\(\int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx\) [79]

Optimal result

Integrand size = 21, antiderivative size = 229 \[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {a^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sin ^{1+m}(c+d x)}{d (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^2 b \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},\sin ^2(c+d x)\right ) \sin ^{2+m}(c+d x)}{d (2+m)}+\frac {3 a b^2 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{3+m}(c+d x)}{d (3+m)}+\frac {b^3 \operatorname {Hypergeometric2F1}\left (2,\frac {4+m}{2},\frac {6+m}{2},\sin ^2(c+d x)\right ) \sin ^{4+m}(c+d x)}{d (4+m)} \] Output:

a^3*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)+3*a^2*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)+3*a*b^2*(cos(d*x+c)^2)^(1/2) 
*hypergeom([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)+b^3*hypergeom([2, 2+1/2*m],[3+1/2*m],sin(d*x+c)^2)*sin(d*x 
+c)^(4+m)/d/(4+m)
 

Mathematica [A] (verified)

Time = 1.73 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.90 \[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\sin ^{1+m}(c+d x) \left (\frac {a^3 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x)}{1+m}+b \sin (c+d x) \left (\frac {3 a^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},\sin ^2(c+d x)\right )}{2+m}+b \left (\frac {b \operatorname {Hypergeometric2F1}\left (2,\frac {4+m}{2},\frac {6+m}{2},\sin ^2(c+d x)\right ) \sin ^2(c+d x)}{4+m}+\frac {3 a \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},\sin ^2(c+d x)\right ) \tan (c+d x)}{3+m}\right )\right )\right )}{d} \] Input:

Integrate[Sin[c + d*x]^m*(a + b*Tan[c + d*x])^3,x]
 

Output:

(Sin[c + d*x]^(1 + m)*((a^3*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] 
*((3*a^2*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[c + d*x]^2])/(2 + 
m) + b*((b*Hypergeometric2F1[2, (4 + m)/2, (6 + m)/2, Sin[c + d*x]^2]*Sin[ 
c + d*x]^2)/(4 + m) + (3*a*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3/2, (3 
+ m)/2, (5 + m)/2, Sin[c + d*x]^2]*Tan[c + d*x])/(3 + m)))))/d
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 229, 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.

\(\displaystyle \int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^m (a+b \tan (c+d x))^3dx\)

\(\Big \downarrow \) 4901

\(\displaystyle \int \left (a^3 \sin ^m(c+d x)+3 a^2 b \sec (c+d x) \sin ^{m+1}(c+d x)+3 a b^2 \sec ^2(c+d x) \sin ^{m+2}(c+d x)+b^3 \sec ^3(c+d x) \sin ^{m+3}(c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 \cos (c+d x) \sin ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d (m+1) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^2 b \sin ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},\sin ^2(c+d x)\right )}{d (m+2)}+\frac {3 a b^2 \sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{m+3}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+3}{2},\frac {m+5}{2},\sin ^2(c+d x)\right )}{d (m+3)}+\frac {b^3 \sin ^{m+4}(c+d x) \operatorname {Hypergeometric2F1}\left (2,\frac {m+4}{2},\frac {m+6}{2},\sin ^2(c+d x)\right )}{d (m+4)}\)

Input:

Int[Sin[c + d*x]^m*(a + b*Tan[c + d*x])^3,x]
 

Output:

(a^3*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]) + (3*a^2*b*Hyp 
ergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2 + 
m))/(d*(2 + m)) + (3*a*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)) + (b^3*Hypergeometric2F1[2, (4 + m)/2, (6 + m)/2, Sin[c + d*x]^2]* 
Sin[c + d*x]^(4 + m))/(d*(4 + 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]

\[\int \sin \left (d x +c \right )^{m} \left (a +b \tan \left (d x +c \right )\right )^{3}d x\]

Input:

int(sin(d*x+c)^m*(a+b*tan(d*x+c))^3,x)
 

Output:

int(sin(d*x+c)^m*(a+b*tan(d*x+c))^3,x)
 

Fricas [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sin ^{m}{\left (c + d x \right )}\, dx \] Input:

integrate(sin(d*x+c)**m*(a+b*tan(d*x+c))**3,x)
 

Output:

Integral((a + b*tan(c + d*x))**3*sin(c + d*x)**m, x)
 

Maxima [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

integrate((b*tan(d*x + c) + a)^3*sin(d*x + c)^m, x)
 

Giac [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate((b*tan(d*x + c) + a)^3*sin(d*x + c)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx=\int {\sin \left (c+d\,x\right )}^m\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^3 \,d x \] Input:

int(sin(c + d*x)^m*(a + b*tan(c + d*x))^3,x)
 

Output:

int(sin(c + d*x)^m*(a + b*tan(c + d*x))^3, x)
 

Reduce [F]

\[ \int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx=\left (\int \sin \left (d x +c \right )^{m}d x \right ) a^{3}+\left (\int \sin \left (d x +c \right )^{m} \tan \left (d x +c \right )^{3}d x \right ) b^{3}+3 \left (\int \sin \left (d x +c \right )^{m} \tan \left (d x +c \right )^{2}d x \right ) a \,b^{2}+3 \left (\int \sin \left (d x +c \right )^{m} \tan \left (d x +c \right )d x \right ) a^{2} b \] Input:

int(sin(d*x+c)^m*(a+b*tan(d*x+c))^3,x)
 

Output:

int(sin(c + d*x)**m,x)*a**3 + int(sin(c + d*x)**m*tan(c + d*x)**3,x)*b**3 
+ 3*int(sin(c + d*x)**m*tan(c + d*x)**2,x)*a*b**2 + 3*int(sin(c + d*x)**m* 
tan(c + d*x),x)*a**2*b