Integrand size = 25, antiderivative size = 183 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {2 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \]
(I*a-b)^(5/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2) )/d+2*b^(5/2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d-( I*a+b)^(5/2)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2) )/d-2*a^2*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)
Time = 2.31 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.33 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\frac {2 \sqrt {a} b^{5/2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}-\frac {\sqrt [4]{-1} (-a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}+\sqrt [4]{-1} (a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}+2 a^2 \sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}}}{d} \]
((2*Sqrt[a]*b^(5/2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/Sqrt[a + b*Tan[c + d*x]] - ((-1)^(1/4)*(-a + I*b)^(5 /2)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Tan[c + d*x]] + (-1)^(1/4)*(a + I*b)^(5/2)*ArcTan[((-1)^(1/ 4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Tan[c + d*x]] + 2*a^2*Sqrt[a + b*Tan[c + d*x]])/Sqrt[Tan[c + d*x]])/d
Time = 0.93 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4048, 27, 3042, 4138, 2035, 2257, 2009}
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.
\(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan (c+d x)^{3/2}}dx\) |
\(\Big \downarrow \) 4048 |
\(\displaystyle 2 \int \frac {\tan ^2(c+d x) b^3+3 a^2 b-a \left (a^2-3 b^2\right ) \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\tan ^2(c+d x) b^3+3 a^2 b-a \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^2 b^3+3 a^2 b-a \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 4138 |
\(\displaystyle \frac {\int \frac {\tan ^2(c+d x) b^3+3 a^2 b-a \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {2 \int \frac {\tan ^2(c+d x) b^3+3 a^2 b-a \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 2257 |
\(\displaystyle \frac {2 \int \left (\frac {b^3}{\sqrt {a+b \tan (c+d x)}}+\frac {-b^3+3 a^2 b-a \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {2 \left (\frac {1}{2} (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {1}{2} (b+i a)^{5/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )}{d}\) |
(2*(((I*a - b)^(5/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b* Tan[c + d*x]]])/2 + b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - ((I*a + b)^(5/2)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d* x]])/Sqrt[a + b*Tan[c + d*x]]])/2))/d - (2*a^2*Sqrt[a + b*Tan[c + d*x]])/( d*Sqrt[Tan[c + d*x]])
3.7.27.3.1 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[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol ] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a , c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 /(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c *(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) *Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( 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] && GtQ[m, 2] && LtQ [n, -1] && IntegerQ[2*m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), 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*((A + B*ff*x + C*ff^2*x^ 2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f , A, B, C, 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 = 0.65 (sec) , antiderivative size = 1344862, normalized size of antiderivative = 7348.97
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 4698 vs. \(2 (147) = 294\).
Time = 1.36 (sec) , antiderivative size = 9395, normalized size of antiderivative = 51.34 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \]