\(\int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx\) [82]

Optimal result

Integrand size = 21, antiderivative size = 765 \[ \int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx =\text {Too large to display} \] Output:

2^(1+m)*hypergeom([1+m, 1/2+1/2*m],[3/2+1/2*m],-tan(1/2*d*x+1/2*c)^2)*tan( 
1/2*d*x+1/2*c)*(tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2))^m*(1+tan(1/2* 
d*x+1/2*c)^2)^m/a/d/(1+m)+2^(1+m)*b*AppellF1(1+1/2*m,1,1+m,2+1/2*m,a^2*tan 
(1/2*d*x+1/2*c)^2/(b-(a^2+b^2)^(1/2))^2,-tan(1/2*d*x+1/2*c)^2)*tan(1/2*d*x 
+1/2*c)^2*(tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2))^m*(1+tan(1/2*d*x+1 
/2*c)^2)^m/(a^2+b^2)^(1/2)/(b-(a^2+b^2)^(1/2))/d/(2+m)-2^(1+m)*b*AppellF1( 
1+1/2*m,1,1+m,2+1/2*m,a^2*tan(1/2*d*x+1/2*c)^2/(b+(a^2+b^2)^(1/2))^2,-tan( 
1/2*d*x+1/2*c)^2)*tan(1/2*d*x+1/2*c)^2*(tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+ 
1/2*c)^2))^m*(1+tan(1/2*d*x+1/2*c)^2)^m/(a^2+b^2)^(1/2)/(b+(a^2+b^2)^(1/2) 
)/d/(2+m)+2^(1+m)*a*b*AppellF1(3/2+1/2*m,1,1+m,5/2+1/2*m,a^2*tan(1/2*d*x+1 
/2*c)^2/(b-(a^2+b^2)^(1/2))^2,-tan(1/2*d*x+1/2*c)^2)*tan(1/2*d*x+1/2*c)^3* 
(tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2))^m*(1+tan(1/2*d*x+1/2*c)^2)^m 
/(a^2+b^2)^(1/2)/(b-(a^2+b^2)^(1/2))^2/d/(3+m)-2^(1+m)*a*b*AppellF1(3/2+1/ 
2*m,1,1+m,5/2+1/2*m,a^2*tan(1/2*d*x+1/2*c)^2/(b+(a^2+b^2)^(1/2))^2,-tan(1/ 
2*d*x+1/2*c)^2)*tan(1/2*d*x+1/2*c)^3*(tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/ 
2*c)^2))^m*(1+tan(1/2*d*x+1/2*c)^2)^m/(a^2+b^2)^(1/2)/(b+(a^2+b^2)^(1/2))^ 
2/d/(3+m)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 2.42 (sec) , antiderivative size = 580, normalized size of antiderivative = 0.76, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4902, 27, 7270, 7279, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4902

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7270

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

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2^{m+1} \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}\right )^m \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^m \left (\frac {b \tan ^{m+2}\left (\frac {1}{2} (c+d x)\right ) \operatorname {AppellF1}\left (\frac {m+2}{2},m+1,1,\frac {m+4}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b-\sqrt {a^2+b^2}\right )^2}\right )}{(m+2) \sqrt {a^2+b^2} \left (b-\sqrt {a^2+b^2}\right )}-\frac {b \tan ^{m+2}\left (\frac {1}{2} (c+d x)\right ) \operatorname {AppellF1}\left (\frac {m+2}{2},m+1,1,\frac {m+4}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b+\sqrt {a^2+b^2}\right )^2}\right )}{(m+2) \sqrt {a^2+b^2} \left (\sqrt {a^2+b^2}+b\right )}+\frac {a b \tan ^{m+3}\left (\frac {1}{2} (c+d x)\right ) \operatorname {AppellF1}\left (\frac {m+3}{2},m+1,1,\frac {m+5}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b-\sqrt {a^2+b^2}\right )^2}\right )}{(m+3) \sqrt {a^2+b^2} \left (b-\sqrt {a^2+b^2}\right )^2}-\frac {a b \tan ^{m+3}\left (\frac {1}{2} (c+d x)\right ) \operatorname {AppellF1}\left (\frac {m+3}{2},m+1,1,\frac {m+5}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b+\sqrt {a^2+b^2}\right )^2}\right )}{(m+3) \sqrt {a^2+b^2} \left (\sqrt {a^2+b^2}+b\right )^2}+\frac {\tan ^{m+1}\left (\frac {1}{2} (c+d x)\right ) \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},m+1,\frac {m+3}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{a (m+1)}\right )}{d}\)

