Integrand size = 21, antiderivative size = 158 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=-\frac {(a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d} \]
-(a-I*b)^(5/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d-(a+I*b)^(5/ 2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d+2*(a^2-b^2)*(a+b*tan(d* x+c))^(1/2)/d+2/3*a*(a+b*tan(d*x+c))^(3/2)/d+2/5*(a+b*tan(d*x+c))^(5/2)/d
Time = 0.81 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.87 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {-15 (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )-15 (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+2 \sqrt {a+b \tan (c+d x)} \left (23 a^2-15 b^2+11 a b \tan (c+d x)+3 b^2 \tan ^2(c+d x)\right )}{15 d} \]
(-15*(a - I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] - 15* (a + I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + 2*Sqrt[a + b*Tan[c + d*x]]*(23*a^2 - 15*b^2 + 11*a*b*Tan[c + d*x] + 3*b^2*Tan[c + d*x]^2))/(15*d)
Time = 0.84 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4011, 3042, 4011, 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.
| \(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int (a \tan (c+d x)-b) (a+b \tan (c+d x))^{3/2}dx+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \tan (c+d x)-b) (a+b \tan (c+d x))^{3/2}dx+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \sqrt {a+b \tan (c+d x)} \left (\left (a^2-b^2\right ) \tan (c+d x)-2 a b\right )dx+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {a+b \tan (c+d x)} \left (\left (a^2-b^2\right ) \tan (c+d x)-2 a b\right )dx+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \frac {a \left (a^2-3 b^2\right ) \tan (c+d x)-b \left (3 a^2-b^2\right )}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a \left (a^2-3 b^2\right ) \tan (c+d x)-b \left (3 a^2-b^2\right )}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle -\frac {1}{2} (-b+i a)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (b+i a)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {1}{2} (-b+i a)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (b+i a)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {i (-b+i a)^3 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {i (b+i a)^3 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {i (-b+i a)^3 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i (b+i a)^3 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {(-b+i a)^3 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(b+i a)^3 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {(-b+i a)^3 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {(b+i a)^3 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\) |
((I*a + b)^3*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) - ((I*a - b)^3*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) + (2*(a^2 - b^2)*Sqrt[a + b*Tan[c + d*x]])/d + (2*a*(a + b*Tan[c + d*x])^(3/2))/(3*d) + (2*(a + b*Tan[c + d*x])^(5/2))/(5*d)
3.6.21.3.1 Defintions of rubi rules used
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]
Leaf count of result is larger than twice the leaf count of optimal. \(1360\) vs. \(2(132)=264\).
Time = 0.10 (sec) , antiderivative size = 1361, normalized size of antiderivative = 8.61
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1361\) |
| default | \(\text {Expression too large to display}\) | \(1361\) |
-3/d/(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^3+3/d/(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+3/4/d*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^2-1/4/d*b^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)-3/4/d*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^2+1/4/d*b^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)+2/5*(a+b*tan(d*x+c))^(5/2)/d+2/3*a* (a+b*tan(d*x+c))^(3/2)/d+2/d*a^2*(a+b*tan(d*x+c))^(1/2)-2*b^2*(a+b*tan(d*x +c))^(1/2)/d+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- 2/d/(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^2+b^2)*a-1/d*b^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+1/2/d*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^2+b^2)^(1/2)*a-1/d/(2*(a^2+b^2)^(1/2)-...
Leaf count of result is larger than twice the leaf count of optimal. 1122 vs. \(2 (128) = 256\).
Time = 0.27 (sec) , antiderivative size = 1122, normalized size of antiderivative = 7.10 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]
1/30*(15*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100* a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4* b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a) + (2*a*d^3*sqrt(-(25*a^8*b ^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) - (5*a^6 - 15*a^4 *b^2 + 11*a^2*b^4 - b^6)*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-( 25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)) - 15*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^ 4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a) - (2*a*d^3*sqrt(-(25*a^8*b^2 - 1 00*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) - (5*a^6 - 15*a^4*b^2 + 11*a^2*b^4 - b^6)*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8 *b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)) - 15*d*s qrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 11 0*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2* b^6 + b^8)*sqrt(b*tan(d*x + c) + a) + (2*a*d^3*sqrt(-(25*a^8*b^2 - 100*a^6 *b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) + (5*a^6 - 15*a^4*b^2 + 11*a^ 2*b^4 - b^6)*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)) + 15*d*sqrt((a ^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4* b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2*b^6...
\[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan {\left (c + d x \right )}\, dx \]
Exception generated. \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Timed out. \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]
Time = 13.32 (sec) , antiderivative size = 2191, normalized size of antiderivative = 13.87 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]
atan(((((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x) )^(1/2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4 *d^2))^(1/2))*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b ^2)/(4*d^2))^(1/2) - (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15 *a^4*b^4 - a^6*b^2))/d^2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10 i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i - (((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 + 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 - a^4*b*5i + a^5 - b ^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x) )^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*((5*a*b^4 - a^4*b* 5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i)/((((8*(4* b^6*d^2 - 4*a^4*b^2*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a* b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))* ((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^( 1/2) - (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6 *b^2))/d^2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2 )/(4*d^2))^(1/2) + (((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 + 64*a*b^2*(a + b *tan(c + d*x))^(1/2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 1 0*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*1 0i - 10*a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 -...