Integrand size = 25, antiderivative size = 255 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{3/2} d}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{b^{3/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{3/2} d}-\frac {2 a^2}{b \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}} \]
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+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)/b^(3/2)/d-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)*t an(d*x+c)^(1/2)/(I*a+b)^(3/2)/d-2*a^2/b/(a^2+b^2)/d/cot(d*x+c)^(1/2)/(a+b* tan(d*x+c))^(1/2)
Time = 2.14 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.34 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=-\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (2 \sqrt {a} \sqrt {-a+i b} \sqrt {a+i b} \left (a^2+b^2\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}+\sqrt {b} \left (\sqrt [4]{-1} (a+i b)^{3/2} b \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt {-a+i b} \left (-2 a^2 \sqrt {a+i b} \sqrt {\tan (c+d x)}-\sqrt [4]{-1} (a-i b) b \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}\right )\right )\right )}{(-a+i b)^{3/2} (a+i b)^{3/2} b^{3/2} d \sqrt {a+b \tan (c+d x)}} \]
-((Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(2*Sqrt[a]*Sqrt[-a + I*b]*Sqrt[a + I*b]*(a^2 + b^2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a] + Sqrt[b]*((-1)^(1/4)*(a + I*b)^(3/2)*b*ArcTan[((-1)^( 1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]] + Sqrt[-a + I*b]*(-2*a^2*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]] - (-1)^(1/4)*(a - I*b)*b*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d* x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]]))))/((-a + I*b)^(3 /2)*(a + I*b)^(3/2)*b^(3/2)*d*Sqrt[a + b*Tan[c + d*x]]))
Time = 1.07 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4729, 3042, 4048, 27, 3042, 4138, 2035, 2257, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cot (c+d x)^{5/2} (a+b \tan (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan (c+d x)^{5/2}}{(a+b \tan (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4048 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 \int \frac {a^2-b \tan (c+d x) a+\left (a^2+b^2\right ) \tan ^2(c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {a^2-b \tan (c+d x) a+\left (a^2+b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {a^2-b \tan (c+d x) a+\left (a^2+b^2\right ) \tan (c+d x)^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {a^2-b \tan (c+d x) a+\left (a^2+b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{b d \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 \int \frac {a^2-b \tan (c+d x) a+\left (a^2+b^2\right ) \tan ^2(c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 2257 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 \int \left (\frac {a^2+b^2}{\sqrt {a+b \tan (c+d x)}}-\frac {b^2+a \tan (c+d x) b}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {2 \left (\frac {\left (a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}-\frac {b (b+i a) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 \sqrt {-b+i a}}+\frac {b (-b+i a) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 \sqrt {b+i a}}\right )}{b d \left (a^2+b^2\right )}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((2*(-1/2*(b*(I*a + b)*ArcTan[(Sqrt[ I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[I*a - b] + (( a^2 + b^2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]) /Sqrt[b] + ((I*a - b)*b*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(2*Sqrt[I*a + b])))/(b*(a^2 + b^2)*d) - (2*a^2*Sqrt[Ta n[c + d*x]])/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]))
3.9.72.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol ] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a , c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(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] - Simp[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] && GtQ[m, 2] && LtQ [n, -1] && IntegerQ[2*m]
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] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[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]
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(2936\) vs. \(2(211)=422\).
Time = 39.78 (sec) , antiderivative size = 2937, normalized size of antiderivative = 11.52
1/4/d*(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^3*((csc(d*x+c)^2*a*(1-cos(d*x+c))^ 2-2*b*(csc(d*x+c)-cot(d*x+c))-a)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1))^(1/2)* (4*2^(1/2)*(a^2+b^2)^(1/2)*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2 *b*(csc(d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)))^(1/2)*(-b+(a^2+b^2)^(1/2))^( 1/2)*arctanh(1/2/(1-cos(d*x+c))*sin(d*x+c)*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1 -cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)))^(1/2)*2^(1/2 )/b^(1/2))*a^2+4*2^(1/2)*(a^2+b^2)^(1/2)*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-c os(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)))^(1/2)*(-b+(a^2 +b^2)^(1/2))^(1/2)*arctanh(1/2/(1-cos(d*x+c))*sin(d*x+c)*(-csc(d*x+c)*(csc (d*x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)) )^(1/2)*2^(1/2)/b^(1/2))*b^2-2*ln(1/(1-cos(d*x+c))*(-csc(d*x+c)*a*(1-cos(d *x+c))^2+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))+2*sin(d*x+c)*(-csc(d*x+c)*(csc(d *x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)))^ (1/2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*b*(1-cos(d*x+c))+sin(d*x+c)*a))*b^(5/2)* (-b+(a^2+b^2)^(1/2))^(1/2)*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2 *b*(csc(d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)))^(1/2)*(b+(a^2+b^2)^(1/2))^(1 /2)+(a^2+b^2)^(1/2)*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc (d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*(-b +(a^2+b^2)^(1/2))^(1/2)*ln(1/(1-cos(d*x+c))*(-csc(d*x+c)*a*(1-cos(d*x+c))^ 2+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))+2*sin(d*x+c)*(-csc(d*x+c)*(csc(d*x+c...
Leaf count of result is larger than twice the leaf count of optimal. 7807 vs. \(2 (207) = 414\).
Time = 2.36 (sec) , antiderivative size = 15647, normalized size of antiderivative = 61.36 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]