3.5.87 \(\int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [487]

3.5.87.1 Optimal result

Integrand size = 33, antiderivative size = 193 \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {a^2 (A+i B) \operatorname {AppellF1}\left (1+m,-\frac {5}{2},1,2+m,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {a^2 (A-i B) \operatorname {AppellF1}\left (1+m,-\frac {5}{2},1,2+m,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]

output

1/2*a^2*(A+I*B)*AppellF1(1+m,1,-5/2,2+m,-I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+ 
b*tan(d*x+c))^(1/2)*tan(d*x+c)^(1+m)/d/(1+m)/(1+b*tan(d*x+c)/a)^(1/2)+1/2* 
a^2*(A-I*B)*AppellF1(1+m,1,-5/2,2+m,I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan 
(d*x+c))^(1/2)*tan(d*x+c)^(1+m)/d/(1+m)/(1+b*tan(d*x+c)/a)^(1/2)
 

3.5.87.2 Mathematica [F]

\[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx \]

input

Integrate[Tan[c + d*x]^m*(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x 
]
 

output

Integrate[Tan[c + d*x]^m*(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]), 
x]
 

3.5.87.3 Rubi [A] (verified)

Time = 0.67 (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.

\(\displaystyle \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^m (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4086

\(\displaystyle \frac {1}{2} (A+i B) \int (1-i \tan (c+d x)) \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2}dx+\frac {1}{2} (A-i B) \int (i \tan (c+d x)+1) \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} (A+i B) \int (1-i \tan (c+d x)) \tan (c+d x)^m (a+b \tan (c+d x))^{5/2}dx+\frac {1}{2} (A-i B) \int (i \tan (c+d x)+1) \tan (c+d x)^m (a+b \tan (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 4085

\(\displaystyle \frac {(A-i B) \int \frac {\tan ^m(c+d x) (a+b \tan (c+d x))^{5/2}}{1-i \tan (c+d x)}d\tan (c+d x)}{2 d}+\frac {(A+i B) \int \frac {\tan ^m(c+d x) (a+b \tan (c+d x))^{5/2}}{i \tan (c+d x)+1}d\tan (c+d x)}{2 d}\)

\(\Big \downarrow \) 152

\(\displaystyle \frac {a^2 (A-i B) \sqrt {a+b \tan (c+d x)} \int \frac {\tan ^m(c+d x) \left (\frac {b \tan (c+d x)}{a}+1\right )^{5/2}}{1-i \tan (c+d x)}d\tan (c+d x)}{2 d \sqrt {\frac {b \tan (c+d x)}{a}+1}}+\frac {a^2 (A+i B) \sqrt {a+b \tan (c+d x)} \int \frac {\tan ^m(c+d x) \left (\frac {b \tan (c+d x)}{a}+1\right )^{5/2}}{i \tan (c+d x)+1}d\tan (c+d x)}{2 d \sqrt {\frac {b \tan (c+d x)}{a}+1}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {a^2 (A+i B) \tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} \operatorname {AppellF1}\left (m+1,-\frac {5}{2},1,m+2,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}}+\frac {a^2 (A-i B) \tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} \operatorname {AppellF1}\left (m+1,-\frac {5}{2},1,m+2,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}}\)

input

Int[Tan[c + d*x]^m*(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]
 

output

(a^2*(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[a + b*Tan[c + d*x]])/(2*d*(1 + m) 
*Sqrt[1 + (b*Tan[c + d*x])/a]) + (a^2*(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[a + 
 b*Tan[c + d*x]])/(2*d*(1 + m)*Sqrt[1 + (b*Tan[c + d*x])/a])
 

3.5.87.3.1 Defintions of rubi rules used

rule 150

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])
 

rule 152

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4085

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]
 

rule 4086

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]
 

3.5.87.4 Maple [F]

\[\int \tan \left (d x +c \right )^{m} \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}} \left (A +B \tan \left (d x +c \right )\right )d x\]

input

int(tan(d*x+c)^m*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)
 

output

int(tan(d*x+c)^m*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)
 

3.5.87.5 Fricas [F]

\[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{m} \,d x } \]

input

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorith 
m="fricas")
 

output

integral((B*b^2*tan(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*tan(d*x + c)^2 
+ (B*a^2 + 2*A*a*b)*tan(d*x + c))*sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^m, 
 x)
 

3.5.87.6 Sympy [F(-1)]

Timed out. \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

input

integrate(tan(d*x+c)**m*(a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)
 

output

Timed out
 

3.5.87.7 Maxima [F]

\[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{m} \,d x } \]

input

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorith 
m="maxima")
 

output

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^m, 
x)
 

3.5.87.8 Giac [F(-1)]

Timed out. \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

input

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorith 
m="giac")
 

output

Timed out
 

3.5.87.9 Mupad [F(-1)]

Timed out. \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \]

input

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2),x)
 

output

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2), x)