Integrand size = 25, antiderivative size = 232 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {\left (3 a^2-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{5/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d} \]
1/4*(3*a^2-8*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2)) /b^(5/2)/d+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)+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)-3/4*a*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/b^2/d+1/ 2*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^(3/2)/b/d
Time = 4.25 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.16 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {-\frac {4 (-1)^{3/4} b^2 \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}-\frac {4 (-1)^{3/4} b^2 \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}-3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+2 b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}+\frac {\sqrt {a} \left (3 a^2-8 b^2\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{4 b^2 d} \]
((-4*(-1)^(3/4)*b^2*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/ Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] - (4*(-1)^(3/4)*b^2*ArcTan[((-1) ^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b] - 3*a*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + 2*b*Tan[c + d* x]^(3/2)*Sqrt[a + b*Tan[c + d*x]] + (Sqrt[a]*(3*a^2 - 8*b^2)*ArcSinh[(Sqrt [b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/(Sqrt[b]*Sq rt[a + b*Tan[c + d*x]]))/(4*b^2*d)
Time = 1.06 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4139, 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.
\(\displaystyle \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^{7/2}}{\sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4049 |
\(\displaystyle \frac {\int -\frac {\sqrt {\tan (c+d x)} \left (3 a \tan ^2(c+d x)+4 b \tan (c+d x)+3 a\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{2 b}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}-\frac {\int \frac {\sqrt {\tan (c+d x)} \left (3 a \tan ^2(c+d x)+4 b \tan (c+d x)+3 a\right )}{\sqrt {a+b \tan (c+d x)}}dx}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}-\frac {\int \frac {\sqrt {\tan (c+d x)} \left (3 a \tan (c+d x)^2+4 b \tan (c+d x)+3 a\right )}{\sqrt {a+b \tan (c+d x)}}dx}{4 b}\) |
\(\Big \downarrow \) 4130 |
\(\displaystyle \frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}-\frac {\frac {\int -\frac {3 a^2+\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b}+\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}}{4 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}-\frac {\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {3 a^2+\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}-\frac {\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {3 a^2+\left (3 a^2-8 b^2\right ) \tan (c+d x)^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}}{4 b}\) |
\(\Big \downarrow \) 4139 |
\(\displaystyle \frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}-\frac {\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {3 a^2+\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 b d}}{4 b}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}-\frac {\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {3 a^2+\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{b d}}{4 b}\) |
\(\Big \downarrow \) 2257 |
\(\displaystyle \frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}-\frac {\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \left (\frac {8 b^2}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}+\frac {3 a^2-8 b^2}{\sqrt {a+b \tan (c+d x)}}\right )d\sqrt {\tan (c+d x)}}{b d}}{4 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}-\frac {\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\frac {\left (3 a^2-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}+\frac {4 b^2 \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}+\frac {4 b^2 \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}}{b d}}{4 b}\) |
(Tan[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]])/(2*b*d) - (-(((4*b^2*ArcTan[ (Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[I*a - b ] + ((3*a^2 - 8*b^2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] + (4*b^2*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqr t[a + b*Tan[c + d*x]]])/Sqrt[I*a + b])/(b*d)) + (3*a*Sqrt[Tan[c + d*x]]*Sq rt[a + b*Tan[c + d*x]])/(b*d))/(4*b)
3.7.32.3.1 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^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_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + b*ff*x )^m*(c + d*ff*x)^n*((A + C*ff^2*x^2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/f f], x]] /; FreeQ[{a, b, c, d, e, f, A, 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 = 0.58 (sec) , antiderivative size = 946881, normalized size of antiderivative = 4081.38
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 4119 vs. \(2 (186) = 372\).
Time = 1.21 (sec) , antiderivative size = 8244, normalized size of antiderivative = 35.53 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{\frac {7}{2}}}{\sqrt {b \tan \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]