Integrand size = 25, antiderivative size = 138 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {(i a+b) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}+\frac {(i a-b) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{3/2} f}+\frac {2 (b c-a d)}{\left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \]
-(I*a+b)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(3/2)/f+(I* a-b)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(3/2)/f+2*(-a*d +b*c)/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.21 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {i \left (\frac {(a-i b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )}{c-i d}-\frac {(a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )}{c+i d}\right )}{f \sqrt {c+d \tan (e+f x)}} \]
(I*(((a - I*b)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I *d)])/(c - I*d) - ((a + I*b)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)])/(c + I*d)))/(f*Sqrt[c + d*Tan[e + f*x]])
Time = 0.60 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {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.
\(\displaystyle \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {\int \frac {a c+b d+(b c-a d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 (b c-a d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a c+b d+(b c-a d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 (b c-a d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {2 (b c-a d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} (a+i b) (c-i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) (c+i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 (b c-a d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} (a+i b) (c-i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) (c+i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {2 (b c-a d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {i (a-i b) (c+i d) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a+i b) (c-i d) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}}{c^2+d^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 (b c-a d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {i (a+i b) (c-i d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {i (a-i b) (c+i d) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}}{c^2+d^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 (b c-a d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(a+i b) (c-i d) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a-i b) (c+i d) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{c^2+d^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 (b c-a d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(a-i b) (c+i d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {(a+i b) (c-i d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{c^2+d^2}\) |
(((a - I*b)*(c + I*d)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f ) + ((a + I*b)*(c - I*d)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d ]*f))/(c^2 + d^2) + (2*(b*c - a*d))/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x] ])
3.13.56.3.1 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. \(3682\) vs. \(2(118)=236\).
Time = 0.91 (sec) , antiderivative size = 3683, normalized size of antiderivative = 26.69
method | result | size |
parts | \(\text {Expression too large to display}\) | \(3683\) |
derivativedivides | \(\text {Expression too large to display}\) | \(7951\) |
default | \(\text {Expression too large to display}\) | \(7951\) |
a*(-2/f*d/(c^2+d^2)/(c+d*tan(f*x+e))^(1/2)+1/4/f/d/(c^2+d^2)^2*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)*c^3+1/4/f*d/(c^2+d^2)^2*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)*c-1/4/f/d/(c^2+d^2)^(5/2)*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)*c^4+1/4/f*d^3/(c^2+d^2)^(5/2)*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)-1/f/d/(c^2+d^2)^(3/2)/(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^3-1/f*d/(c^2+d^2)^(3/2)/(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-1/f*d/(c^2+d^2)^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^2+1/f/d/(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))*c^5-1/f*d^3/(c^2+d^2)^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))+3/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))/(...
Leaf count of result is larger than twice the leaf count of optimal. 4318 vs. \(2 (113) = 226\).
Time = 0.73 (sec) , antiderivative size = 4318, normalized size of antiderivative = 31.29 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
1/2*(((c^2*d + d^3)*f*tan(f*x + e) + (c^3 + c*d^2)*f)*sqrt(-(6*a*b*c^2*d - 2*a*b*d^3 + (a^2 - b^2)*c^3 - 3*(a^2 - b^2)*c*d^2 + (c^6 + 3*c^4*d^2 + 3* c^2*d^4 + d^6)*f^2*sqrt(-(4*a^2*b^2*c^6 - 12*(a^3*b - a*b^3)*c^5*d + 3*(3* a^4 - 14*a^2*b^2 + 3*b^4)*c^4*d^2 + 40*(a^3*b - a*b^3)*c^3*d^3 - 6*(a^4 - 8*a^2*b^2 + b^4)*c^2*d^4 - 12*(a^3*b - a*b^3)*c*d^5 + (a^4 - 2*a^2*b^2 + b ^4)*d^6)/((c^12 + 6*c^10*d^2 + 15*c^8*d^4 + 20*c^6*d^6 + 15*c^4*d^8 + 6*c^ 2*d^10 + d^12)*f^4)))/((c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6)*f^2))*log(-(2*( a^3*b + a*b^3)*c^3 - 3*(a^4 - b^4)*c^2*d - 6*(a^3*b + a*b^3)*c*d^2 + (a^4 - b^4)*d^3)*sqrt(d*tan(f*x + e) + c) + ((a*c^8 + 2*b*c^7*d + 2*a*c^6*d^2 + 6*b*c^5*d^3 + 6*b*c^3*d^5 - 2*a*c^2*d^6 + 2*b*c*d^7 - a*d^8)*f^3*sqrt(-(4 *a^2*b^2*c^6 - 12*(a^3*b - a*b^3)*c^5*d + 3*(3*a^4 - 14*a^2*b^2 + 3*b^4)*c ^4*d^2 + 40*(a^3*b - a*b^3)*c^3*d^3 - 6*(a^4 - 8*a^2*b^2 + b^4)*c^2*d^4 - 12*(a^3*b - a*b^3)*c*d^5 + (a^4 - 2*a^2*b^2 + b^4)*d^6)/((c^12 + 6*c^10*d^ 2 + 15*c^8*d^4 + 20*c^6*d^6 + 15*c^4*d^8 + 6*c^2*d^10 + d^12)*f^4)) + (2*a *b^2*c^5 - (7*a^2*b - 3*b^3)*c^4*d + 2*(3*a^3 - 7*a*b^2)*c^3*d^2 + 4*(4*a^ 2*b - b^3)*c^2*d^3 - 2*(a^3 - 4*a*b^2)*c*d^4 - (a^2*b - b^3)*d^5)*f)*sqrt( -(6*a*b*c^2*d - 2*a*b*d^3 + (a^2 - b^2)*c^3 - 3*(a^2 - b^2)*c*d^2 + (c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6)*f^2*sqrt(-(4*a^2*b^2*c^6 - 12*(a^3*b - a*b^3 )*c^5*d + 3*(3*a^4 - 14*a^2*b^2 + 3*b^4)*c^4*d^2 + 40*(a^3*b - a*b^3)*c^3* d^3 - 6*(a^4 - 8*a^2*b^2 + b^4)*c^2*d^4 - 12*(a^3*b - a*b^3)*c*d^5 + (a...
\[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {a + b \tan {\left (e + f x \right )}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Time = 12.27 (sec) , antiderivative size = 5737, normalized size of antiderivative = 41.57 \[ \int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
(log(- (((c + d*tan(e + f*x))^(1/2)*(16*b^2*d^10*f^3 + 32*b^2*c^2*d^8*f^3 - 32*b^2*c^6*d^4*f^3 - 16*b^2*c^8*d^2*f^3) + ((((96*b^4*c^2*d^4*f^4 - 16*b ^4*d^6*f^4 - 144*b^4*c^4*d^2*f^4)^(1/2) + 4*b^2*c^3*f^2 - 12*b^2*c*d^2*f^2 )/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/2)*(32*b*d^12*f^ 4 + ((((96*b^4*c^2*d^4*f^4 - 16*b^4*d^6*f^4 - 144*b^4*c^4*d^2*f^4)^(1/2) + 4*b^2*c^3*f^2 - 12*b^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3* c^4*d^2*f^4))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(64*c*d^12*f^5 + 320*c^3*d^ 10*f^5 + 640*c^5*d^8*f^5 + 640*c^7*d^6*f^5 + 320*c^9*d^4*f^5 + 64*c^11*d^2 *f^5))/4 + 96*b*c^2*d^10*f^4 + 64*b*c^4*d^8*f^4 - 64*b*c^6*d^6*f^4 - 96*b* c^8*d^4*f^4 - 32*b*c^10*d^2*f^4))/4)*(((96*b^4*c^2*d^4*f^4 - 16*b^4*d^6*f^ 4 - 144*b^4*c^4*d^2*f^4)^(1/2) + 4*b^2*c^3*f^2 - 12*b^2*c*d^2*f^2)/(c^6*f^ 4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/2))/4 - 8*b^3*c*d^8*f^2 - 24*b^3*c^3*d^6*f^2 - 24*b^3*c^5*d^4*f^2 - 8*b^3*c^7*d^2*f^2)*(((96*b^4*c^ 2*d^4*f^4 - 16*b^4*d^6*f^4 - 144*b^4*c^4*d^2*f^4)^(1/2) + 4*b^2*c^3*f^2 - 12*b^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/ 2))/4 + (log(- (((c + d*tan(e + f*x))^(1/2)*(16*b^2*d^10*f^3 + 32*b^2*c^2* d^8*f^3 - 32*b^2*c^6*d^4*f^3 - 16*b^2*c^8*d^2*f^3) + ((-((96*b^4*c^2*d^4*f ^4 - 16*b^4*d^6*f^4 - 144*b^4*c^4*d^2*f^4)^(1/2) - 4*b^2*c^3*f^2 + 12*b^2* c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/2)*(32* b*d^12*f^4 + ((-((96*b^4*c^2*d^4*f^4 - 16*b^4*d^6*f^4 - 144*b^4*c^4*d^2...