*(b+(a^2+b^2)^(1/2))
^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*b^2+3*sin(d*x+c)*ln((cos(d*x+c)*cot(d*x+c)*a-2*a*cot(d*x+c)+2*sin(d*x+c)*(cs
c(d*x+c)*(cot(d*x+c)^2*a-2*cot(d*x+c)*csc(d*x+c)*a+csc(d*x+c)^2*a-2*b*csc(d*x+c)+2*cot(d*x+c)*b-a)*(cos(d*x+c)
-1))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*(a^2+b^2)^(1/2)*cos(d*x+c)+2*b*cos(d*x+c)-sin(d*x+c)*a+csc(d*x+c)*a-2*(
a^2+b^2)^(1/2)-2*b)/(cos(d*x+c)-1))*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*a^2*b-sin(d*x+c)*ln((
cos(d*x+c)*cot(d*x+c)*a-2*a*cot(d*x+c)+2*sin(d*x+c)*(csc(d*x+c)*(cot(d*x+c)^2*a-2*cot(d*x+c)*csc(d*x+c)*a+csc(
d*x+c)^2*a-2*b*csc(d*x+c)+2*cot(d*x+c)*b-a)*(cos(d*x+c)-1))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*(a^2+b^2)^(1/2)*
cos(d*x+c)+2*b*cos(d*x+c)-sin(d*x+c)*a+csc(d*x+c)*a-2*(a^2+b^2)^(1/2)-2*b)/(cos(d*x+c)-1))*(b+(a^2+b^2)^(1/2))
^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*b^3-3*sin(d*x+c)*ln(-(-cos(d*x+c)*cot(d*x+c)*a+2*sin(d*x+c)*(csc(d*x+c)*(cot
(d*x+c)^2*a-2*cot(d*x+c)*csc(d*x+c)*a+csc(d*x+c)^2*a-2*b*csc(d*x+c)+2*cot(d*x+c)*b-a)*(cos(d*x+c)-1))^(1/2)*(b
+(a^2+b^2)^(1/2))^(1/2)+2*a*cot(d*x+c)+sin(d*x+c)*a-2*(a^2+b^2)^(1/2)*cos(d*x+c)-2*b*cos(d*x+c)-csc(d*x+c)*a+2
*(a^2+b^2)^(1/2)+2*b)/(cos(d*x+c)-1))*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*a^2*b+sin(d*x+c)*ln
(-(-cos(d*x+c)*cot(d*x+c)*a+2*sin(d*x+c)*(csc(d*x+c)*(cot(d*x+c)^2*a-2*cot(d*x+c)*csc(d*x+c)*a+csc(d*x+c)^2*a-
2*b*csc(d*x+c)+2*cot(d*x+c)*b-a)*(cos(d*x+c)-1))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*a*cot(d*x+c)+sin(d*x+c)*a-2
*(a^2+b^2)^(1/2)*cos(d*x+c)-2*b*cos(d*x+c)-csc(d*x+c)*a+2*(a^2+b^2)^(1/2)+2*b)/(cos(d*x+c)-1))*(b+(a^2+b^2)^(1
/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*b^3+2*tan(d*x+c)*2^(1/2)*((sin(d*x+c)*cos(d*x+c)*a-cos(d*x+c)^2*b+b)/(co
s(d*x+c)+1)^2)^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*a*b^2+4*sin(d*x+c)*(a^2+b^2)^(1/2)*arctan((2^(1/2)*((sin(d*x+c
)*cos(d*x+c)*a-cos(d*x+c)^2*b+b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2)*cos(d*x+c)+(b+(a
^2+b^2)^(1/2))^(1/2))/(cos(d*x+c)-1)/(-b+(a^2+b^2)^(1/2))^(1/2))*a^2*b+4*sin(d*x+c)*(a^2+b^2)^(1/2)*arctan((2^
(1/2)*((sin(d*x+c)*cos(d*x+c)*a-cos(d*x+c)^2*b+b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)+(b+(a^2+b^2)^(1/2))^(1/2)
*cos(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2))/(cos(d*x+c)-1)/(-b+(a^2+b^2)^(1/2))^(1/2))*a^2*b+2*sin(d*x+c)*arctan((2
^(1/2)*((sin(d*x+c)*cos(d*x+c)*a-cos(d*x+c)^2*b+b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2
)*cos(d*x+c)+(b+(a^2+b^2)^(1/2))^(1/2))/(cos(d*x+c)-1)/(-b+(a^2+b^2)^(1/2))^(1/2))*a^4-6*sin(d*x+c)*arctan((2^
(1/2)*((sin(d*x+c)*cos(d*x+c)*a-cos(d*x+c)^2*b+b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2)
*cos(d*x+c)+(b+(a^2+b^2)^(1/2))^(1/2))/(cos(d*x+c)-1)/(-b+(a^2+b^2)^(1/2))^(1/2))*a^2*b^2+2*sin(d*x+c)*arctan(
(2^(1/2)*((sin(d*x+c)*cos(d*x+c)*a-cos(d*x+c)^2*b+b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)+(b+(a^2+b^2)^(1/2))^(1
/2)*cos(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2))/(cos(d*x+c)-1)/(-b+(a^2+b^2)^(1/2))^(1/2))*a^4-6*sin(d*x+c)*arctan((
2^(1/2)*((sin(d*x+c)*cos(d*x+c)*a-cos(d*x+c)^2*b+b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)+(b+(a^2+b^2)^(1/2))^(1/
2)*cos(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2))/(cos(d*x+c)-1)/(-b+(a^2+b^2)^(1/2))^(1/2))*a^2*b^2)*2^(1/2)/a/(-b+(a^
2+b^2)^(1/2))^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4793 vs. \(2 (196) = 392\).

Time = 1.34 (sec) , antiderivative size = 9618, normalized size of antiderivative = 38.78 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]

[In]

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)**(1/2)*(a+b*tan(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \]

[In]

int(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(5/2),x)

[Out]

int(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(5/2), x)