\(\int (a+b \sin ^n(c+d x)) \tan ^m(c+d x) \, dx\) [507]

Optimal result

Integrand size = 21, antiderivative size = 124 \[ \int \left (a+b \sin ^n(c+d x)\right ) \tan ^m(c+d x) \, dx=\frac {a \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {b \cos ^2(c+d x)^{\frac {1+m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\sin ^2(c+d x)\right ) \sin ^n(c+d x) \tan ^{1+m}(c+d x)}{d (1+m+n)} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 14.45 (sec) , antiderivative size = 1395, normalized size of antiderivative = 11.25 \[ \int \left (a+b \sin ^n(c+d x)\right ) \tan ^m(c+d x) \, dx =\text {Too large to display} \] Input:

Integrate[(a + b*Sin[c + d*x]^n)*Tan[c + d*x]^m,x]
 

Output:

(2*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^m*(a*(1 + m + n)*AppellF1[(1 + m)/2, 
m, 1, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + b*(1 + m)*Appe 
llF1[(1 + m + n)/2, m, 1 + n, (3 + m + n)/2, Tan[(c + d*x)/2]^2, -Tan[(c + 
 d*x)/2]^2]*(Sec[(c + d*x)/2]^2)^n*Sin[c + d*x]^n)*Tan[(c + d*x)/2]*Tan[c 
+ d*x]^m*(a*Tan[c + d*x]^m + b*Sin[c + d*x]^n*Tan[c + d*x]^m))/(d*(1 + m)* 
(1 + m + n)*((2*m*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^m*Sec[c + d*x]^2*(a*(1 
 + m + n)*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c 
 + d*x)/2]^2] + b*(1 + m)*AppellF1[(1 + m + n)/2, m, 1 + n, (3 + m + n)/2, 
 Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Sec[(c + d*x)/2]^2)^n*Sin[c + d 
*x]^n)*Tan[(c + d*x)/2]*Tan[c + d*x]^(-1 + m))/((1 + m)*(1 + m + n)) + (Se 
c[(c + d*x)/2]^2*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^m*(a*(1 + m + n)*Appell 
F1[(1 + m)/2, m, 1, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 
b*(1 + m)*AppellF1[(1 + m + n)/2, m, 1 + n, (3 + m + n)/2, Tan[(c + d*x)/2 
]^2, -Tan[(c + d*x)/2]^2]*(Sec[(c + d*x)/2]^2)^n*Sin[c + d*x]^n)*Tan[c + d 
*x]^m)/((1 + m)*(1 + m + n)) + (2*m*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(-1 
+ m)*(a*(1 + m + n)*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, Tan[(c + d*x)/2]^ 
2, -Tan[(c + d*x)/2]^2] + b*(1 + m)*AppellF1[(1 + m + n)/2, m, 1 + n, (3 + 
 m + n)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Sec[(c + d*x)/2]^2)^n 
*Sin[c + d*x]^n)*Tan[(c + d*x)/2]*(-(Sec[(c + d*x)/2]^2*Sin[c + d*x]) + Co 
s[c + d*x]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])*Tan[c + d*x]^m)/((1 + m...
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3713, 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.

\(\displaystyle \int \tan ^m(c+d x) \left (a+b \sin ^n(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^m \left (a+b \sin (c+d x)^n\right )dx\)

\(\Big \downarrow \) 3713

\(\displaystyle \int \left (a \tan ^m(c+d x)+b \tan ^m(c+d x) \sin ^n(c+d x)\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(a + b*Sin[c + d*x]^n)*Tan[c + d*x]^m,x]
 

Output:

(a*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x 
]^(1 + m))/(d*(1 + m)) + (b*(Cos[c + d*x]^2)^((1 + m)/2)*Hypergeometric2F1 
[(1 + m)/2, (1 + m + n)/2, (3 + m + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^n*T 
an[c + d*x]^(1 + m))/(d*(1 + m + n))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3713

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*((d_.)*tan[(e 
_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(d*tan[e + f*x])^m*(a 
 + b*(c*sin[e + f*x])^n)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && 
 IGtQ[p, 0]
 

Maple [F]

\[\int \left (a +b \sin \left (d x +c \right )^{n}\right ) \tan \left (d x +c \right )^{m}d x\]

Input:

int((a+b*sin(d*x+c)^n)*tan(d*x+c)^m,x)
 

Output:

int((a+b*sin(d*x+c)^n)*tan(d*x+c)^m,x)
 

Fricas [F]

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

integrate((a+b*sin(d*x+c)^n)*tan(d*x+c)^m,x, algorithm="fricas")
 

Output:

integral((b*sin(d*x + c)^n + a)*tan(d*x + c)^m, x)
 

Sympy [F]

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

integrate((a+b*sin(d*x+c)**n)*tan(d*x+c)**m,x)
 

Output:

Integral((a + b*sin(c + d*x)**n)*tan(c + d*x)**m, x)
 

Maxima [F]

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

integrate((a+b*sin(d*x+c)^n)*tan(d*x+c)^m,x, algorithm="maxima")
 

Output:

integrate((b*sin(d*x + c)^n + a)*tan(d*x + c)^m, x)
 

Giac [F]

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

integrate((a+b*sin(d*x+c)^n)*tan(d*x+c)^m,x, algorithm="giac")
 

Output:

integrate((b*sin(d*x + c)^n + a)*tan(d*x + c)^m, x)
 

Mupad [F(-1)]

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

int(tan(c + d*x)^m*(a + b*sin(c + d*x)^n),x)
 

Output:

int(tan(c + d*x)^m*(a + b*sin(c + d*x)^n), x)
 

Reduce [F]

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

int((a+b*sin(d*x+c)^n)*tan(d*x+c)^m,x)
 

Output:

int(tan(c + d*x)**m,x)*a + int(tan(c + d*x)**m*sin(c + d*x)**n,x)*b