\(\int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx\) [80]

Optimal result

Integrand size = 21, antiderivative size = 179 \[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sin ^{1+m}(c+d x)}{d (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {2 a b \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},\sin ^2(c+d x)\right ) \sin ^{2+m}(c+d x)}{d (2+m)}+\frac {b^2 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{3+m}(c+d x)}{d (3+m)} \] Output:

a^2*cos(d*x+c)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],sin(d*x+c)^2)*sin(d* 
x+c)^(1+m)/d/(1+m)/(cos(d*x+c)^2)^(1/2)+2*a*b*hypergeom([1, 1+1/2*m],[2+1/ 
2*m],sin(d*x+c)^2)*sin(d*x+c)^(2+m)/d/(2+m)+b^2*(cos(d*x+c)^2)^(1/2)*hyper 
geom([3/2, 3/2+1/2*m],[5/2+1/2*m],sin(d*x+c)^2)*sec(d*x+c)*sin(d*x+c)^(3+m 
)/d/(3+m)
 

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.93 \[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\sin ^{1+m}(c+d x) \left (\frac {a^2 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x)}{1+m}+\frac {b \sin (c+d x) \left (2 a (3+m) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},\sin ^2(c+d x)\right )+b (2+m) \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},\sin ^2(c+d x)\right ) \tan (c+d x)\right )}{(2+m) (3+m)}\right )}{d} \] Input:

Integrate[Sin[c + d*x]^m*(a + b*Tan[c + d*x])^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 179, 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, 4901, 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.

\(\displaystyle \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^m (a+b \tan (c+d x))^2dx\)

\(\Big \downarrow \) 4901

\(\displaystyle \int \left (a^2 \sin ^m(c+d x)+2 a b \sec (c+d x) \sin ^{m+1}(c+d x)+b^2 \sec ^2(c+d x) \sin ^{m+2}(c+d x)\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[Sin[c + d*x]^m*(a + b*Tan[c + d*x])^2,x]
 

Output:

(a^2*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x 
]^2]*Sin[c + d*x]^(1 + m))/(d*(1 + m)*Sqrt[Cos[c + d*x]^2]) + (2*a*b*Hyper 
geometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2 + m) 
)/(d*(2 + m)) + (b^2*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3/2, (3 + m)/2 
, (5 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*Sin[c + d*x]^(3 + m))/(d*(3 + m) 
)
 

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 4901

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /; 
 !InertTrigFreeQ[u]
 

Maple [F]

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

Input:

int(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x)
 

Output:

int(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x)
 

Fricas [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

integral((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)*sin(d*x + c)^m, x 
)
 

Sympy [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sin ^{m}{\left (c + d x \right )}\, dx \] Input:

integrate(sin(d*x+c)**m*(a+b*tan(d*x+c))**2,x)
 

Output:

Integral((a + b*tan(c + d*x))**2*sin(c + d*x)**m, x)
 

Maxima [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x, algorithm="maxima")
 

Output:

integrate((b*tan(d*x + c) + a)^2*sin(d*x + c)^m, x)
 

Giac [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate((b*tan(d*x + c) + a)^2*sin(d*x + c)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\int {\sin \left (c+d\,x\right )}^m\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2 \,d x \] Input:

int(sin(c + d*x)^m*(a + b*tan(c + d*x))^2,x)
 

Output:

int(sin(c + d*x)^m*(a + b*tan(c + d*x))^2, x)
 

Reduce [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx=\left (\int \sin \left (d x +c \right )^{m}d x \right ) a^{2}+\left (\int \sin \left (d x +c \right )^{m} \tan \left (d x +c \right )^{2}d x \right ) b^{2}+2 \left (\int \sin \left (d x +c \right )^{m} \tan \left (d x +c \right )d x \right ) a b \] Input:

int(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x)
 

Output:

int(sin(c + d*x)**m,x)*a**2 + int(sin(c + d*x)**m*tan(c + d*x)**2,x)*b**2 
+ 2*int(sin(c + d*x)**m*tan(c + d*x),x)*a*b