Integrand size = 29, antiderivative size = 339 \[ \int \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx=-\frac {i \sqrt {a-i b} (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {i \sqrt {a+i b} (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {\sqrt {d} \left (10 a b c d-a^2 d^2+b^2 \left (15 c^2-8 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 b^{3/2} f}+\frac {d (9 b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 b f}+\frac {d^2 (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}{2 b f} \]
-I*(c-I*d)^(5/2)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2 )/(c+d*tan(f*x+e))^(1/2))*(a-I*b)^(1/2)/f+I*(c+I*d)^(5/2)*arctanh((c+I*d)^ (1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))*(a+I*b) ^(1/2)/f+1/4*(10*a*b*c*d-a^2*d^2+b^2*(15*c^2-8*d^2))*arctanh(d^(1/2)*(a+b* tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan(f*x+e))^(1/2))*d^(1/2)/b^(3/2)/f+1/4*d* (-a*d+9*b*c)*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/b/f+1/2*d^2*(c+ d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^(3/2)/b/f
Time = 6.92 (sec) , antiderivative size = 550, normalized size of antiderivative = 1.62 \[ \int \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx=\frac {\frac {4 \left (a \sqrt {-b^2} c \left (c^2-3 d^2\right )+b \left (-a+\sqrt {-b^2}\right ) d \left (-3 c^2+d^2\right )+b^2 \left (c^3-3 c d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {-c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+\sqrt {-b^2}} \sqrt {-c+\frac {\sqrt {-b^2} d}{b}}}-\frac {4 \left (-a \sqrt {-b^2} c \left (c^2-3 d^2\right )-b \left (a+\sqrt {-b^2}\right ) d \left (-3 c^2+d^2\right )+b^2 \left (c^3-3 c d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+\frac {\sqrt {-b^2} d}{b}}}+d (9 b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}+2 d^2 (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}+\frac {\sqrt {d} \sqrt {c-\frac {a d}{b}} \left (10 a b c d-a^2 d^2+b^2 \left (15 c^2-8 d^2\right )\right ) \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}}{4 b f} \]
((4*(a*Sqrt[-b^2]*c*(c^2 - 3*d^2) + b*(-a + Sqrt[-b^2])*d*(-3*c^2 + d^2) + b^2*(c^3 - 3*c*d^2))*ArcTanh[(Sqrt[-c + (Sqrt[-b^2]*d)/b]*Sqrt[a + b*Tan[ e + f*x]])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[-a + S qrt[-b^2]]*Sqrt[-c + (Sqrt[-b^2]*d)/b]) - (4*(-(a*Sqrt[-b^2]*c*(c^2 - 3*d^ 2)) - b*(a + Sqrt[-b^2])*d*(-3*c^2 + d^2) + b^2*(c^3 - 3*c*d^2))*ArcTanh[( Sqrt[c + (Sqrt[-b^2]*d)/b]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + Sqrt[-b^2]] *Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[c + (Sqrt[-b^2]*d) /b]) + d*(9*b*c - a*d)*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]] + 2*d^2*(a + b*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]] + (Sqrt[d]*Sqrt [c - (a*d)/b]*(10*a*b*c*d - a^2*d^2 + b^2*(15*c^2 - 8*d^2))*ArcSinh[(Sqrt[ d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c - (a*d)/b])]*Sqrt[(b*(c + d*T an[e + f*x]))/(b*c - a*d)])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]]))/(4*b*f)
Time = 1.84 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4138, 2348, 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.
