Integrand size = 33, antiderivative size = 220 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a} \left (8 a^2 A-15 A b^2-20 a b B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {(a-i b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {a (7 A b+4 a B) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d} \] Output:
1/4*a^(1/2)*(8*A*a^2-15*A*b^2-20*B*a*b)*arctanh((a+b*tan(d*x+c))^(1/2)/a^( 1/2))/d-(a-I*b)^(5/2)*(A-I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2) )/d-(a+I*b)^(5/2)*(A+I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d- 1/4*a*(7*A*b+4*B*a)*cot(d*x+c)*(a+b*tan(d*x+c))^(1/2)/d-1/2*a*A*cot(d*x+c) ^2*(a+b*tan(d*x+c))^(3/2)/d
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(448\) vs. \(2(220)=440\).
Time = 1.63 (sec) , antiderivative size = 448, normalized size of antiderivative = 2.04 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {-\sqrt {a} \left (8 a^2 A-15 A b^2-20 a b B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )+4 (a-i b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+4 a^2 A \sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+8 i a A \sqrt {a+i b} b \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )-4 A \sqrt {a+i b} b^2 \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+4 i a^2 \sqrt {a+i b} B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )-8 a \sqrt {a+i b} b B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )-4 i \sqrt {a+i b} b^2 B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+9 a A b \cot (c+d x) \sqrt {a+b \tan (c+d x)}+4 a^2 B \cot (c+d x) \sqrt {a+b \tan (c+d x)}+2 a^2 A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{4 d} \] Input:
Integrate[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x ]
Output:
-1/4*(-(Sqrt[a]*(8*a^2*A - 15*A*b^2 - 20*a*b*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]]) + 4*(a - I*b)^(5/2)*(A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + 4*a^2*A*Sqrt[a + I*b]*ArcTanh[Sqrt[a + b*Tan[c + d* x]]/Sqrt[a + I*b]] + (8*I)*a*A*Sqrt[a + I*b]*b*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] - 4*A*Sqrt[a + I*b]*b^2*ArcTanh[Sqrt[a + b*Tan[c + d* x]]/Sqrt[a + I*b]] + (4*I)*a^2*Sqrt[a + I*b]*B*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] - 8*a*Sqrt[a + I*b]*b*B*ArcTanh[Sqrt[a + b*Tan[c + d* x]]/Sqrt[a + I*b]] - (4*I)*Sqrt[a + I*b]*b^2*B*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + 9*a*A*b*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]] + 4*a ^2*B*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]] + 2*a^2*A*Cot[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]])/d
Time = 1.87 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.97, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 4088, 27, 3042, 4128, 27, 3042, 4136, 27, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 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.
\(\displaystyle \int \cot ^3(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)^3}dx\) |
\(\Big \downarrow \) 4088 |
\(\displaystyle \frac {1}{2} \int \frac {1}{2} \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)} \left (-b (a A-4 b B) \tan ^2(c+d x)-4 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (7 A b+4 a B)\right )dx-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)} \left (-b (a A-4 b B) \tan ^2(c+d x)-4 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (7 A b+4 a B)\right )dx-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (a A-4 b B) \tan (c+d x)^2-4 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (7 A b+4 a B)\right )}{\tan (c+d x)^2}dx-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 4128 |
\(\displaystyle \frac {1}{4} \left (\int -\frac {\cot (c+d x) \left (b \left (4 B a^2+9 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-20 b B a-15 A b^2\right )\right )}{2 \sqrt {a+b \tan (c+d x)}}dx-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \frac {\cot (c+d x) \left (b \left (4 B a^2+9 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-20 b B a-15 A b^2\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \frac {b \left (4 B a^2+9 A b a-8 b^2 B\right ) \tan (c+d x)^2+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-20 b B a-15 A b^2\right )}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 4136 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-a \left (8 a^2 A-20 a b B-15 A b^2\right ) \int \frac {\cot (c+d x) \left (\tan ^2(c+d x)+1\right )}{\sqrt {a+b \tan (c+d x)}}dx-\int \frac {8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3-\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)}}dx\right )-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-a \left (8 a^2 A-20 a b B-15 A b^2\right ) \int \frac {\cot (c+d x) \left (\tan ^2(c+d x)+1\right )}{\sqrt {a+b \tan (c+d x)}}dx-8 \int \frac {B a^3+3 A b a^2-3 b^2 B a-A b^3-\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-a \left (8 a^2 A-20 a b B-15 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-8 \int \frac {B a^3+3 A b a^2-3 b^2 B a-A b^3-\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle -\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (-a \left (8 a^2 A-20 a b B-15 