Integrand size = 25, antiderivative size = 186 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {(i a+b) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}+\frac {(i a-b) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{5/2} f}+\frac {2 (b c-a d)}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{\left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \]
-(I*a+b)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(5/2)/f+(I* a-b)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(5/2)/f-2*(2*a* c*d-b*(c^2-d^2))/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)+2/3*(-a*d+b*c)/(c^2+ d^2)/f/(c+d*tan(f*x+e))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.62 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i \left (-\frac {(a-i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )}{c-i d}+\frac {(a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )}{c+i d}\right )}{3 f (c+d \tan (e+f x))^{3/2}} \]
((-1/3*I)*(-(((a - I*b)*Hypergeometric2F1[-3/2, 1, -1/2, (c + d*Tan[e + f* x])/(c - I*d)])/(c - I*d)) + ((a + I*b)*Hypergeometric2F1[-3/2, 1, -1/2, ( c + d*Tan[e + f*x])/(c + I*d)])/(c + I*d)))/(f*(c + d*Tan[e + f*x])^(3/2))
Time = 0.91 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4012, 3042, 4012, 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.
\(\displaystyle \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {\int \frac {a c+b d+(b c-a d) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx}{c^2+d^2}+\frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a c+b d+(b c-a d) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx}{c^2+d^2}+\frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {\frac {\int \frac {2 b c d+a \left (c^2-d^2\right )-\left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{c^2+d^2}+\frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {2 b c d+a \left (c^2-d^2\right )-\left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{c^2+d^2}+\frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} (a+i b) (c-i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{c^2+d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} (a+i b) (c-i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{c^2+d^2}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {i (a-i b) (c+i d)^2 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a+i b) (c-i d)^2 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}}{c^2+d^2}}{c^2+d^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {i (a+i b) (c-i d)^2 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {i (a-i b) (c+i d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}}{c^2+d^2}}{c^2+d^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(a+i b) (c-i d)^2 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a-i b) (c+i d)^2 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{c^2+d^2}}{c^2+d^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(a+i b) (c-i d)^2 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {(a-i b) (c+i d)^2 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}}{c^2+d^2}}{c^2+d^2}\) |
(2*(b*c - a*d))/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + ((((a - I*b )*(c + I*d)^2*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + ((a + I*b)*(c - I*d)^2*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f))/ (c^2 + d^2) - (2*(2*a*c*d - b*(c^2 - d^2)))/((c^2 + d^2)*f*Sqrt[c + d*Tan[ e + f*x]]))/(c^2 + d^2)
3.13.62.3.1 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[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
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. \(4465\) vs. \(2(162)=324\).
Time = 1.11 (sec) , antiderivative size = 4466, normalized size of antiderivative = 24.01
method | result | size |
parts | \(\text {Expression too large to display}\) | \(4466\) |
derivativedivides | \(\text {Expression too large to display}\) | \(12836\) |
default | \(\text {Expression too large to display}\) | \(12836\) |
a*(-4/f*d*c/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)+1/f*d^3/(c^2+d^2)^(5/2)/(2* (c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^( 1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-2/f*d^5/(c^2+d^2)^(7/2)/(2 *(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^ (1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+1/f*d^3/(c^2+d^2)^(5/2)/( 2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2) ^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-2/f*d^5/(c^2+d^2)^(7/2)/ (2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2 )^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+1/4/f*d^3/(c^2+d^2)^3*l n(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2 +d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-1/4/f*d^3/(c^2+d^2)^3*ln(d*tan( f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1 /2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2/3/f*d/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2 )+4/f*d/(c^2+d^2)^(7/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f *x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)) *c^4-2/f*d/(c^2+d^2)^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f* x+e))^(1/2)-(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))* c^3-2/f*d/(c^2+d^2)^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x +e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c ^3-1/4/f/d/(c^2+d^2)^(7/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*...
Leaf count of result is larger than twice the leaf count of optimal. 7181 vs. \(2 (157) = 314\).
Time = 2.10 (sec) , antiderivative size = 7181, normalized size of antiderivative = 38.61 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {a + b \tan {\left (e + f x \right )}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(d-c>0)', see `assume?` for more details)Is
Timed out. \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Time = 26.51 (sec) , antiderivative size = 9459, normalized size of antiderivative = 50.85 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
(log(8*b^3*d^16*f^2 - ((((320*b^4*c^2*d^8*f^4 - 16*b^4*d^10*f^4 - 1760*b^4 *c^4*d^6*f^4 + 1600*b^4*c^6*d^4*f^4 - 400*b^4*c^8*d^2*f^4)^(1/2) + 4*b^2*c ^5*f^2 + 20*b^2*c*d^4*f^2 - 40*b^2*c^3*d^2*f^2)/(c^10*f^4 + d^10*f^4 + 5*c ^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2)*((((( 320*b^4*c^2*d^8*f^4 - 16*b^4*d^10*f^4 - 1760*b^4*c^4*d^6*f^4 + 1600*b^4*c^ 6*d^4*f^4 - 400*b^4*c^8*d^2*f^4)^(1/2) + 4*b^2*c^5*f^2 + 20*b^2*c*d^4*f^2 - 40*b^2*c^3*d^2*f^2)/(c^10*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^ 4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2)*(((((320*b^4*c^2*d^8*f^4 - 16*b ^4*d^10*f^4 - 1760*b^4*c^4*d^6*f^4 + 1600*b^4*c^6*d^4*f^4 - 400*b^4*c^8*d^ 2*f^4)^(1/2) + 4*b^2*c^5*f^2 + 20*b^2*c*d^4*f^2 - 40*b^2*c^3*d^2*f^2)/(c^1 0*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8 *d^2*f^4))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(64*c*d^22*f^5 + 640*c^3*d^20* f^5 + 2880*c^5*d^18*f^5 + 7680*c^7*d^16*f^5 + 13440*c^9*d^14*f^5 + 16128*c ^11*d^12*f^5 + 13440*c^13*d^10*f^5 + 7680*c^15*d^8*f^5 + 2880*c^17*d^6*f^5 + 640*c^19*d^4*f^5 + 64*c^21*d^2*f^5))/4 + 736*b*c^3*d^18*f^4 + 2432*b*c^ 5*d^16*f^4 + 4480*b*c^7*d^14*f^4 + 4928*b*c^9*d^12*f^4 + 3136*b*c^11*d^10* f^4 + 896*b*c^13*d^8*f^4 - 128*b*c^15*d^6*f^4 - 160*b*c^17*d^4*f^4 - 32*b* c^19*d^2*f^4 + 96*b*c*d^20*f^4))/4 + (c + d*tan(e + f*x))^(1/2)*(320*b^2*c ^4*d^14*f^3 - 16*b^2*d^18*f^3 + 1024*b^2*c^6*d^12*f^3 + 1440*b^2*c^8*d^10* f^3 + 1024*b^2*c^10*d^8*f^3 + 320*b^2*c^12*d^6*f^3 - 16*b^2*c^16*d^2*f^...