\(\int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) [483]

Optimal result

Integrand size = 31, antiderivative size = 185 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {(a A+b B) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{\left (a^2+b^2\right ) d (1+m)}+\frac {b (A b-a B) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (1+m)}-\frac {(A b-a B) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{\left (a^2+b^2\right ) d (2+m)} \] Output:

(A*a+B*b)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^( 
1+m)/(a^2+b^2)/d/(1+m)+b*(A*b-B*a)*hypergeom([1, 1+m],[2+m],-b*tan(d*x+c)/ 
a)*tan(d*x+c)^(1+m)/a/(a^2+b^2)/d/(1+m)-(A*b-B*a)*hypergeom([1, 1+1/2*m],[ 
2+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^(2+m)/(a^2+b^2)/d/(2+m)
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.78 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\tan ^{1+m}(c+d x) \left ((a A+b B) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right )+\frac {(A b-a B) \left (b (2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b \tan (c+d x)}{a}\right )-a (1+m) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)\right )}{a (2+m)}\right )}{\left (a^2+b^2\right ) d (1+m)} \] Input:

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

Output:

(Tan[c + d*x]^(1 + m)*((a*A + b*B)*Hypergeometric2F1[1, (1 + m)/2, (3 + m) 
/2, -Tan[c + d*x]^2] + ((A*b - a*B)*(b*(2 + m)*Hypergeometric2F1[1, 1 + m, 
 2 + m, -((b*Tan[c + d*x])/a)] - a*(1 + m)*Hypergeometric2F1[1, (2 + m)/2, 
 (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]))/(a*(2 + m))))/((a^2 + b^2)*d*( 
1 + m))
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4096, 3042, 4021, 3042, 3957, 278, 4117, 74}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4096

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4021

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 278

\(\displaystyle \frac {b (A b-a B) \int \frac {\tan (c+d x)^m \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {\frac {(a A+b B) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}-\frac {(A b-a B) \tan ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2)}}{a^2+b^2}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {b (A b-a B) \int \frac {\tan ^m(c+d x)}{a+b \tan (c+d x)}d\tan (c+d x)}{d \left (a^2+b^2\right )}+\frac {\frac {(a A+b B) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}-\frac {(A b-a B) \tan ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2)}}{a^2+b^2}\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {b (A b-a B) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {b \tan (c+d x)}{a}\right )}{a d (m+1) \left (a^2+b^2\right )}+\frac {\frac {(a A+b B) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}-\frac {(A b-a B) \tan ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2)}}{a^2+b^2}\)

Input:

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

Output:

(b*(A*b - a*B)*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*Tan[c + d*x])/a)]*T 
an[c + d*x]^(1 + m))/(a*(a^2 + b^2)*d*(1 + m)) + (((a*A + b*B)*Hypergeomet 
ric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m))/(d* 
(1 + m)) - ((A*b - a*B)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c 
+ d*x]^2]*Tan[c + d*x]^(2 + m))/(d*(2 + m)))/(a^2 + b^2)
 

Defintions of rubi rules used

rule 74

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x 
)^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] 
/; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] 
 &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 3042

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

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4021

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Tan[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ[c^ 
2 + d^2, 0] &&  !IntegerQ[2*m]
 

rule 4096

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_ 
.)*(x_)])^(n_))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*Simp[a*A + b*B - (A*b - a*B)*Tan 
[e + f*x], x], x], x] + Simp[b*((A*b - a*B)/(a^2 + b^2))   Int[(c + d*Tan[e 
 + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, 
 b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N 
eQ[c^2 + d^2, 0]
 

rule 4117

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="fri 
cas")
 

Output:

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

Sympy [F]

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

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

Output:

Integral((A + B*tan(c + d*x))*tan(c + d*x)**m/(a + b*tan(c + d*x)), x)
 

Maxima [F]

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

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="max 
ima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="gia 
c")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:Unable to divide, perhaps due to rounding error%%%{1,[0 
,1,0]%%%} / %%%{1,[0,0,1]%%%} Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(tan(c + d*x)**m,x)