Integrand size = 38, antiderivative size = 150 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=-\frac {i B \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {i B \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {2 B \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}} \]
-I*B*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(I*a- b)^(1/2)+I*B*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2) )/d/(I*a+b)^(1/2)-2*B*(a+b*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(1/2)
Time = 0.42 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.05 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\frac {B \left (\frac {\sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}-\frac {\sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}-\frac {2 \sqrt {a+b \tan (c+d x)}}{a \sqrt {\tan (c+d x)}}\right )}{d} \]
(B*(((-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt [a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] - ((-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqr t[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b] - (2*Sqrt[a + b*Tan[c + d*x]])/(a*Sqrt[Tan[c + d*x]])))/d
Time = 0.46 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2011, 3042, 4052, 27, 3042, 4058, 613, 104, 218, 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.
\(\displaystyle \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 2011 |
\(\displaystyle B \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \int \frac {1}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4052 |
\(\displaystyle B \left (-\frac {2 \int \frac {a \sqrt {\tan (c+d x)}}{2 \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle B \left (-\int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \left (-\int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 4058 |
\(\displaystyle B \left (-\frac {\int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 613 |
\(\displaystyle B \left (-\frac {2 \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {\frac {1}{2} \int \frac {1}{\sqrt {\tan (c+d x)} (\tan (c+d x)+i) \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)-\frac {1}{2} \int \frac {1}{(i-\tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{d}\right )\) |
\(\Big \downarrow \) 104 |
\(\displaystyle B \left (-\frac {2 \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {\int \frac {1}{\frac {(a-i b) \tan (c+d x)}{a+b \tan (c+d x)}+i}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}-\int \frac {1}{i-\frac {(a+i b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle B \left (-\frac {2 \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {\int \frac {1}{\frac {(a-i b) \tan (c+d x)}{a+b \tan (c+d x)}+i}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}+\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}}{d}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle B \left (-\frac {2 \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}-\frac {i \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}}{d}\right )\) |
B*(-(((I*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x] ]])/Sqrt[I*a - b] - (I*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[I*a + b])/d) - (2*Sqrt[a + b*Tan[c + d*x]])/(a*d*S qrt[Tan[c + d*x]]))
3.5.72.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[Sqrt[(e_.)*(x_)]/(Sqrt[(c_) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^2)), x_Sym bol] :> Simp[e/(2*b) Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] + x)), x ], x] - Simp[e/(2*b) Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] - x)), x ], x] /; FreeQ[{a, b, c, d, e}, x]
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[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 /((m + 1)*(a^2 + b^2)*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x], x], x] / ; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || Integ erQ[m]) && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[ff/f Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a *d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.58 (sec) , antiderivative size = 943929, normalized size of antiderivative = 6292.86
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 4608 vs. \(2 (118) = 236\).
Time = 0.76 (sec) , antiderivative size = 4608, normalized size of antiderivative = 30.72 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
1/8*(a*d*sqrt(((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^ 2 - B^2*b)/((a^2 + b^2)*d^2))*log((2*(2*B^3*a^4*b + 4*B^3*a^2*b^3 + (B^3*a ^5 + 3*B^3*a^3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) + (2*(B*a^5*b + 3*B*a^3*b^3 + 2*B*a*b^5)*d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B* b^6)*d^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(b*tan(d*x + c ) + a)*sqrt(tan(d*x + c)) + ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4)*d* tan(d*x + c)^2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a^2*b^4)*d - 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*t an(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-B^4*a^2/((a^4 + 2 *a^2*b^2 + b^4)*d^4)))*sqrt(((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 - B^2*b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1))*tan(d*x + c) + a*d*sqrt(((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) *d^2 - B^2*b)/((a^2 + b^2)*d^2))*log(-(2*(2*B^3*a^4*b + 4*B^3*a^2*b^3 + (B ^3*a^5 + 3*B^3*a^3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) + (2*(B*a^5*b + 3*B*a^3 *b^3 + 2*B*a*b^5)*d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B*b^6)*d^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4 )*d*tan(d*x + c)^2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a^2*b^4)*d - 2*((a^4*b^3 + 5*a^2...
\[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=B \int \frac {1}{\sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Timed out. \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {B\,a+B\,b\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]