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-8 \left (\frac {1}{2} (a-i b)^3 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (a+i b)^3 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )\right )\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (-a \left (8 a^2 A-20 a b B-15 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-8 \left (\frac {1}{2} (a-i b)^3 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (a+i b)^3 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )\right )\right )\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle -\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (-a \left (8 a^2 A-20 a b B-15 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-8 \left (\frac {i (a-i b)^3 (B+i A) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (a+i b)^3 (-B+i A) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}\right )\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (-a \left (8 a^2 A-20 a b B-15 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-8 \left (-\frac {i (a-i b)^3 (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a+i b)^3 (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}\right )\right )\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (-a \left (8 a^2 A-20 a b B-15 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-8 \left (\frac {(a-i b)^3 (B+i A) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}-\frac {(a+i b)^3 (-B+i A) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}\right )\right )\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (-a \left (8 a^2 A-20 a b B-15 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-8 \left (\frac {(a-i b)^{5/2} (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )\right )\right )\) |
\(\Big \downarrow \) 4117 |
\(\displaystyle -\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (-\frac {a \left (8 a^2 A-20 a b B-15 A b^2\right ) \int \frac {\cot (c+d x)}{\sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{d}-8 \left (\frac {(a-i b)^{5/2} (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )\right )\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (-\frac {2 a \left (8 a^2 A-20 a b B-15 A b^2\right ) \int \frac {1}{\frac {a+b \tan (c+d x)}{b}-\frac {a}{b}}d\sqrt {a+b \tan (c+d x)}}{b d}-8 \left (\frac {(a-i b)^{5/2} (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )\right )\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (-\frac {a (4 a B+7 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (\frac {2 \sqrt {a} \left (8 a^2 A-20 a b B-15 A b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{d}-8 \left (\frac {(a-i b)^{5/2} (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )\right )\right )\) |
Input:
Int[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]
Output:
-1/2*(a*A*Cot[c + d*x]^2*(a + b*Tan[c + d*x])^(3/2))/d + ((-8*(((a - I*b)^ (5/2)*(I*A + B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d - ((a + I*b)^(5/2)*( I*A - B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d) + (2*Sqrt[a]*(8*a^2*A - 15 *A*b^2 - 20*a*b*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/d)/2 - (a*(7 *A*b + 4*a*B)*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/d)/4
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_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
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[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^ n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ (A*b^2 - a*b*B + a^2*C)/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ e + f*x]^2)/(a + b*Tan[e + f*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[n, 0] && !LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(2487\) vs. \(2(186)=372\).
Time = 0.17 (sec) , antiderivative size = 2488, normalized size of antiderivative = 11.31
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(2488\) |
default | \(\text {Expression too large to display}\) | \(2488\) |
Input:
int(cot(d*x+c)^3*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x,method=_RETURNV ERBOSE)
Output:
1/4/d/b*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))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a^2- 1/4/d/b*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))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a^2+ 2/d*b/(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))*B*(a^2+b^2)^(1/2) *a-2/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))*B*(a^2+b^2)^(1 /2)*a-3/d*b/(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))*B*a^2+1/4/d *b*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))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)-3/4/d*b*l n((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))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-1/d*b^2/(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*(a^2+b^2)^(1/2)+3/d*b^2/(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*a+3/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))*B*a^2+1/d*b^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arcta...
Leaf count of result is larger than twice the leaf count of optimal. 4932 vs. \(2 (180) = 360\).
