Integrand size = 35, antiderivative size = 196 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(i a-b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(i a+b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (4 A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \]
(I*a-b)^(3/2)*(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c ))^(1/2))/d+(I*a+b)^(3/2)*(A-I*B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/( a+b*tan(d*x+c))^(1/2))/d-2/3*(4*A*b+3*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))^(1/2)/d/tan(d*x+c)^(3/2)
Time = 1.00 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {3 \sqrt [4]{-1} \left ((-a+i b)^{3/2} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+i (a+i b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right ) \tan ^{\frac {3}{2}}(c+d x)-3 b B \sqrt {a+b \tan (c+d x)}+(-2 a A+3 b B) \sqrt {a+b \tan (c+d x)}-2 (4 A b+3 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]
(3*(-1)^(1/4)*((-a + I*b)^(3/2)*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b ]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + I*(a + I*b)^(3/2)*(A + I *B)*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) - 3*b*B*Sqrt[a + b*Tan[c + d*x]] + (-2*a*A + 3*b*B)*Sqrt[a + b*Tan[c + d*x]] - 2*(4*A*b + 3*a*B)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2))
Time = 1.34 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.15, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4088, 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.
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan (c+d x)^{5/2}}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {2}{3} \int \frac {-b (2 a A-3 b B) \tan ^2(c+d x)-3 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (4 A b+3 a B)}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {-b (2 a A-3 b B) \tan ^2(c+d x)-3 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (4 A b+3 a B)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {-b (2 a A-3 b B) \tan (c+d x)^2-3 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (4 A b+3 a B)}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {1}{3} \left (-\frac {2 \int \frac {3 \left (a \left (A a^2-2 b B a-A b^2\right )+a \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (-\frac {3 \int \frac {a \left (A a^2-2 b B a-A b^2\right )+a \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (-\frac {3 \int \frac {a \left (A a^2-2 b B a-A b^2\right )+a \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4099 |
| \(\displaystyle -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {1}{2} a (a-i b)^2 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b)^2 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {1}{2} a (a-i b)^2 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b)^2 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}\right )\) |
| \(\Big \downarrow \) 4098 |
| \(\displaystyle -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a-i b)^2 (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+\frac {a (a+i b)^2 (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a}\right )\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a-i b)^2 (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a (a+i b)^2 (A+i B) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a-i b)^2 (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a (a+i b)^2 (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a+i b)^2 (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {a (a-i b)^2 (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\right )}{a}\right )\) |
(-2*a*A*Sqrt[a + b*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) + ((-3*((a*(a + I*b)^2*(A + I*B)*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)^2*(A - I*B)*ArcTanh[(Sqrt[I* a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d)))/ a - (2*(4*A*b + 3*a*B)*Sqrt[a + b*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/3
3.5.38.3.1 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[(((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_)*((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_)*((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 = 0.91 (sec) , antiderivative size = 2398858, normalized size of antiderivative = 12239.07
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 12024 vs. \(2 (156) = 312\).
Time = 2.46 (sec) , antiderivative size = 12024, normalized size of antiderivative = 61.35 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]