Integrand size = 25, antiderivative size = 139 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \]
-arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a-b)^(1/ 2)/d+arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a+b )^(1/2)/d-2*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)
Time = 0.40 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\sqrt [4]{-1} \sqrt {-a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt [4]{-1} \sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\frac {2 \sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}}}{d} \]
-(((-1)^(1/4)*Sqrt[-a + I*b]*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + (-1)^(1/4)*Sqrt[a + I*b]*ArcTan[((-1) ^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + (2*Sq rt[a + b*Tan[c + d*x]])/Sqrt[Tan[c + d*x]])/d)
Time = 0.77 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4051, 27, 3042, 4099, 3042, 4098, 104, 216, 219}
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.
\(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan (c+d x)^{3/2}}dx\) |
\(\Big \downarrow \) 4051 |
\(\displaystyle -2 \int -\frac {b-a \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {b-a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {b-a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 4099 |
\(\displaystyle -\frac {1}{2} (-b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{2} (-b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 4098 |
\(\displaystyle \frac {(b+i a) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {(-b+i a) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle -\frac {(-b+i a) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {(b+i a) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {(b+i a) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {\sqrt {-b+i a} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\sqrt {-b+i a} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b+i a} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
-((Sqrt[I*a - b]*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[ c + d*x]]])/d) + (Sqrt[I*a + b]*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]]) /Sqrt[a + b*Tan[c + d*x]]])/d - (2*Sqrt[a + b*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]])
3.7.12.3.1 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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^n/(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(a^2 + b^2 )) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c *(m + 1) - b*d*n - (b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(m + n + 1)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && GtQ[n, 0] && Int egerQ[2*m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[A^2/f Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(A + I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T an[e + f*x]), x], x] + Simp[(A - I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d *Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A , B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 0]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.51 (sec) , antiderivative size = 1089773, normalized size of antiderivative = 7840.09
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 2696 vs. \(2 (111) = 222\).
Time = 0.50 (sec) , antiderivative size = 2696, normalized size of antiderivative = 19.40 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
-1/8*(d*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2)*log((2*(2*a^4*b + 4*a^2*b^3 + ( a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + (a^5*d - (a^5 + 6*a^3*b^2 + 8*a*b^4)*d*tan(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3)*d*tan(d*x + c) - ((a^3*b + 4*a*b^3)*d^3*tan (d*x + c)^2 - 2*(a^4 + 3*a^2*b^2 + 4*b^4)*d^3*tan(d*x + c) - (3*a^3*b + 4* a*b^3)*d^3)*sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c) + d*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2)*log(-(2*(2 *a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4 ))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + (a^5*d - (a^5 + 6*a^3*b^2 + 8*a*b^4)*d*tan(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3)*d*tan(d*x + c) - ((a^ 3*b + 4*a*b^3)*d^3*tan(d*x + c)^2 - 2*(a^4 + 3*a^2*b^2 + 4*b^4)*d^3*tan(d* x + c) - (3*a^3*b + 4*a*b^3)*d^3)*sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c) - d*sqrt((d^2*sqrt(-a^2/d^4 ) + b)/d^2)*log((2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan( d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^ 4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) - (a^5 *d - (a^5 + 6*a^3*b^2 + 8*a*b^4)*d*tan(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3)* d*tan(d*x + c) - ((a^3*b + 4*a*b^3)*d^3*tan(d*x + c)^2 - 2*(a^4 + 3*a^2...
\[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {\sqrt {b \tan \left (d x + c\right ) + a}}{\tan \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \]