Integrand size = 33, antiderivative size = 261 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}+\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}+\frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (a^2 A b+7 A b^3-4 a^3 B-10 a b^2 B\right )}{3 b^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {2 \left (a A b-4 a^2 B-3 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d} \]
(A-I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(5/2)/d+(A+I *B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(5/2)/d-2/3*a^2* (A*a^2*b+7*A*b^3-4*B*a^3-10*B*a*b^2)/b^3/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1 /2)-2/3*(A*a*b-4*B*a^2-3*B*b^2)*(a+b*tan(d*x+c))^(1/2)/b^3/(a^2+b^2)/d+2/3 *a*(A*b-B*a)*tan(d*x+c)^2/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 3.64 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.18 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {-2 (a-i b) (a+i b) \left (-2 a A b+8 a^2 B+b^2 B\right )-b^2 (a A+b B) \left (i (a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )-(i a+b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )\right )-6 (a-i b) (a+i b) b (-A b+4 a B) \tan (c+d x)-6 (a-i b) (a+i b) b^2 B \tan ^2(c+d x)+3 A b^2 \left (i (a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )-(i a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )\right ) (a+b \tan (c+d x))}{3 b^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}} \]
-1/3*(-2*(a - I*b)*(a + I*b)*(-2*a*A*b + 8*a^2*B + b^2*B) - b^2*(a*A + b*B )*(I*(a + I*b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a - I*b)] - (I*a + b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a + I*b)]) - 6*(a - I*b)*(a + I*b)*b*(-(A*b) + 4*a*B)*Tan[c + d*x] - 6*(a - I*b)*(a + I*b)*b^2*B*Tan[c + d*x]^2 + 3*A*b^2*(I*(a + I*b)*Hypergeometric 2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I*b)] - (I*a + b)*Hypergeometr ic2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)])*(a + b*Tan[c + d*x])) /(b^3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2))
Time = 1.45 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.14, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4088, 27, 3042, 4118, 3042, 4113, 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.
| \(\displaystyle \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^3 (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {2 \int -\frac {\tan (c+d x) \left (\left (-4 B a^2+A b a-3 b^2 B\right ) \tan ^2(c+d x)-3 b (A b-a B) \tan (c+d x)+4 a (A b-a B)\right )}{2 (a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}+\frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan (c+d x) \left (\left (-4 B a^2+A b a-3 b^2 B\right ) \tan ^2(c+d x)-3 b (A b-a B) \tan (c+d x)+4 a (A b-a B)\right )}{(a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan (c+d x) \left (\left (-4 B a^2+A b a-3 b^2 B\right ) \tan (c+d x)^2-3 b (A b-a B) \tan (c+d x)+4 a (A b-a B)\right )}{(a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4118 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {3 \left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2+\left (a^2+b^2\right ) \left (-4 B a^2+A b a-3 b^2 B\right ) \tan ^2(c+d x)+a \left (-4 B a^3+A b a^2-10 b^2 B a+7 A b^3\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}+\frac {2 a^2 \left (-4 a^3 B+a^2 A b-10 a b^2 B+7 A b^3\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {3 \left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2+\left (a^2+b^2\right ) \left (-4 B a^2+A b a-3 b^2 B\right ) \tan (c+d x)^2+a \left (-4 B a^3+A b a^2-10 b^2 B a+7 A b^3\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}+\frac {2 a^2 \left (-4 a^3 B+a^2 A b-10 a b^2 B+7 A b^3\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {3 \left (-B a^2+2 A b a+b^2 B\right ) b^2+3 \left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2+b^2\right ) \left (-4 a^2 B+a A b-3 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}+\frac {2 a^2 \left (-4 a^3 B+a^2 A b-10 a b^2 B+7 A b^3\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {3 \left (-B a^2+2 A b a+b^2 B\right ) b^2+3 \left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2+b^2\right ) \left (-4 a^2 B+a A b-3 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}+\frac {2 a^2 \left (-4 a^3 B+a^2 A b-10 a b^2 B+7 A b^3\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 a^2 \left (-4 a^3 B+a^2 A b-10 a b^2 B+7 A b^3\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {3}{2} b^2 (a-i b)^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {3}{2} b^2 (a+i b)^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2+b^2\right ) \left (-4 a^2 B+a A b-3 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 a^2 \left (-4 a^3 B+a^2 A b-10 a b^2 B+7 A b^3\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {3}{2} b^2 (a-i b)^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {3}{2} b^2 (a+i b)^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2+b^2\right ) \left (-4 a^2 B+a A b-3 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 a^2 \left (-4 a^3 B+a^2 A b-10 a b^2 B+7 A b^3\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {-\frac {3 i b^2 (a+i b)^2 (B+i A) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {3 i b^2 (a-i b)^2 (-B+i A) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (a^2+b^2\right ) \left (-4 a^2 B+a A b-3 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 a^2 \left (-4 a^3 B+a^2 A b-10 a b^2 B+7 A b^3\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {3 i b^2 (a+i b)^2 (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {3 i b^2 (a-i b)^2 (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (a^2+b^2\right ) \left (-4 a^2 B+a A b-3 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 a^2 \left (-4 a^3 B+a^2 A b-10 a b^2 B+7 A b^3\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {3 b (a-i b)^2 (-B+i A) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}-\frac {3 b (a+i b)^2 (B+i A) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {2 \left (a^2+b^2\right ) \left (-4 a^2 B+a A b-3 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 a^2 \left (-4 a^3 B+a^2 A b-10 a b^2 B+7 A b^3\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (a^2+b^2\right ) \left (-4 a^2 B+a A b-3 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {3 b^2 (a+i b)^2 (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {3 b^2 (a-i b)^2 (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
(2*a*(A*b - a*B)*Tan[c + d*x]^2)/(3*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^( 3/2)) - ((2*a^2*(a^2*A*b + 7*A*b^3 - 4*a^3*B - 10*a*b^2*B))/(b^2*(a^2 + b^ 2)*d*Sqrt[a + b*Tan[c + d*x]]) + ((-3*(a + I*b)^2*b^2*(I*A + B)*ArcTan[Tan [c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + (3*(a - I*b)^2*b^2*(I*A - B) *ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) + (2*(a^2 + b^2)*(a *A*b - 4*a^2*B - 3*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(b*d))/(b*(a^2 + b^2)) )/(3*b*(a^2 + b^2))
3.4.58.3.1 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[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*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_.)*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_)])*((c_.) + (d_.)*tan[(e_.) + (f_. )*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(c^2*C - B*c*d + A*d^2)*((c + d*Tan[e + f*x])^(n + 1)/(d^2*f*(n + 1)*(c^2 + d^2))), x] + Simp[1/(d*(c^2 + d^2)) Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b* (c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d + a*C*d) *Tan[e + f*x] + b*C*(c^2 + d^2)*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[c^2 + d^2, 0] && LtQ[n , -1]
Leaf count of result is larger than twice the leaf count of optimal. \(4547\) vs. \(2(231)=462\).
