Integrand size = 33, antiderivative size = 193 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {(A+i B) \operatorname {AppellF1}\left (1+m,\frac {5}{2},1,2+m,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 a^2 d (1+m) \sqrt {a+b \tan (c+d x)}}+\frac {(A-i B) \operatorname {AppellF1}\left (1+m,\frac {5}{2},1,2+m,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 a^2 d (1+m) \sqrt {a+b \tan (c+d x)}} \]
1/2*(A+I*B)*AppellF1(1+m,1,5/2,2+m,-I*tan(d*x+c),-b*tan(d*x+c)/a)*(1+b*tan (d*x+c)/a)^(1/2)*tan(d*x+c)^(1+m)/a^2/d/(1+m)/(a+b*tan(d*x+c))^(1/2)+1/2*( A-I*B)*AppellF1(1+m,1,5/2,2+m,I*tan(d*x+c),-b*tan(d*x+c)/a)*(1+b*tan(d*x+c )/a)^(1/2)*tan(d*x+c)^(1+m)/a^2/d/(1+m)/(a+b*tan(d*x+c))^(1/2)
\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx \]
Time = 0.64 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 4086, 3042, 4085, 152, 150}
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.
\(\displaystyle \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^m (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 4086 |
\(\displaystyle \frac {1}{2} (A+i B) \int \frac {(1-i \tan (c+d x)) \tan ^m(c+d x)}{(a+b \tan (c+d x))^{5/2}}dx+\frac {1}{2} (A-i B) \int \frac {(i \tan (c+d x)+1) \tan ^m(c+d x)}{(a+b \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} (A+i B) \int \frac {(1-i \tan (c+d x)) \tan (c+d x)^m}{(a+b \tan (c+d x))^{5/2}}dx+\frac {1}{2} (A-i B) \int \frac {(i \tan (c+d x)+1) \tan (c+d x)^m}{(a+b \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 4085 |
\(\displaystyle \frac {(A-i B) \int \frac {\tan ^m(c+d x)}{(1-i \tan (c+d x)) (a+b \tan (c+d x))^{5/2}}d\tan (c+d x)}{2 d}+\frac {(A+i B) \int \frac {\tan ^m(c+d x)}{(i \tan (c+d x)+1) (a+b \tan (c+d x))^{5/2}}d\tan (c+d x)}{2 d}\) |
\(\Big \downarrow \) 152 |
\(\displaystyle \frac {(A-i B) \sqrt {\frac {b \tan (c+d x)}{a}+1} \int \frac {\tan ^m(c+d x)}{(1-i \tan (c+d x)) \left (\frac {b \tan (c+d x)}{a}+1\right )^{5/2}}d\tan (c+d x)}{2 a^2 d \sqrt {a+b \tan (c+d x)}}+\frac {(A+i B) \sqrt {\frac {b \tan (c+d x)}{a}+1} \int \frac {\tan ^m(c+d x)}{(i \tan (c+d x)+1) \left (\frac {b \tan (c+d x)}{a}+1\right )^{5/2}}d\tan (c+d x)}{2 a^2 d \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {(A+i B) \tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} \operatorname {AppellF1}\left (m+1,\frac {5}{2},1,m+2,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 a^2 d (m+1) \sqrt {a+b \tan (c+d x)}}+\frac {(A-i B) \tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} \operatorname {AppellF1}\left (m+1,\frac {5}{2},1,m+2,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 a^2 d (m+1) \sqrt {a+b \tan (c+d x)}}\) |
((A + I*B)*AppellF1[1 + m, 5/2, 1, 2 + m, -((b*Tan[c + d*x])/a), (-I)*Tan[ c + d*x]]*Tan[c + d*x]^(1 + m)*Sqrt[1 + (b*Tan[c + d*x])/a])/(2*a^2*d*(1 + m)*Sqrt[a + b*Tan[c + d*x]]) + ((A - I*B)*AppellF1[1 + m, 5/2, 1, 2 + m, -((b*Tan[c + d*x])/a), I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*Sqrt[1 + (b*Ta n[c + d*x])/a])/(2*a^2*d*(1 + m)*Sqrt[a + b*Tan[c + d*x]])
3.5.92.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[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] && !IntegerQ[m] && !IntegerQ[n] && !IntegersQ[2*m, 2*n ] && 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] && !IntegerQ[m] & & !IntegerQ[n] && !IntegersQ[2*m, 2*n] && NeQ[A^2 + B^2, 0]
\[\int \frac {\tan \left (d x +c \right )^{m} \left (A +B \tan \left (d x +c \right )\right )}{\left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}d x\]
\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral((B*tan(d*x + c) + A)*sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^m/(b^3 *tan(d*x + c)^3 + 3*a*b^2*tan(d*x + c)^2 + 3*a^2*b*tan(d*x + c) + a^3), x)
\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]