Integrand size = 33, antiderivative size = 233 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a-i b} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 \left (14 a A b-8 a^2 B+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d} \]
(A-I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))*(a-I*b)^(1/2)/d+(A+I *B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))*(a+I*b)^(1/2)/d-2*A*(a+b *tan(d*x+c))^(1/2)/d-2/105*(14*A*a*b-8*B*a^2+35*B*b^2)*(a+b*tan(d*x+c))^(3 /2)/b^3/d+2/35*(7*A*b-4*B*a)*tan(d*x+c)*(a+b*tan(d*x+c))^(3/2)/b^2/d+2/7*B *tan(d*x+c)^2*(a+b*tan(d*x+c))^(3/2)/b/d
Time = 2.97 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.91 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a-i b} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-14 a^2 A b-105 A b^3+8 a^3 B-35 a b^2 B-b \left (-7 a A b+4 a^2 B+35 b^2 B\right ) \tan (c+d x)+3 b^2 (7 A b+a B) \tan ^2(c+d x)+15 b^3 B \tan ^3(c+d x)\right )}{105 b^3 d} \]
(Sqrt[a - I*b]*(A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/ d + (Sqrt[a + I*b]*(A + I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b ]])/d + (2*Sqrt[a + b*Tan[c + d*x]]*(-14*a^2*A*b - 105*A*b^3 + 8*a^3*B - 3 5*a*b^2*B - b*(-7*a*A*b + 4*a^2*B + 35*b^2*B)*Tan[c + d*x] + 3*b^2*(7*A*b + a*B)*Tan[c + d*x]^2 + 15*b^3*B*Tan[c + d*x]^3))/(105*b^3*d)
Time = 1.46 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.12, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 4090, 27, 3042, 4130, 27, 3042, 4113, 3042, 4011, 3042, 4022, 3042, 4020, 25, 73, 221}
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.
\(\displaystyle \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^3 \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))dx\) |
\(\Big \downarrow \) 4090 |
\(\displaystyle \frac {2 \int -\frac {1}{2} \tan (c+d x) \sqrt {a+b \tan (c+d x)} \left (-\left ((7 A b-4 a B) \tan ^2(c+d x)\right )+7 b B \tan (c+d x)+4 a B\right )dx}{7 b}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \left (-\left ((7 A b-4 a B) \tan ^2(c+d x)\right )+7 b B \tan (c+d x)+4 a B\right )dx}{7 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \left (-\left ((7 A b-4 a B) \tan (c+d x)^2\right )+7 b B \tan (c+d x)+4 a B\right )dx}{7 b}\) |
\(\Big \downarrow \) 4130 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {2 \int \frac {1}{2} \sqrt {a+b \tan (c+d x)} \left (35 A \tan (c+d x) b^2+\left (-8 B a^2+14 A b a+35 b^2 B\right ) \tan ^2(c+d x)+2 a (7 A b-4 a B)\right )dx}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \sqrt {a+b \tan (c+d x)} \left (35 A \tan (c+d x) b^2+\left (-8 B a^2+14 A b a+35 b^2 B\right ) \tan ^2(c+d x)+2 a (7 A b-4 a B)\right )dx}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \sqrt {a+b \tan (c+d x)} \left (35 A \tan (c+d x) b^2+\left (-8 B a^2+14 A b a+35 b^2 B\right ) \tan (c+d x)^2+2 a (7 A b-4 a B)\right )dx}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\) |
\(\Big \downarrow \) 4113 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \sqrt {a+b \tan (c+d x)} \left (35 A b^2 \tan (c+d x)-35 b^2 B\right )dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \sqrt {a+b \tan (c+d x)} \left (35 A b^2 \tan (c+d x)-35 b^2 B\right )dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \frac {35 b^2 (a A-b B) \tan (c+d x)-35 b^2 (A b+a B)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \frac {35 b^2 (a A-b B) \tan (c+d x)-35 b^2 (A b+a B)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {\frac {35}{2} b^2 (-b+i a) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {35}{2} b^2 (b+i a) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {\frac {35}{2} b^2 (-b+i a) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {35}{2} b^2 (b+i a) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {-\frac {35 i b^2 (b+i a) (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {35 i b^2 (-b+i a) (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {\frac {35 i b^2 (b+i a) (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {35 i b^2 (-b+i a) (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {\frac {35 b (-b+i a) (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}-\frac {35 b (b+i a) (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}-\frac {35 b^2 (b+i a) (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {35 b^2 (-b+i a) (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\) |
(2*B*Tan[c + d*x]^2*(a + b*Tan[c + d*x])^(3/2))/(7*b*d) - ((-2*(7*A*b - 4* a*B)*Tan[c + d*x]*(a + b*Tan[c + d*x])^(3/2))/(5*b*d) + ((-35*b^2*(I*a + b )*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + (35*(I *a - b)*b^2*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d ) + (70*A*b^2*Sqrt[a + b*Tan[c + d*x]])/d + (2*(14*a*A*b - 8*a^2*B + 35*b^ 2*B)*(a + b*Tan[c + d*x])^(3/2))/(3*b*d))/(5*b))/(7*b)
3.4.17.3.1 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_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[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[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Ta n[e + f*x])^n*Simp[a^2*A*d*(m + n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m - 1) - b *(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 , 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -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] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C *(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[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] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ c, 0] && NeQ[a, 0])))
Leaf count of result is larger than twice the leaf count of optimal. \(936\) vs. \(2(201)=402\).
