\(\int \sec ^5(c+d x) (a+b \sin (c+d x))^m \, dx\) [630]

Optimal result

Integrand size = 21, antiderivative size = 305 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^m \, dx=-\frac {\left (3 a^2-3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a-b}\right ) (a+b \sin (c+d x))^{1+m}}{16 (a-b)^3 d (1+m)}+\frac {\left (3 a^2+3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a+b}\right ) (a+b \sin (c+d x))^{1+m}}{16 (a+b)^3 d (1+m)}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{4 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^{1+m} \left (b \left (b^2 (3-m)-a^2 (1+m)\right )+a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \] Output:

-1/16*(3*a^2-3*a*b*(2-m)+b^2*(m^2-4*m+3))*hypergeom([1, 1+m],[2+m],(a+b*si 
n(d*x+c))/(a-b))*(a+b*sin(d*x+c))^(1+m)/(a-b)^3/d/(1+m)+1/16*(3*a^2+3*a*b* 
(2-m)+b^2*(m^2-4*m+3))*hypergeom([1, 1+m],[2+m],(a+b*sin(d*x+c))/(a+b))*(a 
+b*sin(d*x+c))^(1+m)/(a+b)^3/d/(1+m)-1/4*sec(d*x+c)^4*(b-a*sin(d*x+c))*(a+ 
b*sin(d*x+c))^(1+m)/(a^2-b^2)/d+1/8*sec(d*x+c)^2*(a+b*sin(d*x+c))^(1+m)*(b 
*(b^2*(3-m)-a^2*(1+m))+a*(3*a^2-b^2*(5-2*m))*sin(d*x+c))/(a^2-b^2)^2/d
 

Mathematica [A] (verified)

Time = 4.12 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.85 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^m \, dx=\frac {(a+b \sin (c+d x))^{1+m} \left (\frac {(a+b)^3 \left (3 a^2+3 a b (-2+m)+b^2 \left (3-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a-b}\right )-(a-b)^3 \left (3 a^2-3 a b (-2+m)+b^2 \left (3-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a+b}\right )}{(a-b) (a+b) \left (a^2-b^2\right ) (1+m)}+4 \sec ^4(c+d x) (b-a \sin (c+d x))+\frac {2 \sec ^2(c+d x) \left (b^3 (-3+m)+a^2 b (1+m)-a \left (3 a^2+b^2 (-5+2 m)\right ) \sin (c+d x)\right )}{a^2-b^2}\right )}{16 \left (-a^2+b^2\right ) d} \] Input:

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

Output:

((a + b*Sin[c + d*x])^(1 + m)*(((a + b)^3*(3*a^2 + 3*a*b*(-2 + m) + b^2*(3 
 - 4*m + m^2))*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Sin[c + d*x])/(a 
- b)] - (a - b)^3*(3*a^2 - 3*a*b*(-2 + m) + b^2*(3 - 4*m + m^2))*Hypergeom 
etric2F1[1, 1 + m, 2 + m, (a + b*Sin[c + d*x])/(a + b)])/((a - b)*(a + b)* 
(a^2 - b^2)*(1 + m)) + 4*Sec[c + d*x]^4*(b - a*Sin[c + d*x]) + (2*Sec[c + 
d*x]^2*(b^3*(-3 + m) + a^2*b*(1 + m) - a*(3*a^2 + b^2*(-5 + 2*m))*Sin[c + 
d*x]))/(a^2 - b^2)))/(16*(-a^2 + b^2)*d)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.28, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3147, 496, 686, 25, 657, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3147

