Integrand size = 25, antiderivative size = 138 \[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}+\frac {(i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}-\frac {2 (A b-a B)}{\left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]
-(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(3/2)/d+(I* A-B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(3/2)/d-2*(A*b- B*a)/(a^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.23 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.82 \[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {i \left (\frac {(A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )}{a-i b}-\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )}{a+i b}\right )}{d \sqrt {a+b \tan (c+d x)}} \]
(I*(((A - I*B)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I *b)])/(a - I*b) - ((A + I*B)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)])/(a + I*b)))/(d*Sqrt[a + b*Tan[c + d*x]])
Time = 0.61 (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.
\(\displaystyle \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {\int \frac {a A+b B-(A b-a B) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 (A b-a B)}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a A+b B-(A b-a B) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 (A b-a B)}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle -\frac {2 (A b-a B)}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (a-i b) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 (A b-a B)}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (a-i b) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle -\frac {2 (A b-a B)}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i (a+i b) (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a-i b) (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}}{a^2+b^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 (A b-a B)}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i (a-i b) (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i (a+i b) (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}}{a^2+b^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {2 (A b-a B)}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(a-i b) (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b) (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}}{a^2+b^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 (A b-a B)}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(a+i b) (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {(a-i b) (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{a^2+b^2}\) |
(((a + I*b)*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d ) + ((a - I*b)*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b ]*d))/(a^2 + b^2) - (2*(A*b - a*B))/((a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x] ])
3.4.53.3.1 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. \(3683\) vs. \(2(118)=236\).
Time = 0.09 (sec) , antiderivative size = 3684, normalized size of antiderivative = 26.70
method | result | size |
parts | \(\text {Expression too large to display}\) | \(3684\) |
derivativedivides | \(\text {Expression too large to display}\) | \(7951\) |
default | \(\text {Expression too large to display}\) | \(7951\) |
A*(1/4/d/b/(a^2+b^2)^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)*a^3+1/4 /d*b/(a^2+b^2)^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)*a-1/4/d/b/(a^ 2+b^2)^(5/2)*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)*a^4+1/4/d*b^3/(a^ 2+b^2)^(5/2)*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)-1/d/b/(a^2+b^2)^( 3/2)/(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^3-1/d*b/(a^2+b^2 )^(3/2)/(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/d*b/(a^2+b^ 2)^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^2+1/d/b/(a^2+b^2 )^(5/2)/(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^5-1/d*b^3/(a^ 2+b^2)^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))+3/d*b^3/(a^2+b ^2)^(5/2)/(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+4/d*b/(a...
Leaf count of result is larger than twice the leaf count of optimal. 4318 vs. \(2 (113) = 226\).
Time = 0.69 (sec) , antiderivative size = 4318, normalized size of antiderivative = 31.29 \[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
1/2*(((a^2*b + b^3)*d*tan(d*x + c) + (a^3 + a*b^2)*d)*sqrt(-(6*A*B*a^2*b - 2*A*B*b^3 + (A^2 - B^2)*a^3 - 3*(A^2 - B^2)*a*b^2 + (a^6 + 3*a^4*b^2 + 3* a^2*b^4 + b^6)*d^2*sqrt(-(4*A^2*B^2*a^6 - 12*(A^3*B - A*B^3)*a^5*b + 3*(3* A^4 - 14*A^2*B^2 + 3*B^4)*a^4*b^2 + 40*(A^3*B - A*B^3)*a^3*b^3 - 6*(A^4 - 8*A^2*B^2 + B^4)*a^2*b^4 - 12*(A^3*B - A*B^3)*a*b^5 + (A^4 - 2*A^2*B^2 + B ^4)*b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^ 2*b^10 + b^12)*d^4)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*log(-(2*( A^3*B + A*B^3)*a^3 - 3*(A^4 - B^4)*a^2*b - 6*(A^3*B + A*B^3)*a*b^2 + (A^4 - B^4)*b^3)*sqrt(b*tan(d*x + c) + a) + ((A*a^8 + 2*B*a^7*b + 2*A*a^6*b^2 + 6*B*a^5*b^3 + 6*B*a^3*b^5 - 2*A*a^2*b^6 + 2*B*a*b^7 - A*b^8)*d^3*sqrt(-(4 *A^2*B^2*a^6 - 12*(A^3*B - A*B^3)*a^5*b + 3*(3*A^4 - 14*A^2*B^2 + 3*B^4)*a ^4*b^2 + 40*(A^3*B - A*B^3)*a^3*b^3 - 6*(A^4 - 8*A^2*B^2 + B^4)*a^2*b^4 - 12*(A^3*B - A*B^3)*a*b^5 + (A^4 - 2*A^2*B^2 + B^4)*b^6)/((a^12 + 6*a^10*b^ 2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + (2*A *B^2*a^5 - (7*A^2*B - 3*B^3)*a^4*b + 2*(3*A^3 - 7*A*B^2)*a^3*b^2 + 4*(4*A^ 2*B - B^3)*a^2*b^3 - 2*(A^3 - 4*A*B^2)*a*b^4 - (A^2*B - B^3)*b^5)*d)*sqrt( -(6*A*B*a^2*b - 2*A*B*b^3 + (A^2 - B^2)*a^3 - 3*(A^2 - B^2)*a*b^2 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(4*A^2*B^2*a^6 - 12*(A^3*B - A*B^3 )*a^5*b + 3*(3*A^4 - 14*A^2*B^2 + 3*B^4)*a^4*b^2 + 40*(A^3*B - A*B^3)*a^3* b^3 - 6*(A^4 - 8*A^2*B^2 + B^4)*a^2*b^4 - 12*(A^3*B - A*B^3)*a*b^5 + (A...
\[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Timed out. \[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Time = 14.92 (sec) , antiderivative size = 5737, normalized size of antiderivative = 41.57 \[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
(log((((a + b*tan(c + d*x))^(1/2)*(16*A^2*b^10*d^3 + 32*A^2*a^2*b^8*d^3 - 32*A^2*a^6*b^4*d^3 - 16*A^2*a^8*b^2*d^3) - ((((96*A^4*a^2*b^4*d^4 - 16*A^4 *b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 12*A^2*a*b^2*d^2)/ (a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2)*(((((96*A^4*a^2 *b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 1 2*A^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2 )*(a + b*tan(c + d*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10*d^5 + 640*a^5*b ^8*d^5 + 640*a^7*b^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2*d^5))/4 + 64*A*a* b^11*d^4 + 256*A*a^3*b^9*d^4 + 384*A*a^5*b^7*d^4 + 256*A*a^7*b^5*d^4 + 64* A*a^9*b^3*d^4))/4)*(((96*A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^ 2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 12*A^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^ 2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2))/4 + 8*A^3*b^9*d^2 + 24*A^3*a^2*b^7*d^2 + 24*A^3*a^4*b^5*d^2 + 8*A^3*a^6*b^3*d^2)*(((96*A^4*a^2*b^4*d^4 - 16*A^4*b ^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 12*A^2*a*b^2*d^2)/(a ^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2))/4 + (log((((a + b*tan(c + d*x))^(1/2)*(16*A^2*b^10*d^3 + 32*A^2*a^2*b^8*d^3 - 32*A^2*a^6*b ^4*d^3 - 16*A^2*a^8*b^2*d^3) - ((-((96*A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) + 4*A^2*a^3*d^2 - 12*A^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2)*(((-((96*A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) + 4*A^2*a^3*d^2 - 12*A^2*a...