Time = 0.40 (sec) , antiderivative size = 937, normalized size of antiderivative = 4.02
method | result | size |
parts | \(\frac {2 A \left (\frac {\left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {a \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\sqrt {a +b \tan \left (d x +c \right )}\, b^{2}-b^{2} \left (-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )\right )}{d \,b^{2}}+B \left (\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d \,b^{3}}-\frac {4 \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}} a}{5 d \,b^{3}}+\frac {2 a^{2} \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d \,b^{3}}-\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d b}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 d b}\right )\) | \(937\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1092\) |
default | \(\text {Expression too large to display}\) | \(1092\) |
2*A/d/b^2*(1/5*(a+b*tan(d*x+c))^(5/2)-1/3*a*(a+b*tan(d*x+c))^(3/2)-(a+b*ta n(d*x+c))^(1/2)*b^2-b^2*(-1/8*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln(b*tan(d*x+c )+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))+ 1/2*(a-(a^2+b^2)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d *x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)) +1/8*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^ (1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))+1/2*((a^2+b^2)^(1/2)-a)/( 2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b* tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))))+B*(2/7/d/b^3*(a+b*tan( d*x+c))^(7/2)-4/5/d/b^3*(a+b*tan(d*x+c))^(5/2)*a+2/3/d/b^3*a^2*(a+b*tan(d* x+c))^(3/2)-2/3/d/b*(a+b*tan(d*x+c))^(3/2)-1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a) ^(1/2)*(a^2+b^2)^(1/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^ 2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))+1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*a rctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2 )^(1/2)-2*a)^(1/2))+1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a*ln(b*tan(d*x+c )+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))+ 1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*ln((a+b*tan(d*x+c))^ (1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))-1/d*b/ (2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b *tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))-1/4/d/b*(2*(a^2+b^2)...
Leaf count of result is larger than twice the leaf count of optimal. 1312 vs. \(2 (195) = 390\).
