3.9.60 \(\int \frac {(a+b \tan (c+d x))^{5/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\) [860]

Optimal. Leaf size=337 \[ -\frac {i (i a-b)^{5/2} \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {5 a \left (a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{8 \sqrt {b} d}-\frac {i (i a+b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {b^2 \sqrt {a+b \tan (c+d x)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \tan (c+d x)}}{12 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (11 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}} \]

[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-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+5/8*a*(a^2-8*b^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/b^(1/2)+1/3*b^2*(a+b*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(5/2)+13/12*a*b*(a+b*tan(d*x+c))^(1/2)/d/cot(d*
x+c)^(3/2)+1/8*(11*a^2-8*b^2)*(a+b*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(1/2)

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Rubi [A]
time = 1.59, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {4326, 3647, 3728, 3736, 6857, 65, 223, 212, 95, 211, 214} \begin {gather*} \frac {\left (11 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {5 a \left (a^2-8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 \sqrt {b} d}-\frac {i (-b+i a)^{5/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {ArcTan}\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)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \tan (c+d x)}}{12 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {i (b+i a)^{5/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(5/2)/Cot[c + d*x]^(3/2),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]]*S
qrt[Tan[c + d*x]])/d + (5*a*(a^2 - 8*b^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]])/(8*Sqrt[b]*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]])/(3*d*Cot
[c + d*x]^(5/2)) + (13*a*b*Sqrt[a + b*Tan[c + d*x]])/(12*d*Cot[c + d*x]^(3/2)) + ((11*a^2 - 8*b^2)*Sqrt[a + b*
Tan[c + d*x]])/(8*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 3728

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((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 - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[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*} \int \frac {(a+b \tan (c+d x))^{5/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\\ &=\frac {b^2 \sqrt {a+b \tan (c+d x)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {1}{3} \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (\frac {1}{2} a \left (6 a^2-5 b^2\right )+3 b \left (3 a^2-b^2\right ) \tan (c+d x)+\frac {13}{2} a b^2 \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {b^2 \sqrt {a+b \tan (c+d x)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \tan (c+d x)}}{12 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {\tan (c+d x)} \left (-\frac {39}{4} a^2 b^2+6 a b \left (a^2-3 b^2\right ) \tan (c+d x)+\frac {3}{4} b^2 \left (11 a^2-8 b^2\right ) \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx}{6 b}\\ &=\frac {b^2 \sqrt {a+b \tan (c+d x)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \tan (c+d x)}}{12 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (11 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {-\frac {3}{8} a b^2 \left (11 a^2-8 b^2\right )-6 b^3 \left (3 a^2-b^2\right ) \tan (c+d x)+\frac {15}{8} a b^2 \left (a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{6 b^2}\\ &=\frac {b^2 \sqrt {a+b \tan (c+d x)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \tan (c+d x)}}{12 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (11 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{8 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 {3}{8} a b^2 \left (11 a^2-8 b^2\right )-6 b^3 \left (3 a^2-b^2\right ) x+\frac {15}{8} a b^2 \left (a^2-8 b^2\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 b^2 d}\\ &=\frac {b^2 \sqrt {a+b \tan (c+d x)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \tan (c+d x)}}{12 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (11 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{8 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 {15 a b^2 \left (a^2-8 b^2\right )}{8 \sqrt {x} \sqrt {a+b x}}-\frac {6 \left (a b^2 \left (a^2-3 b^2\right )+b^3 \left (3 a^2-b^2\right ) x\right )}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{6 b^2 d}\\ &=\frac {b^2 \sqrt {a+b \tan (c+d x)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \tan (c+d x)}}{12 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (11 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{8 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 b^2 \left (a^2-3 b^2\right )+b^3 \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 )}{b^2 d}+\frac {\left (5 a \left (a^2-8 b^2\right ) \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 )}{16 d}\\ &=\frac {b^2 \sqrt {a+b \tan (c+d x)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \tan (c+d x)}}{12 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (11 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{8 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 {i a b^2 \left (a^2-3 b^2\right )-b^3 \left (3 a^2-b^2\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {i a b^2 \left (a^2-3 b^2\right )+b^3 \left (3 a^2-b^2\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b^2 d}+\frac {\left (5 a \left (a^2-8 b^2\right ) \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 )}{8 d}\\ &=\frac {b^2 \sqrt {a+b \tan (c+d x)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \tan (c+d x)}}{12 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (11 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{8 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 ((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 {\left (5 a \left (a^2-8 b^2\right ) \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 )}{8 d}\\ &=\frac {5 a \left (a^2-8 b^2\right ) \tanh ^{-1}\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)}}{8 \sqrt {b} d}+\frac {b^2 \sqrt {a+b \tan (c+d x)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \tan (c+d x)}}{12 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (11 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{8 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} \tan ^{-1}\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 \left (a^2-8 b^2\right ) \tanh ^{-1}\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)}}{8 \sqrt {b} d}-\frac {i (i a+b)^{5/2} \tanh ^{-1}\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)}}{3 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \tan (c+d x)}}{12 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (11 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 4.00, size = 320, normalized size = 0.95 \begin {gather*} \frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-24 (-1)^{3/4} (-a+i b)^{5/2} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+24 (-1)^{3/4} (a+i b)^{5/2} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+3 \left (11 a^2-8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+26 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}+8 b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}+\frac {15 a^{3/2} \left (a^2-8 b^2\right ) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}\right )}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(5/2)/Cot[c + d*x]^(3/2),x]

[Out]

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-24*(-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]]] + 24*(-1)^(3/4)*(a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqr
t[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 3*(11*a^2 - 8*b^2)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] +
26*a*b*Tan[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]] + 8*b^2*Tan[c + d*x]^(5/2)*Sqrt[a + b*Tan[c + d*x]] + (15*a
^(3/2)*(a^2 - 8*b^2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt
[a + b*Tan[c + d*x]])))/(24*d)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 43.03, size = 18203, normalized size = 54.01

method	result	size
default	\(\text {Expression too large to display}\)	\(18203\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(5/2)/cot(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(5/2)/cot(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^(5/2)/cot(d*x + c)^(3/2), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(5/2)/cot(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**(5/2)/cot(d*x+c)**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(5/2)/cot(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^(5/2)/cot(c + d*x)^(3/2),x)

[Out]

int((a + b*tan(c + d*x))^(5/2)/cot(c + d*x)^(3/2), x)

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