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

\(\Big \downarrow \) 496

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

\(\Big \downarrow \) 686

\(\displaystyle \frac {b^5 \left (\frac {\frac {(a+b \sin (c+d x))^{m+1} \left (a b \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)+b^2 \left (b^2 (3-m)-a^2 (m+1)\right )\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int -\frac {(a+b \sin (c+d x))^m \left (3 a^4-b^2 \left (-m^2-2 m+6\right ) a^2-b \left (3 a^2-b^2 (5-2 m)\right ) m \sin (c+d x) a+b^4 \left (m^2-4 m+3\right )\right )}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2 \left (a^2-b^2\right )}}{4 b^2 \left (a^2-b^2\right )}-\frac {\left (b^2-a b \sin (c+d x)\right ) (a+b \sin (c+d x))^{m+1}}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b^5 \left (\frac {\frac {\int \frac {(a+b \sin (c+d x))^m \left (3 a^4-b^2 \left (-m^2-2 m+6\right ) a^2-b \left (3 a^2-b^2 (5-2 m)\right ) m \sin (c+d x) a+b^4 \left (m^2-4 m+3\right )\right )}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2 \left (a^2-b^2\right )}+\frac {\left (a b \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)+b^2 \left (b^2 (3-m)-a^2 (m+1)\right )\right ) (a+b \sin (c+d x))^{m+1}}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2 \left (a^2-b^2\right )}-\frac {\left (b^2-a b \sin (c+d x)\right ) (a+b \sin (c+d x))^{m+1}}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 657

\(\displaystyle \frac {b^5 \left (\frac {\frac {\int \left (\frac {\left (b \left (3 a^4-b^2 \left (-m^2-2 m+6\right ) a^2+b^4 \left (m^2-4 m+3\right )\right )-a b^2 \left (3 a^2-b^2 (5-2 m)\right ) m\right ) (a+b \sin (c+d x))^m}{2 b^2 (b-b \sin (c+d x))}+\frac {\left (a \left (3 a^2-b^2 (5-2 m)\right ) m b^2+\left (3 a^4-b^2 \left (-m^2-2 m+6\right ) a^2+b^4 \left (m^2-4 m+3\right )\right ) b\right ) (a+b \sin (c+d x))^m}{2 b^2 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 b^2 \left (a^2-b^2\right )}+\frac {\left (a b \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)+b^2 \left (b^2 (3-m)-a^2 (m+1)\right )\right ) (a+b \sin (c+d x))^{m+1}}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2 \left (a^2-b^2\right )}-\frac {\left (b^2-a b \sin (c+d x)\right ) (a+b \sin (c+d x))^{m+1}}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(b^5*(-1/4*((a + b*Sin[c + d*x])^(1 + m)*(b^2 - a*b*Sin[c + d*x]))/(b^2*(a 
^2 - b^2)*(b^2 - b^2*Sin[c + d*x]^2)^2) + (((a + b*Sin[c + d*x])^(1 + m)*( 
b^2*(b^2*(3 - m) - a^2*(1 + m)) + a*b*(3*a^2 - b^2*(5 - 2*m))*Sin[c + d*x] 
))/(2*b^2*(a^2 - b^2)*(b^2 - b^2*Sin[c + d*x]^2)) + (-1/2*((a + b)^2*(3*a^ 
2 - 3*a*b*(2 - m) + b^2*(3 - 4*m + m^2))*Hypergeometric2F1[1, 1 + m, 2 + m 
, (a + b*Sin[c + d*x])/(a - b)]*(a + b*Sin[c + d*x])^(1 + m))/((a - b)*b*( 
1 + m)) + ((a - b)^2*(3*a^2 + 3*a*b*(2 - m) + b^2*(3 - 4*m + m^2))*Hyperge 
ometric2F1[1, 1 + m, 2 + m, (a + b*Sin[c + d*x])/(a + b)]*(a + b*Sin[c + d 
*x])^(1 + m))/(2*b*(a + b)*(1 + m)))/(2*b^2*(a^2 - b^2)))/(4*b^2*(a^2 - b^ 
2))))/d
 

Defintions of rubi rules used

rule 25

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

rule 496

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 
 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a 
 + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 
*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad 
raticQ[a, 0, b, c, d, n, p, x]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 686

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + 
a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 
1/(2*a*c*(p + 1)*(c*d^2 + a*e^2))   Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim 
p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f 
+ a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ 
[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

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 3147

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^m,x, algorithm="fricas")
 

Output:

integral((b*sin(d*x + c) + a)^m*sec(d*x + c)^5, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^m,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sin(c + d*x)*b + a)**m*sec(c + d*x)**5,x)