Time = 0.27 (sec) , antiderivative size = 1312, normalized size of antiderivative = 5.63 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
-1/210*(105*b^3*d*sqrt(-(2*A*B*b + d^2*sqrt(-(4*A^2*B^2*a^2 + 4*(A^3*B - A *B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/d^4) - (A^2 - B^2)*a)/d^2)*log(-( 2*(A^3*B + A*B^3)*a + (A^4 - B^4)*b)*sqrt(b*tan(d*x + c) + a) + (B*d^3*sqr t(-(4*A^2*B^2*a^2 + 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/d ^4) + (2*A^2*B*a + (A^3 - A*B^2)*b)*d)*sqrt(-(2*A*B*b + d^2*sqrt(-(4*A^2*B ^2*a^2 + 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/d^4) - (A^2 - B^2)*a)/d^2)) - 105*b^3*d*sqrt(-(2*A*B*b + d^2*sqrt(-(4*A^2*B^2*a^2 + 4* (A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/d^4) - (A^2 - B^2)*a)/d ^2)*log(-(2*(A^3*B + A*B^3)*a + (A^4 - B^4)*b)*sqrt(b*tan(d*x + c) + a) - (B*d^3*sqrt(-(4*A^2*B^2*a^2 + 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B ^4)*b^2)/d^4) + (2*A^2*B*a + (A^3 - A*B^2)*b)*d)*sqrt(-(2*A*B*b + d^2*sqrt (-(4*A^2*B^2*a^2 + 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/d^ 4) - (A^2 - B^2)*a)/d^2)) - 105*b^3*d*sqrt(-(2*A*B*b - d^2*sqrt(-(4*A^2*B^ 2*a^2 + 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/d^4) - (A^2 - B^2)*a)/d^2)*log(-(2*(A^3*B + A*B^3)*a + (A^4 - B^4)*b)*sqrt(b*tan(d*x + c) + a) + (B*d^3*sqrt(-(4*A^2*B^2*a^2 + 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A ^2*B^2 + B^4)*b^2)/d^4) - (2*A^2*B*a + (A^3 - A*B^2)*b)*d)*sqrt(-(2*A*B*b - d^2*sqrt(-(4*A^2*B^2*a^2 + 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^ 4)*b^2)/d^4) - (A^2 - B^2)*a)/d^2)) + 105*b^3*d*sqrt(-(2*A*B*b - d^2*sqrt( -(4*A^2*B^2*a^2 + 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/...
\[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )}\, dx \]
\[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
Time = 66.52 (sec) , antiderivative size = 1093, normalized size of antiderivative = 4.69 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {2\,A\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,b^2\,d}-\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (2\,a\,\left (\frac {2\,B\,\left (a^2+b^2\right )}{b^3\,d}-\frac {4\,B\,a^2}{b^3\,d}\right )+\frac {8\,B\,a^3}{b^3\,d}-\frac {4\,B\,a\,\left (a^2+b^2\right )}{b^3\,d}\right )-\left (\frac {2\,A\,\left (a^2+b^2\right )}{b^2\,d}-\frac {2\,A\,a^2}{b^2\,d}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}-\left (\frac {2\,B\,\left (a^2+b^2\right )}{3\,b^3\,d}-\frac {4\,B\,a^2}{3\,b^3\,d}\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}+\frac {2\,B\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{7/2}}{7\,b^3\,d}-\frac {2\,A\,a\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{3\,b^2\,d}-\frac {4\,B\,a\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,b^3\,d}+\mathrm {atan}\left (\frac {d^3\,\left (\frac {16\,\left (B^2\,b^4-B^2\,a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}+\frac {16\,a\,b^2\,\left (\sqrt {-B^4\,b^2\,d^4}+B^2\,a\,d^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^4}\right )\,\sqrt {-\frac {\sqrt {-B^4\,b^2\,d^4}+B^2\,a\,d^2}{4\,d^4}}\,1{}\mathrm {i}}{8\,\left (B^3\,a^2\,b^3+B^3\,b^5\right )}\right )\,\sqrt {-\frac {\sqrt {-B^4\,b^2\,d^4}+B^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {d^3\,\left (\frac {16\,\left (B^2\,b^4-B^2\,a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}-\frac {16\,a\,b^2\,\left (\sqrt {-B^4\,b^2\,d^4}-B^2\,a\,d^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^4}\right )\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}-B^2\,a\,d^2}{4\,d^4}}\,1{}\mathrm {i}}{8\,\left (B^3\,a^2\,b^3+B^3\,b^5\right )}\right )\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}-B^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {A^2\,b^4\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}}{4\,d^4}+\frac {A^2\,a}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {16\,A\,b^4\,\sqrt {-A^4\,b^2\,d^4}}{d^3}+\frac {16\,A\,a^2\,b^2\,\sqrt {-A^4\,b^2\,d^4}}{d^3}}+\frac {a\,b^2\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}}{4\,d^4}+\frac {A^2\,a}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-A^4\,b^2\,d^4}\,32{}\mathrm {i}}{\frac {16\,A\,b^4\,\sqrt {-A^4\,b^2\,d^4}}{d}+\frac {16\,A\,a^2\,b^2\,\sqrt {-A^4\,b^2\,d^4}}{d}}\right )\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}+A^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {A^2\,b^4\,\sqrt {\frac {A^2\,a}{4\,d^2}-\frac {\sqrt {-A^4\,b^2\,d^4}}{4\,d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {16\,A\,b^4\,\sqrt {-A^4\,b^2\,d^4}}{d^3}+\frac {16\,A\,a^2\,b^2\,\sqrt {-A^4\,b^2\,d^4}}{d^3}}-\frac {a\,b^2\,\sqrt {\frac {A^2\,a}{4\,d^2}-\frac {\sqrt {-A^4\,b^2\,d^4}}{4\,d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-A^4\,b^2\,d^4}\,32{}\mathrm {i}}{\frac {16\,A\,b^4\,\sqrt {-A^4\,b^2\,d^4}}{d}+\frac {16\,A\,a^2\,b^2\,\sqrt {-A^4\,b^2\,d^4}}{d}}\right )\,\sqrt {-\frac {\sqrt {-A^4\,b^2\,d^4}-A^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i} \]
atan((d^3*((16*(B^2*b^4 - B^2*a^2*b^2)*(a + b*tan(c + d*x))^(1/2))/d^2 + ( 16*a*b^2*((-B^4*b^2*d^4)^(1/2) + B^2*a*d^2)*(a + b*tan(c + d*x))^(1/2))/d^ 4)*(-((-B^4*b^2*d^4)^(1/2) + B^2*a*d^2)/(4*d^4))^(1/2)*1i)/(8*(B^3*b^5 + B ^3*a^2*b^3)))*(-((-B^4*b^2*d^4)^(1/2) + B^2*a*d^2)/(4*d^4))^(1/2)*2i - ((2 *B*(a^2 + b^2))/(3*b^3*d) - (4*B*a^2)/(3*b^3*d))*(a + b*tan(c + d*x))^(3/2 ) + atan((d^3*((16*(B^2*b^4 - B^2*a^2*b^2)*(a + b*tan(c + d*x))^(1/2))/d^2 - (16*a*b^2*((-B^4*b^2*d^4)^(1/2) - B^2*a*d^2)*(a + b*tan(c + d*x))^(1/2) )/d^4)*(((-B^4*b^2*d^4)^(1/2) - B^2*a*d^2)/(4*d^4))^(1/2)*1i)/(8*(B^3*b^5 + B^3*a^2*b^3)))*(((-B^4*b^2*d^4)^(1/2) - B^2*a*d^2)/(4*d^4))^(1/2)*2i - ( a + b*tan(c + d*x))^(1/2)*(2*a*((2*B*(a^2 + b^2))/(b^3*d) - (4*B*a^2)/(b^3 *d)) + (8*B*a^3)/(b^3*d) - (4*B*a*(a^2 + b^2))/(b^3*d)) - atan((A^2*b^4*(( -A^4*b^2*d^4)^(1/2)/(4*d^4) + (A^2*a)/(4*d^2))^(1/2)*(a + b*tan(c + d*x))^ (1/2)*32i)/((16*A*b^4*(-A^4*b^2*d^4)^(1/2))/d^3 + (16*A*a^2*b^2*(-A^4*b^2* d^4)^(1/2))/d^3) + (a*b^2*((-A^4*b^2*d^4)^(1/2)/(4*d^4) + (A^2*a)/(4*d^2)) ^(1/2)*(a + b*tan(c + d*x))^(1/2)*(-A^4*b^2*d^4)^(1/2)*32i)/((16*A*b^4*(-A ^4*b^2*d^4)^(1/2))/d + (16*A*a^2*b^2*(-A^4*b^2*d^4)^(1/2))/d))*(((-A^4*b^2 *d^4)^(1/2) + A^2*a*d^2)/(4*d^4))^(1/2)*2i + atan((A^2*b^4*((A^2*a)/(4*d^2 ) - (-A^4*b^2*d^4)^(1/2)/(4*d^4))^(1/2)*(a + b*tan(c + d*x))^(1/2)*32i)/(( 16*A*b^4*(-A^4*b^2*d^4)^(1/2))/d^3 + (16*A*a^2*b^2*(-A^4*b^2*d^4)^(1/2))/d ^3) - (a*b^2*((A^2*a)/(4*d^2) - (-A^4*b^2*d^4)^(1/2)/(4*d^4))^(1/2)*(a ...