Integrand size = 14, antiderivative size = 134 \[ \int (a+b \tan (c+d x))^{5/2} \, dx=-\frac {i (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {i (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d} \]
-I*(a-I*b)^(5/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+I*(a+I*b) ^(5/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d+4*a*b*(a+b*tan(d*x+ c))^(1/2)/d+2/3*b*(a+b*tan(d*x+c))^(3/2)/d
Time = 0.42 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90 \[ \int (a+b \tan (c+d x))^{5/2} \, dx=\frac {-3 i (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+3 i (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+2 b \sqrt {a+b \tan (c+d x)} (7 a+b \tan (c+d x))}{3 d} \]
((-3*I)*(a - I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + (3*I)*(a + I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + 2* b*Sqrt[a + b*Tan[c + d*x]]*(7*a + b*Tan[c + d*x]))/(3*d)
Time = 0.72 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.84, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {3042, 3963, 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.
\(\displaystyle \int (a+b \tan (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \tan (c+d x))^{5/2}dx\) |
\(\Big \downarrow \) 3963 |
\(\displaystyle \int \sqrt {a+b \tan (c+d x)} \left (a^2+2 b \tan (c+d x) a-b^2\right )dx+\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a+b \tan (c+d x)} \left (a^2+2 b \tan (c+d x) a-b^2\right )dx+\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \frac {a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {1}{2} (a-i b)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} (a-i b)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {i (a-i b)^3 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a+i b)^3 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i (a-i b)^3 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (a+i b)^3 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a-i b)^3 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b)^3 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a-i b)^{5/2} \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}+\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\) |
((a - I*b)^(5/2)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d + ((a + I*b)^(5/2)* ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d + (4*a*b*Sqrt[a + b*Tan[c + d*x]])/d + (2*b*(a + b*Tan[c + d*x])^(3/2))/(3*d)
3.6.22.3.1 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[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d *x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[n, 1]
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. \(1173\) vs. \(2(110)=220\).
Time = 0.08 (sec) , antiderivative size = 1174, normalized size of antiderivative = 8.76
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1174\) |
default | \(\text {Expression too large to display}\) | \(1174\) |
2/3*b*(a+b*tan(d*x+c))^(3/2)/d+4*a*b*(a+b*tan(d*x+c))^(1/2)/d+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))*(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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)-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) )*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3+3/4/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))*(2*(a^2+b^2)^(1/ 2)+2*a)^(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))*( a^2+b^2)^(1/2)*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))* a^2+1/d*b^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))-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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a^2+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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)+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 ))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3-3/4/d*b*ln(b*tan(d*x+c)+a+(a+b*tan...
Leaf count of result is larger than twice the leaf count of optimal. 1125 vs. \(2 (104) = 208\).
Time = 0.26 (sec) , antiderivative size = 1125, normalized size of antiderivative = 8.40 \[ \int (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]
-1/6*(3*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*b - 14*a^ 4*b^5 - 8*a^2*b^7 + b^9)*sqrt(b*tan(d*x + c) + a) + ((a^2 - b^2)*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) + 2*(5* a^5*b^2 - 10*a^3*b^4 + a*b^6)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*s qrt(-(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)) - 3*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*b - 14*a^ 4*b^5 - 8*a^2*b^7 + b^9)*sqrt(b*tan(d*x + c) + a) - ((a^2 - b^2)*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) + 2*(5* a^5*b^2 - 10*a^3*b^4 + a*b^6)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*s qrt(-(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)) - 3*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*b - 14*a^ 4*b^5 - 8*a^2*b^7 + b^9)*sqrt(b*tan(d*x + c) + a) + ((a^2 - b^2)*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) - 2*(5* a^5*b^2 - 10*a^3*b^4 + a*b^6)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*s qrt(-(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)) + 3*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*b - 14...
\[ \int (a+b \tan (c+d x))^{5/2} \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]
Exception generated. \[ \int (a+b \tan (c+d x))^{5/2} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]
Time = 7.76 (sec) , antiderivative size = 2100, normalized size of antiderivative = 15.67 \[ \int (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]
(2*b*(a + b*tan(c + d*x))^(3/2))/(3*d) - atan(((((8*(8*a*b^5*d^2 + 8*a^3*b ^3*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*(8*a*b^5*d^2 + 8*a^3*b^3*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*(8*a*b^5*d^2 + 8*a^3*b^3*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*5 i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*t an(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) - (16*(6*a^4*b^7 - b^11 + 8*a^6*b^5 + 3*a^8*b^3))/d^3 + (((8*(8*a*b^5*d^2 + 8*a^3*b^3*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*...