| \(\displaystyle \int \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 4049 |
| \(\displaystyle \frac {\int \frac {\sqrt {a+b \tan (e+f x)} \left (4 b c^3-3 b d^2 c-a d^3+d^2 (9 b c-a d) \tan ^2(e+f x)+4 b d \left (3 c^2-d^2\right ) \tan (e+f x)\right )}{2 \sqrt {c+d \tan (e+f x)}}dx}{2 b}+\frac {d^2 (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}{2 b f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {a+b \tan (e+f x)} \left (4 b c^3-3 b d^2 c-a d^3+d^2 (9 b c-a d) \tan ^2(e+f x)+4 b d \left (3 c^2-d^2\right ) \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{4 b}+\frac {d^2 (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}{2 b f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {a+b \tan (e+f x)} \left (4 b c^3-3 b d^2 c-a d^3+d^2 (9 b c-a d) \tan (e+f x)^2+4 b d \left (3 c^2-d^2\right ) \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{4 b}+\frac {d^2 (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}{2 b f}\) |
| \(\Big \downarrow \) 4130 |
| \(\displaystyle \frac {\frac {\int -\frac {-d^2 \left (\left (15 c^2-8 d^2\right ) b^2+10 a c d b-a^2 d^2\right ) \tan ^2(e+f x)-8 b d \left (b c^3+3 a d c^2-3 b d^2 c-a d^3\right ) \tan (e+f x)+d \left (a^2 d^3+9 b^2 c^2 d-a b \left (8 c^3-14 c d^2\right )\right )}{2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{d}+\frac {d (9 b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{4 b}+\frac {d^2 (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}{2 b f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {d (9 b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}-\frac {\int \frac {-d^2 \left (\left (15 c^2-8 d^2\right ) b^2+10 a c d b-a^2 d^2\right ) \tan ^2(e+f x)-8 b d \left (b c^3+3 a d c^2-3 b d^2 c-a d^3\right ) \tan (e+f x)+d \left (a^2 d^3+9 b^2 c^2 d-a b \left (8 c^3-14 c d^2\right )\right )}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{2 d}}{4 b}+\frac {d^2 (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}{2 b f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {d (9 b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}-\frac {\int \frac {-d^2 \left (\left (15 c^2-8 d^2\right ) b^2+10 a c d b-a^2 d^2\right ) \tan (e+f x)^2-8 b d \left (b c^3+3 a d c^2-3 b d^2 c-a d^3\right ) \tan (e+f x)+d \left (a^2 d^3+9 b^2 c^2 d-a b \left (8 c^3-14 c d^2\right )\right )}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{2 d}}{4 b}+\frac {d^2 (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}{2 b f}\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \frac {\frac {d (9 b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}-\frac {\int \frac {-d^2 \left (\left (15 c^2-8 d^2\right ) b^2+10 a c d b-a^2 d^2\right ) \tan ^2(e+f x)-8 b d \left (b c^3+3 a d c^2-3 b d^2 c-a d^3\right ) \tan (e+f x)+d \left (a^2 d^3+9 b^2 c^2 d-a b \left (8 c^3-14 c d^2\right )\right )}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (\tan ^2(e+f x)+1\right )}d\tan (e+f x)}{2 d f}}{4 b}+\frac {d^2 (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}{2 b f}\) |
| \(\Big \downarrow \) 2348 |
| \(\displaystyle \frac {d^2 (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}{2 b f}+\frac {\frac {d (9 b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}-\frac {\int \left (\frac {\left (-\left (\left (15 c^2-8 d^2\right ) b^2\right )-10 a c d b+a^2 d^2\right ) d^2}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-8 a b d^4-24 b^2 c d^3+24 a b c^2 d^2+8 b^2 c^3 d+i \left (-8 b^2 d^4+24 a b c d^3+24 b^2 c^2 d^2-8 a b c^3 d\right )}{2 (i-\tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {8 a b d^4+24 b^2 c d^3-24 a b c^2 d^2-8 b^2 c^3 d+i \left (-8 b^2 d^4+24 a b c d^3+24 b^2 c^2 d^2-8 a b c^3 d\right )}{2 (\tan (e+f x)+i) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\right )d\tan (e+f x)}{2 d f}}{4 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}{2 b f}+\frac {\frac {d (9 b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}-\frac {-\frac {2 d^{3/2} \left (-a^2 d^2+10 a b c d+b^2 \left (15 c^2-8 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {b}}+8 i b d \sqrt {a-i b} (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )-8 i b d \sqrt {a+i b} (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{2 d f}}{4 b}\) |
(d^2*(a + b*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]])/(2*b*f) + (-1/2* ((8*I)*Sqrt[a - I*b]*b*(c - I*d)^(5/2)*d*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b *Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])] - (8*I)*Sqrt[a + I*b]*b*(c + I*d)^(5/2)*d*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]]) /(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])] - (2*d^(3/2)*(10*a*b*c*d - a^2* d^2 + b^2*(15*c^2 - 8*d^2))*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sq rt[b]*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[b])/(d*f) + (d*(9*b*c - a*d)*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/f)/(4*b)
3.13.78.3.1 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[(Px_)*((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_. )*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(c + d*x)^m*(e + f*x)^ n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[P x, x] && (IntegerQ[p] || (IntegerQ[2*p] && IntegerQ[m] && ILtQ[n, 0])) && !(IGtQ[m, 0] && IGtQ[n, 0])
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 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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] && GtQ[m, 2] && (GeQ[n, -1] || I ntegerQ[m]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) )
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] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C *(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ c, 0] && NeQ[a, 0])))
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]
Timed out.
\[\int \sqrt {a +b \tan \left (f x +e \right )}\, \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 10572 vs. \(2 (269) = 538\).
Time = 11.11 (sec) , antiderivative size = 21171, normalized size of antiderivative = 62.45 \[ \int \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \]
\[ \int \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx=\int \sqrt {a + b \tan {\left (e + f x \right )}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \]
\[ \int \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx=\int { \sqrt {b \tan \left (f x + e\right ) + a} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]
\[ \int \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx=\int { \sqrt {b \tan \left (f x + e\right ) + a} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx=\int \sqrt {a+b\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2} \,d x \]