3.873 \(\int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=310 \[ -\frac {2 a^2}{b d \left (a^2+b^2\right ) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {\left (3 a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b^2 d \left (a^2+b^2\right ) \sqrt {\cot (c+d x)}}-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d}+\frac {i \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}+\frac {i \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 1.57, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {4241, 3565, 3647, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ -\frac {2 a^2}{b d \left (a^2+b^2\right ) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {\left (3 a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b^2 d \left (a^2+b^2\right ) \sqrt {\cot (c+d x)}}-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d}+\frac {i \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}+\frac {i \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cot[c + d*x]^(7/2)*(a + b*Tan[c + d*x])^(3/2)),x]

[Out]

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

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 93

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 205

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

Rule 206

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

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 217

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 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 3647

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 3655

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 4241

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 6725

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

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Mathematica [C]  time = 6.24, size = 319, normalized size = 1.03 \[ \frac {2 \sqrt {\frac {b \tan (c+d x)}{a}+1} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};-\frac {b \tan (c+d x)}{a}\right )}{5 a d \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {(-1)^{3/4} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-a+i b)^{3/2}}-\frac {(-1)^{3/4} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (a+i b)^{3/2}}-\frac {1}{d (a-i b) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {1}{d (a+i b) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cot[c + d*x]^(7/2)*(a + b*Tan[c + d*x])^(3/2)),x]

[Out]

((-1)^(3/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]
*Sqrt[Tan[c + d*x]])/((-a + I*b)^(3/2)*d) - ((-1)^(3/4)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/S
qrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/((a + I*b)^(3/2)*d) - 1/((a - I*b)*d*Sqrt[Cot[
c + d*x]]*Sqrt[a + b*Tan[c + d*x]]) - 1/((a + I*b)*d*Sqrt[Cot[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]) + (2*Hyperge
ometric2F1[3/2, 5/2, 7/2, -((b*Tan[c + d*x])/a)]*Sqrt[1 + (b*Tan[c + d*x])/a])/(5*a*d*Cot[c + d*x]^(5/2)*Sqrt[
a + b*Tan[c + d*x]])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 1.68, size = 15848, normalized size = 51.12 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*tan(d*x + c) + a)^(3/2)*cot(d*x + c)^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cot(c + d*x)^(7/2)*(a + b*tan(c + d*x))^(3/2)),x)

[Out]

int(1/(cot(c + d*x)^(7/2)*(a + b*tan(c + d*x))^(3/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)**(7/2)/(a+b*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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