Integrand size = 25, antiderivative size = 182 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {i (i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {i (i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \]
-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))/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))/d-14/3*a*b*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/3*a^2*( a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)
Time = 0.94 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-3 (-1)^{3/4} (-a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+3 (-1)^{3/4} (a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {2 a \sqrt {a+b \tan (c+d x)} (a+7 b \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)}}{3 d} \]
(-3*(-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]]] + 3*(-1)^(3/4)*(a + I*b)^(5/2)*ArcTa n[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - (2*a*Sqrt[a + b*Tan[c + d*x]]*(a + 7*b*Tan[c + d*x]))/Tan[c + d*x]^(3/2) )/(3*d)
Time = 1.23 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.13, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4048, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 219}
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.
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan (c+d x)^{5/2}}dx\) |
| \(\Big \downarrow \) 4048 |
| \(\displaystyle \frac {2}{3} \int \frac {7 b a^2-3 \left (a^2-3 b^2\right ) \tan (c+d x) a-b \left (2 a^2-3 b^2\right ) \tan ^2(c+d x)}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {7 b a^2-3 \left (a^2-3 b^2\right ) \tan (c+d x) a-b \left (2 a^2-3 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {7 b a^2-3 \left (a^2-3 b^2\right ) \tan (c+d x) a-b \left (2 a^2-3 b^2\right ) \tan (c+d x)^2}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {1}{3} \left (-\frac {2 \int \frac {3 \left (\left (a^2-3 b^2\right ) a^2+b \left (3 a^2-b^2\right ) \tan (c+d x) a\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (-\frac {3 \int \frac {\left (a^2-3 b^2\right ) a^2+b \left (3 a^2-b^2\right ) \tan (c+d x) a}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (-\frac {3 \int \frac {\left (a^2-3 b^2\right ) a^2+b \left (3 a^2-b^2\right ) \tan (c+d x) a}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4099 |
| \(\displaystyle -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {1}{2} a (a-i b)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {1}{2} a (a-i b)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}\right )\) |
| \(\Big \downarrow \) 4098 |
| \(\displaystyle -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a-i b)^3 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+\frac {a (a+i b)^3 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a}\right )\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a-i b)^3 \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a (a+i b)^3 \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a-i b)^3 \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a (a+i b)^3 \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a+i b)^3 \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {a (a-i b)^3 \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\right )}{a}\right )\) |
(-2*a^2*Sqrt[a + b*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) + ((-3*((a*(a + I*b)^3*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]] ])/(Sqrt[I*a - b]*d) + (a*(a - I*b)^3*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d)))/a - (14*a*b*Sqrt[a + b*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/3
3.7.28.3.1 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[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[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] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
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_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[A^2/f Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(A + I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T an[e + f*x]), x], x] + Simp[(A - I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d *Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A , B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^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 = 0.65 (sec) , antiderivative size = 1308720, normalized size of antiderivative = 7190.77
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 5026 vs. \(2 (142) = 284\).
Time = 0.84 (sec) , antiderivative size = 5026, normalized size of antiderivative = 27.62 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]