\(\int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1506]

Optimal result

Integrand size = 29, antiderivative size = 186 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a \left (a^2 (3-n)-3 b^2 (1+n)\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{4 d (1+n)}+\frac {b \left (3 a^2 (2-n)-b^2 (2+n)\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{4 d (2+n)}+\frac {\sec ^4(c+d x) \sin ^{1+n}(c+d x) \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.85 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {\left (6 a (a-b) (a+b) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )+3 (a-b)^2 (a+b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,-\sin (c+d x))+3 (a-b) (a+b)^2 \operatorname {Hypergeometric2F1}(2,1+n,2+n,\sin (c+d x))+2 (a-b)^3 \operatorname {Hypergeometric2F1}(3,1+n,2+n,-\sin (c+d x))+2 (a+b)^3 \operatorname {Hypergeometric2F1}(3,1+n,2+n,\sin (c+d x))\right ) \sin ^{1+n}(c+d x)}{16 d (1+n)} \] Input:

Integrate[Sec[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^3,x]
 

Output:

((6*a*(a - b)*(a + b)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, Sin[c + d 
*x]^2] + 3*(a - b)^2*(a + b)*Hypergeometric2F1[2, 1 + n, 2 + n, -Sin[c + d 
*x]] + 3*(a - b)*(a + b)^2*Hypergeometric2F1[2, 1 + n, 2 + n, Sin[c + d*x] 
] + 2*(a - b)^3*Hypergeometric2F1[3, 1 + n, 2 + n, -Sin[c + d*x]] + 2*(a + 
 b)^3*Hypergeometric2F1[3, 1 + n, 2 + n, Sin[c + d*x]])*Sin[c + d*x]^(1 + 
n))/(16*d*(1 + n))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 3316, 558, 25, 557, 278}

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.

\(\displaystyle \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^n (a+b \sin (c+d x))^3}{\cos (c+d x)^5}dx\)

\(\Big \downarrow \) 3316

\(\displaystyle \frac {b^5 \int \frac {\sin ^n(c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 558

\(\displaystyle \frac {b^5 \left (\frac {\sin ^{n+1}(c+d x) \left (b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )\right )}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {\sin ^n(c+d x) \left (a \left (a^2 (3-n)-3 b^2 (n+1)\right )+b \left (3 a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x)\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b^5 \left (\frac {\int \frac {\sin ^n(c+d x) \left (a \left (a^2 (3-n)-3 b^2 (n+1)\right )+b \left (3 a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x)\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {\sin ^{n+1}(c+d x) \left (b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )\right )}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 557

\(\displaystyle \frac {b^5 \left (\frac {b \left (3 a^2 (2-n)-b^2 (n+2)\right ) \int \frac {\sin ^{n+1}(c+d x)}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))+a \left (a^2 (3-n)-3 b^2 (n+1)\right ) \int \frac {\sin ^n(c+d x)}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {\sin ^{n+1}(c+d x) \left (b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )\right )}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 278

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

Input:

Int[Sec[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^3,x]
 

Output:

(b^5*((Sin[c + d*x]^(1 + n)*(a*(a^2 + 3*b^2) + b*(3*a^2 + b^2)*Sin[c + d*x 
]))/(4*b*(b^2 - b^2*Sin[c + d*x]^2)^2) + ((a*(a^2*(3 - n) - 3*b^2*(1 + n)) 
*Hypergeometric2F1[2, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^( 
1 + n))/(b^3*(1 + n)) + ((3*a^2*(2 - n) - b^2*(2 + n))*Hypergeometric2F1[2 
, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2 + n))/(b^2*(2 + n) 
))/(4*b^2)))/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 557

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

rule 558

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, a + b*x^2, x], f = 
 Coeff[PolynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 0], g = Coeff[Pol 
ynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(-(e*x)^(m + 1))* 
(f + g*x)*((a + b*x^2)^(p + 1)/(2*a*e*(p + 1))), x] + Simp[1/(2*a*(p + 1)) 
  Int[(e*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + f*(m + 2*p + 
 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, m}, x] && IGt 
Q[n, 1] &&  !IntegerQ[m] && LtQ[p, -1]
 

rule 3042

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

rule 3316

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

Maple [F]

\[\int \sec \left (d x +c \right )^{5} \sin \left (d x +c \right )^{n} \left (a +b \sin \left (d x +c \right )\right )^{3}d x\]

Input:

int(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^3,x)
 

Output:

int(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^3,x)
 

Fricas [F]

\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \] Input:

integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^3,x, algorithm="frica 
s")
 

Output:

integral(-((b^3*cos(d*x + c)^2 - 3*a^2*b - b^3)*sec(d*x + c)^5*sin(d*x + c 
) + (3*a*b^2*cos(d*x + c)^2 - a^3 - 3*a*b^2)*sec(d*x + c)^5)*sin(d*x + c)^ 
n, x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**5*sin(d*x+c)**n*(a+b*sin(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \] Input:

integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^3,x, algorithm="maxim 
a")
 

Output:

integrate((b*sin(d*x + c) + a)^3*sin(d*x + c)^n*sec(d*x + c)^5, x)
 

Giac [F(-2)]

Exception generated. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^3,x, algorithm="giac" 
)
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:Unable to divide, perhaps due to rounding error%%%{1,[0 
,1,12,0]%%%}+%%%{6,[0,1,10,0]%%%}+%%%{15,[0,1,8,0]%%%}+%%%{20,[0,1,6,0]%%% 
}+%%%{15,[0,
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\int \frac {{\sin \left (c+d\,x\right )}^n\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^3}{{\cos \left (c+d\,x\right )}^5} \,d x \] Input:

int((sin(c + d*x)^n*(a + b*sin(c + d*x))^3)/cos(c + d*x)^5,x)
 

Output:

int((sin(c + d*x)^n*(a + b*sin(c + d*x))^3)/cos(c + d*x)^5, x)
 

Reduce [F]

\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\left (\int \sin \left (d x +c \right )^{n} \sec \left (d x +c \right )^{5} \sin \left (d x +c \right )^{3}d x \right ) b^{3}+3 \left (\int \sin \left (d x +c \right )^{n} \sec \left (d x +c \right )^{5} \sin \left (d x +c \right )^{2}d x \right ) a \,b^{2}+3 \left (\int \sin \left (d x +c \right )^{n} \sec \left (d x +c \right )^{5} \sin \left (d x +c \right )d x \right ) a^{2} b +\left (\int \sin \left (d x +c \right )^{n} \sec \left (d x +c \right )^{5}d x \right ) a^{3} \] Input:

int(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^3,x)
                                                                                    
                                                                                    
 

Output:

int(sin(c + d*x)**n*sec(c + d*x)**5*sin(c + d*x)**3,x)*b**3 + 3*int(sin(c 
+ d*x)**n*sec(c + d*x)**5*sin(c + d*x)**2,x)*a*b**2 + 3*int(sin(c + d*x)** 
n*sec(c + d*x)**5*sin(c + d*x),x)*a**2*b + int(sin(c + d*x)**n*sec(c + d*x 
)**5,x)*a**3