Integrand size = 28, antiderivative size = 123 \[ \int \frac {a B+b B \tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {i B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}+\frac {i B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}-\frac {2 b B}{\left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]
-I*B*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(3/2)/d+I*B*arc tanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(3/2)/d-2*b*B/(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 = 0.14 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.86 \[ \int \frac {a B+b B \tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {B \left (i (a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )+(-i a-b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )\right )}{\left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]
(B*(I*(a + I*b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I*b)] + ((-I)*a - b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/ (a + I*b)]))/((a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]])
Time = 0.59 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {2011, 3042, 3964, 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.
\(\displaystyle \int \frac {a B+b B \tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2011 |
\(\displaystyle B \int \frac {1}{(a+b \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \int \frac {1}{(a+b \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3964 |
\(\displaystyle B \left (\frac {\int \frac {a-b \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \left (\frac {\int \frac {a-b \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle B \left (-\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (a-i b) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \left (-\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (a-i b) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle B \left (-\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i (a+i b) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a-i b) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle B \left (-\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i (a-i b) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i (a+i b) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle B \left (-\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(a-i b) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle B \left (-\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(a+i b) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {(a-i b) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{a^2+b^2}\right )\) |
B*((((a + I*b)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + ((a - I*b)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d))/(a^2 + b^2) - (2*b)/((a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]))
3.4.68.3.1 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[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -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. \(1954\) vs. \(2(103)=206\).
Time = 0.09 (sec) , antiderivative size = 1955, normalized size of antiderivative = 15.89
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1955\) |
default | \(\text {Expression too large to display}\) | \(1955\) |
parts | \(\text {Expression too large to display}\) | \(4475\) |
1/4/d*B/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*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/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*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)+4/d*B*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^3+3/d*B*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-1/ d*B/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*b/(a^2+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*ta n(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*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/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...
Leaf count of result is larger than twice the leaf count of optimal. 2174 vs. \(2 (97) = 194\).
Time = 0.31 (sec) , antiderivative size = 2174, normalized size of antiderivative = 17.67 \[ \int \frac {a B+b B \tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
-1/2*(4*sqrt(b*tan(d*x + c) + a)*B*b + ((a^2*b + b^3)*d*tan(d*x + c) + (a^ 3 + a*b^2)*d)*sqrt(-(B^2*a^3 - 3*B^2*a*b^2 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*B^4*a^4*b^2 - 6*B^4*a^2*b^4 + 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(-(3*B^3*a^2*b - B^3*b^3)*sqrt (b*tan(d*x + c) + a) + ((a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d^3*sqrt(-(9*B ^4*a^4*b^2 - 6*B^4*a^2*b^4 + B^4*b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 2 0*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + 2*(3*B^2*a^3*b^2 - B^2 *a*b^4)*d)*sqrt(-(B^2*a^3 - 3*B^2*a*b^2 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b ^6)*d^2*sqrt(-(9*B^4*a^4*b^2 - 6*B^4*a^2*b^4 + 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))) - ((a^2*b + b^3)*d*tan(d*x + c) + ( a^3 + a*b^2)*d)*sqrt(-(B^2*a^3 - 3*B^2*a*b^2 + (a^6 + 3*a^4*b^2 + 3*a^2*b^ 4 + b^6)*d^2*sqrt(-(9*B^4*a^4*b^2 - 6*B^4*a^2*b^4 + 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(-(3*B^3*a^2*b - B^3*b^3)*sq rt(b*tan(d*x + c) + a) - ((a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d^3*sqrt(-(9 *B^4*a^4*b^2 - 6*B^4*a^2*b^4 + 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*(3*B^2*a^3*b^2 - B ^2*a*b^4)*d)*sqrt(-(B^2*a^3 - 3*B^2*a*b^2 + (a^6 + 3*a^4*b^2 + 3*a^2*b^...
\[ \int \frac {a B+b B \tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=B \int \frac {1}{a \sqrt {a + b \tan {\left (c + d x \right )}} + b \sqrt {a + b \tan {\left (c + d x \right )}} \tan {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {a B+b B \tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int \frac {a B+b B \tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Time = 21.21 (sec) , antiderivative size = 9618, normalized size of antiderivative = 78.20 \[ \int \frac {a B+b B \tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
(log(16*B^3*a^4*b^15*d^2 - ((((320*B^4*a^6*b^8*d^4 - 16*B^4*a^4*b^10*d^4 - 1760*B^4*a^8*b^6*d^4 + 1600*B^4*a^10*b^4*d^4 - 400*B^4*a^12*b^2*d^4)^(1/2 ) - 4*B^2*a^7*d^2 - 20*B^2*a^3*b^4*d^2 + 40*B^2*a^5*b^2*d^2)/(a^10*d^4 + b ^10*d^4 + 5*a^2*b^8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4) )^(1/2)*(((((320*B^4*a^6*b^8*d^4 - 16*B^4*a^4*b^10*d^4 - 1760*B^4*a^8*b^6* d^4 + 1600*B^4*a^10*b^4*d^4 - 400*B^4*a^12*b^2*d^4)^(1/2) - 4*B^2*a^7*d^2 - 20*B^2*a^3*b^4*d^2 + 40*B^2*a^5*b^2*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b^ 8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2)*(((((320*B ^4*a^6*b^8*d^4 - 16*B^4*a^4*b^10*d^4 - 1760*B^4*a^8*b^6*d^4 + 1600*B^4*a^1 0*b^4*d^4 - 400*B^4*a^12*b^2*d^4)^(1/2) - 4*B^2*a^7*d^2 - 20*B^2*a^3*b^4*d ^2 + 40*B^2*a^5*b^2*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 10*a^4*b^6 *d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2)*(a + b*tan(c + d*x))^(1/2)*( 64*a*b^22*d^5 + 640*a^3*b^20*d^5 + 2880*a^5*b^18*d^5 + 7680*a^7*b^16*d^5 + 13440*a^9*b^14*d^5 + 16128*a^11*b^12*d^5 + 13440*a^13*b^10*d^5 + 7680*a^1 5*b^8*d^5 + 2880*a^17*b^6*d^5 + 640*a^19*b^4*d^5 + 64*a^21*b^2*d^5))/4 - 3 2*B*a*b^21*d^4 - 160*B*a^3*b^19*d^4 - 128*B*a^5*b^17*d^4 + 896*B*a^7*b^15* d^4 + 3136*B*a^9*b^13*d^4 + 4928*B*a^11*b^11*d^4 + 4480*B*a^13*b^9*d^4 + 2 432*B*a^15*b^7*d^4 + 736*B*a^17*b^5*d^4 + 96*B*a^19*b^3*d^4))/4 - (a + b*t an(c + d*x))^(1/2)*(320*B^2*a^6*b^14*d^3 - 16*B^2*a^2*b^18*d^3 + 1024*B^2* a^8*b^12*d^3 + 1440*B^2*a^10*b^10*d^3 + 1024*B^2*a^12*b^8*d^3 + 320*B^2...