Integrand size = 25, antiderivative size = 180 \[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {\text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 b \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\tan (c+d x)}} \]
-arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(I*a-b)^( 1/2)-arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(I*a +b)^(1/2)+4/3*b*(a+b*tan(d*x+c))^(1/2)/a^2/d/tan(d*x+c)^(1/2)-2/3*(a+b*tan (d*x+c))^(1/2)/a/d/tan(d*x+c)^(3/2)
Time = 1.78 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {\frac {3 (-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {3 (-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}-\frac {2 (a-2 b \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}{a^2 \tan ^{\frac {3}{2}}(c+d x)}}{3 d} \]
((3*(-1)^(3/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[ a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] + (3*(-1)^(3/4)*ArcTan[((-1)^(1/4)*Sq rt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b] - (2*(a - 2*b*Tan[c + d*x])*Sqrt[a + b*Tan[c + d*x]])/(a^2*Tan[c + d*x]^(3/ 2)))/(3*d)
Time = 0.72 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4052, 27, 3042, 4132, 27, 3042, 4058, 615, 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.
| \(\displaystyle \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (c+d x)^{5/2} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4052 |
| \(\displaystyle -\frac {2 \int \frac {2 b \tan ^2(c+d x)+3 a \tan (c+d x)+2 b}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {2 b \tan ^2(c+d x)+3 a \tan (c+d x)+2 b}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {2 b \tan (c+d x)^2+3 a \tan (c+d x)+2 b}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle -\frac {-\frac {2 \int -\frac {3 a^2}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {3 a \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 a \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4058 |
| \(\displaystyle -\frac {\frac {3 a \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {3 a \int \left (\frac {i}{2 (i-\tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {i}{2 \sqrt {\tan (c+d x)} (\tan (c+d x)+i) \sqrt {a+b \tan (c+d x)}}\right )d\tan (c+d x)}{d}}{3 a}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {4 b \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {3 a \left (\frac {\arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}+\frac {\text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}\right )}{d}}{3 a}\) |
(-2*Sqrt[a + b*Tan[c + d*x]])/(3*a*d*Tan[c + d*x]^(3/2)) - ((3*a*(ArcTan[( Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/Sqrt[I*a - b] + ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/Sqr t[I*a + b]))/d - (4*b*Sqrt[a + b*Tan[c + d*x]])/(a*d*Sqrt[Tan[c + d*x]]))/ (3*a)
3.7.38.3.1 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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 /((m + 1)*(a^2 + b^2)*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*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] && LtQ[m, -1] && (LtQ[n, 0] || Integ erQ[m]) && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), 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/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a *d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
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[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* (m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d )*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta n[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] && LtQ[m, -1] && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 4.65 (sec) , antiderivative size = 945613, normalized size of antiderivative = 5253.41
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 4096 vs. \(2 (144) = 288\).
Time = 0.73 (sec) , antiderivative size = 4096, normalized size of antiderivative = 22.76 \[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]
-1/24*(3*a^2*d*sqrt(-((a^2 + b^2)*d^2*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d ^4)) - b)/((a^2 + b^2)*d^2))*log(1/2*(2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^ 3*b^2 + 4*a*b^4)*tan(d*x + c) + (2*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*d^2*tan(d *x + c) - (a^6 + 4*a^4*b^2 + 7*a^2*b^4 + 4*b^6)*d^2)*sqrt(-a^2/((a^4 + 2*a ^2*b^2 + b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + (2*(a^3 *b^3 + 4*a*b^5)*d*tan(d*x + c)^2 + 2*(a^6 + 5*a^4*b^2 + 8*a^2*b^4)*d*tan(d *x + c) + 2*(a^5*b + 2*a^3*b^3)*d + ((a^7 + 8*a^5*b^2 + 19*a^3*b^4 + 12*a* b^6)*d^3*tan(d*x + c)^2 + 2*(a^6*b + 2*a^4*b^3 - 3*a^2*b^5 - 4*b^7)*d^3*ta n(d*x + c) - (a^7 + 4*a^5*b^2 + 7*a^3*b^4 + 4*a*b^6)*d^3)*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(-((a^2 + b^2)*d^2*sqrt(-a^2/((a^4 + 2*a^2*b ^2 + b^4)*d^4)) - b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1))*tan(d*x + c )^2 + 3*a^2*d*sqrt(-((a^2 + b^2)*d^2*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^ 4)) - b)/((a^2 + b^2)*d^2))*log(-1/2*(2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^ 3*b^2 + 4*a*b^4)*tan(d*x + c) + (2*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*d^2*tan(d *x + c) - (a^6 + 4*a^4*b^2 + 7*a^2*b^4 + 4*b^6)*d^2)*sqrt(-a^2/((a^4 + 2*a ^2*b^2 + b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + (2*(a^3 *b^3 + 4*a*b^5)*d*tan(d*x + c)^2 + 2*(a^6 + 5*a^4*b^2 + 8*a^2*b^4)*d*tan(d *x + c) + 2*(a^5*b + 2*a^3*b^3)*d + ((a^7 + 8*a^5*b^2 + 19*a^3*b^4 + 12*a* b^6)*d^3*tan(d*x + c)^2 + 2*(a^6*b + 2*a^4*b^3 - 3*a^2*b^5 - 4*b^7)*d^3*ta n(d*x + c) - (a^7 + 4*a^5*b^2 + 7*a^3*b^4 + 4*a*b^6)*d^3)*sqrt(-a^2/((a...
\[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {1}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]