Time = 15.97 (sec) , antiderivative size = 9885, normalized size of antiderivative = 44.93 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:
integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorith m="fricas")
Output:
Too large to include
Timed out. \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)**3*(a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)
Output:
Timed out
Timed out. \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorith m="maxima")
Output:
Timed out
Exception generated. \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorith m="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Time = 7.35 (sec) , antiderivative size = 32561, normalized size of antiderivative = 148.00 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:
int(cot(c + d*x)^3*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2),x)
Output:
atan(-((((3708*A^3*a^2*b^16*d^2 - 6912*A^3*a^4*b^14*d^2 - 5820*A^3*a^6*b^1 2*d^2 + 4608*A^3*a^8*b^10*d^2 - 192*A^3*a^10*b^8*d^2 - 6400*B^3*a^3*b^15*d ^2 + 2816*B^3*a^5*b^13*d^2 + 7808*B^3*a^7*b^11*d^2 - 1472*B^3*a^9*b^9*d^2 + 64*B^3*a*b^17*d^2 - 1856*A^2*B*a*b^17*d^2 - 7552*A*B^2*a^2*b^16*d^2 + 23 488*A*B^2*a^4*b^14*d^2 + 16256*A*B^2*a^6*b^12*d^2 - 14208*A*B^2*a^8*b^10*d ^2 + 576*A*B^2*a^10*b^8*d^2 + 20504*A^2*B*a^3*b^15*d^2 - 5000*A^2*B*a^5*b^ 13*d^2 - 22944*A^2*B*a^7*b^11*d^2 + 4416*A^2*B*a^9*b^9*d^2)/(2*d^5) - (((1 664*A*a*b^12*d^4 + 896*A*a^3*b^10*d^4 - 768*A*a^5*b^8*d^4 + 2048*B*a^2*b^1 1*d^4 + 2048*B*a^4*b^9*d^4)/(2*d^5) - ((512*b^10*d^4 + 768*a^2*b^8*d^4)*(a + b*tan(c + d*x))^(1/2)*((((8*B^2*a^5*d^2 - 8*A^2*a^5*d^2 + 80*A^2*a^3*b^ 2*d^2 - 80*B^2*a^3*b^2*d^2 + 16*A*B*b^5*d^2 - 40*A^2*a*b^4*d^2 + 40*B^2*a* b^4*d^2 - 160*A*B*a^2*b^3*d^2 + 80*A*B*a^4*b*d^2)^2/64 - d^4*(A^4*a^10 + A ^4*b^10 + B^4*a^10 + B^4*b^10 + 2*A^2*B^2*a^10 + 2*A^2*B^2*b^10 + 5*A^4*a^ 2*b^8 + 10*A^4*a^4*b^6 + 10*A^4*a^6*b^4 + 5*A^4*a^8*b^2 + 5*B^4*a^2*b^8 + 10*B^4*a^4*b^6 + 10*B^4*a^6*b^4 + 5*B^4*a^8*b^2 + 10*A^2*B^2*a^2*b^8 + 20* A^2*B^2*a^4*b^6 + 20*A^2*B^2*a^6*b^4 + 10*A^2*B^2*a^8*b^2))^(1/2) + A^2*a^ 5*d^2 - B^2*a^5*d^2 - 10*A^2*a^3*b^2*d^2 + 10*B^2*a^3*b^2*d^2 - 2*A*B*b^5* d^2 + 5*A^2*a*b^4*d^2 - 5*B^2*a*b^4*d^2 + 20*A*B*a^2*b^3*d^2 - 10*A*B*a^4* b*d^2)/(4*d^4))^(1/2))/d^4)*((((8*B^2*a^5*d^2 - 8*A^2*a^5*d^2 + 80*A^2*a^3 *b^2*d^2 - 80*B^2*a^3*b^2*d^2 + 16*A*B*b^5*d^2 - 40*A^2*a*b^4*d^2 + 40*...
\[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int \cot \left (d x +c \right )^{3} \left (a +\tan \left (d x +c \right ) b \right )^{\frac {5}{2}} \left (A +B \tan \left (d x +c \right )\right )d x \] Input:
int(cot(d*x+c)^3*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)
Output:
int(cot(d*x+c)^3*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)