Input:

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

Output:

(2^(1 + m)*(Tan[(c + d*x)/2]/(1 + Tan[(c + d*x)/2]^2))^m*(1 + Tan[(c + d*x 
)/2]^2)^m*((Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/ 
2]^2]*Tan[(c + d*x)/2]^(1 + m))/(a*(1 + m)) + (b*AppellF1[(2 + m)/2, 1 + m 
, 1, (4 + m)/2, -Tan[(c + d*x)/2]^2, (a^2*Tan[(c + d*x)/2]^2)/(b - Sqrt[a^ 
2 + b^2])^2]*Tan[(c + d*x)/2]^(2 + m))/(Sqrt[a^2 + b^2]*(b - Sqrt[a^2 + b^ 
2])*(2 + m)) - (b*AppellF1[(2 + m)/2, 1 + m, 1, (4 + m)/2, -Tan[(c + d*x)/ 
2]^2, (a^2*Tan[(c + d*x)/2]^2)/(b + Sqrt[a^2 + b^2])^2]*Tan[(c + d*x)/2]^( 
2 + m))/(Sqrt[a^2 + b^2]*(b + Sqrt[a^2 + b^2])*(2 + m)) + (a*b*AppellF1[(3 
 + m)/2, 1 + m, 1, (5 + m)/2, -Tan[(c + d*x)/2]^2, (a^2*Tan[(c + d*x)/2]^2 
)/(b - Sqrt[a^2 + b^2])^2]*Tan[(c + d*x)/2]^(3 + m))/(Sqrt[a^2 + b^2]*(b - 
 Sqrt[a^2 + b^2])^2*(3 + m)) - (a*b*AppellF1[(3 + m)/2, 1 + m, 1, (5 + m)/ 
2, -Tan[(c + d*x)/2]^2, (a^2*Tan[(c + d*x)/2]^2)/(b + Sqrt[a^2 + b^2])^2]* 
Tan[(c + d*x)/2]^(3 + m))/(Sqrt[a^2 + b^2]*(b + Sqrt[a^2 + b^2])^2*(3 + m) 
)))/(d*Tan[(c + d*x)/2]^m)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 4902

Int[u_, x_Symbol] :> With[{w = Block[{$ShowSteps = False, $StepCounter = Nu 
ll}, Int[SubstFor[1/(1 + FreeFactors[Tan[FunctionOfTrig[u, x]/2], x]^2*x^2) 
, Tan[FunctionOfTrig[u, x]/2]/FreeFactors[Tan[FunctionOfTrig[u, x]/2], x], 
u, x], x]]}, Module[{v = FunctionOfTrig[u, x], d}, Simp[d = FreeFactors[Tan 
[v/2], x]; 2*(d/Coefficient[v, x, 1])   Subst[Int[SubstFor[1/(1 + d^2*x^2), 
 Tan[v/2]/d, u, x], x], x, Tan[v/2]/d], x]] /; CalculusFreeQ[w, x]] /; Inve 
rseFunctionFreeQ[u, x] &&  !FalseQ[FunctionOfTrig[u, x]]
 

rule 7270

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Simp[a^IntPart[p 
]*((a*v^m*w^n)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[p])))   Int[u*v 
^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !Free 
Q[v, x] &&  !FreeQ[w, x]
 

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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