Integrand size = 35, antiderivative size = 240 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {(i a-b)^{5/2} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {2 b^{5/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (2 A b+a B) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]
-(I*a-b)^(5/2)*(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+ c))^(1/2))/d+2*b^(5/2)*B*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c)) ^(1/2))/d-(I*a+b)^(5/2)*(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+ b*tan(d*x+c))^(1/2))/d-2*a*(2*A*b+B*a)*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c) ^(1/2)-2/3*a*A*(a+b*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 4.47 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.74 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {i a b (A+i B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},-\frac {b \tan (c+d x)}{a}\right ) \sqrt {a+b \tan (c+d x)}-a b (i A+B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},-\frac {b \tan (c+d x)}{a}\right ) \sqrt {a+b \tan (c+d x)}+(i a+b) (A-i B) \sqrt {1+\frac {b \tan (c+d x)}{a}} \left (3 \sqrt [4]{-1} (-a+i b)^{3/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \tan ^{\frac {3}{2}}(c+d x)+(i a+(-3 a+4 i b) \tan (c+d x)) \sqrt {a+b \tan (c+d x)}\right )+(i a-b) (A+i B) \sqrt {1+\frac {b \tan (c+d x)}{a}} \left (3 \sqrt [4]{-1} (a+i b)^{3/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \tan ^{\frac {3}{2}}(c+d x)+(i a+(3 a+4 i b) \tan (c+d x)) \sqrt {a+b \tan (c+d x)}\right )}{3 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]
(I*a*b*(A + I*B)*Hypergeometric2F1[-3/2, -3/2, -1/2, -((b*Tan[c + d*x])/a) ]*Sqrt[a + b*Tan[c + d*x]] - a*b*(I*A + B)*Hypergeometric2F1[-3/2, -3/2, - 1/2, -((b*Tan[c + d*x])/a)]*Sqrt[a + b*Tan[c + d*x]] + (I*a + b)*(A - I*B) *Sqrt[1 + (b*Tan[c + d*x])/a]*(3*(-1)^(1/4)*(-a + I*b)^(3/2)*ArcTan[((-1)^ (1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Tan[c + d*x]^(3/2) + (I*a + (-3*a + (4*I)*b)*Tan[c + d*x])*Sqrt[a + b*Tan[c + d*x ]]) + (I*a - b)*(A + I*B)*Sqrt[1 + (b*Tan[c + d*x])/a]*(3*(-1)^(1/4)*(a + I*b)^(3/2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b *Tan[c + d*x]]]*Tan[c + d*x]^(3/2) + (I*a + (3*a + (4*I)*b)*Tan[c + d*x])* Sqrt[a + b*Tan[c + d*x]]))/(3*d*Tan[c + d*x]^(3/2)*Sqrt[1 + (b*Tan[c + d*x ])/a])
Time = 1.43 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 4088, 27, 3042, 4128, 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.
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)^{5/2}}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {2}{3} \int \frac {3 \sqrt {a+b \tan (c+d x)} \left (b^2 B \tan ^2(c+d x)-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (2 A b+a B)\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)} \left (b^2 B \tan ^2(c+d x)-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (2 A b+a B)\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)} \left (b^2 B \tan (c+d x)^2-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (2 A b+a B)\right )}{\tan (c+d x)^{3/2}}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4128 |
| \(\displaystyle 2 \int -\frac {-B \tan ^2(c+d x) b^3+a \left (A a^2-3 b B a-3 A b^2\right )+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (a B+2 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\int \frac {-B \tan ^2(c+d x) b^3+a \left (A a^2-3 b B a-3 A b^2\right )+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (a B+2 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\int \frac {-B \tan (c+d x)^2 b^3+a \left (A a^2-3 b B a-3 A b^2\right )+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (a B+2 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle -\frac {\int \frac {-B \tan ^2(c+d x) b^3+a \left (A a^2-3 b B a-3 A b^2\right )+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {2 a (a B+2 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {2 \int \frac {-B \tan ^2(c+d x) b^3+a \left (A a^2-3 b B a-3 A b^2\right )+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}-\frac {2 a (a B+2 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 2257 |
| \(\displaystyle -\frac {2 \int \left (\frac {A a^3-3 b B a^2-3 A b^2 a+b^3 B+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}-\frac {b^3 B}{\sqrt {a+b \tan (c+d x)}}\right )d\sqrt {\tan (c+d x)}}{d}-\frac {2 a (a B+2 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \left (\frac {1}{2} (-b+i a)^{5/2} (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\frac {1}{2} (b+i a)^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-b^{5/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )}{d}-\frac {2 a (a B+2 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
(-2*(((I*a - b)^(5/2)*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/ Sqrt[a + b*Tan[c + d*x]]])/2 - b^(5/2)*B*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x ]])/Sqrt[a + b*Tan[c + d*x]]] + ((I*a + b)^(5/2)*(I*A + B)*ArcTanh[(Sqrt[I *a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/2))/d - (2*a*(2*A*b + a*B)*Sqrt[a + b*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + b*T an[c + d*x])^(3/2))/(3*d*Tan[c + d*x]^(3/2))
3.5.46.3.1 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_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((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 - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d* (b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1) + a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[ e + f*x] - b*(d*(A*b*c + a*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] & & LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
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*d^2 + c*(c*C - B*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim p[1/(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* (n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b *(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ [a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
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]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.02 (sec) , antiderivative size = 2654078, normalized size of antiderivative = 11058.66
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 17675 vs. \(2 (194) = 388\).
Time = 7.33 (sec) , antiderivative size = 35349, normalized size of antiderivative = 147.29 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]