Time = 0.14 (sec) , antiderivative size = 4548, normalized size of antiderivative = 17.43
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(4548\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(12907\) |
| default | \(\text {Expression too large to display}\) | \(12907\) |
A*(-6/d*a^2/(a^2+b^2)^2/(a+b*tan(d*x+c))^(1/2)+2/3/d/b^2*a^3/(a^2+b^2)/(a+ b*tan(d*x+c))^(3/2)+1/d/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arct an(((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))*a^5+3/4/d/(a^2+b^2)^(7/2)*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))*(2*(a^2+b^2)^(1/2 )+2*a)^(1/2)*a^4-1/2/d/(a^2+b^2)^3*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))*(2*(a^2+b^2)^(1/2)+2*a)^( 1/2)*a^3-1/d/(a^2+b^2)^3/(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) )*a^4+1/2/d/(a^2+b^2)^3*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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3+1/ d/(a^2+b^2)^3/(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))*a^4+2/d/( a^2+b^2)^(5/2)/(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))*a^3-1/d/ (a^2+b^2)^(7/2)/(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))*a^5-3/4 /d/(a^2+b^2)^(7/2)*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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^4-1/4/d*b ^4/(a^2+b^2)^(7/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2...
Leaf count of result is larger than twice the leaf count of optimal. 7284 vs. \(2 (227) = 454\).
Time = 2.20 (sec) , antiderivative size = 7284, normalized size of antiderivative = 27.91 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Time = 37.09 (sec) , antiderivative size = 9498, normalized size of antiderivative = 36.39 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
(log(((((320*B^4*a^2*b^8*d^4 - 16*B^4*b^10*d^4 - 1760*B^4*a^4*b^6*d^4 + 16 00*B^4*a^6*b^4*d^4 - 400*B^4*a^8*b^2*d^4)^(1/2) - 4*B^2*a^5*d^2 + 40*B^2*a ^3*b^2*d^2 - 20*B^2*a*b^4*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 10*a ^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2)*((a + b*tan(c + d*x))^ (1/2)*(320*B^2*a^4*b^14*d^3 - 16*B^2*b^18*d^3 + 1024*B^2*a^6*b^12*d^3 + 14 40*B^2*a^8*b^10*d^3 + 1024*B^2*a^10*b^8*d^3 + 320*B^2*a^12*b^6*d^3 - 16*B^ 2*a^16*b^2*d^3) - ((((320*B^4*a^2*b^8*d^4 - 16*B^4*b^10*d^4 - 1760*B^4*a^4 *b^6*d^4 + 1600*B^4*a^6*b^4*d^4 - 400*B^4*a^8*b^2*d^4)^(1/2) - 4*B^2*a^5*d ^2 + 40*B^2*a^3*b^2*d^2 - 20*B^2*a*b^4*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b ^8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2)*(((((320* B^4*a^2*b^8*d^4 - 16*B^4*b^10*d^4 - 1760*B^4*a^4*b^6*d^4 + 1600*B^4*a^6*b^ 4*d^4 - 400*B^4*a^8*b^2*d^4)^(1/2) - 4*B^2*a^5*d^2 + 40*B^2*a^3*b^2*d^2 - 20*B^2*a*b^4*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^ 22*d^5 + 640*a^3*b^20*d^5 + 2880*a^5*b^18*d^5 + 7680*a^7*b^16*d^5 + 13440* a^9*b^14*d^5 + 16128*a^11*b^12*d^5 + 13440*a^13*b^10*d^5 + 7680*a^15*b^8*d ^5 + 2880*a^17*b^6*d^5 + 640*a^19*b^4*d^5 + 64*a^21*b^2*d^5))/4 - 32*B*b^2 1*d^4 - 160*B*a^2*b^19*d^4 - 128*B*a^4*b^17*d^4 + 896*B*a^6*b^15*d^4 + 313 6*B*a^8*b^13*d^4 + 4928*B*a^10*b^11*d^4 + 4480*B*a^12*b^9*d^4 + 2432*B*a^1 4*b^7*d^4 + 736*B*a^16*b^5*d^4 + 96*B*a^18*b^3*d^4))/4))/4 + 96*B^3*a^3...