\(\int \sec ^5(c+d x) (a+b \sin (c+d x))^m \, dx\) [636]

Optimal result

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

[Out]

-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)+1/16*(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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2747, 755, 837, 845, 70} \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^m \, dx=-\frac {\left (3 a^2-3 a b (2-m)+b^2 \left (m^2-4 m+3\right )\right ) (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \sin (c+d x)}{a-b}\right )}{16 d (m+1) (a-b)^3}+\frac {\left (3 a^2+3 a b (2-m)+b^2 \left (m^2-4 m+3\right )\right ) (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \sin (c+d x)}{a+b}\right )}{16 d (m+1) (a+b)^3}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{m+1}}{4 d \left (a^2-b^2\right )}+\frac {\sec ^2(c+d x) \left (a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)+b \left (b^2 (3-m)-a^2 (m+1)\right )\right ) (a+b \sin (c+d x))^{m+1}}{8 d \left (a^2-b^2\right )^2} \]

[In]

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

[Out]

-1/16*((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)^3*d*(1 + m)) + ((3*a^2 + 3*a*b*(2 - m) + b^2*(3 - 4*m + m^2))*H
ypergeometric2F1[1, 1 + m, 2 + m, (a + b*Sin[c + d*x])/(a + b)]*(a + b*Sin[c + d*x])^(1 + m))/(16*(a + b)^3*d*
(1 + m)) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(1 + m))/(4*(a^2 - b^2)*d) + (Sec[c + d*x
]^2*(a + b*Sin[c + d*x])^(1 + m)*(b*(b^2*(3 - m) - a^2*(1 + m)) + a*(3*a^2 - b^2*(5 - 2*m))*Sin[c + d*x]))/(8*
(a^2 - b^2)^2*d)

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 755

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(2*a*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^
m*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[
{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 837

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] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[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] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 845

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {(a+x)^m}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\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 {b^3 \text {Subst}\left (\int \frac {(a+x)^m \left (3 a^2-b^2 (3-m)+a (2-m) x\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d} \\ & = -\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}-\frac {b \text {Subst}\left (\int \frac {(a+x)^m \left (-3 a^4+a^2 b^2 \left (6-2 m-m^2\right )-b^4 \left (3-4 m+m^2\right )+a \left (3 a^2-b^2 (5-2 m)\right ) m x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \\ & = -\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}-\frac {b \text {Subst}\left (\int \left (\frac {\left (a b^2 \left (3 a^2-b^2 (5-2 m)\right ) m+b \left (-3 a^4+a^2 b^2 \left (6-2 m-m^2\right )-b^4 \left (3-4 m+m^2\right )\right )\right ) (a+x)^m}{2 b^2 (b-x)}+\frac {\left (-a b^2 \left (3 a^2-b^2 (5-2 m)\right ) m+b \left (-3 a^4+a^2 b^2 \left (6-2 m-m^2\right )-b^4 \left (3-4 m+m^2\right )\right )\right ) (a+x)^m}{2 b^2 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \\ & = -\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}+\frac {\left (3 a^2-3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \text {Subst}\left (\int \frac {(a+x)^m}{b+x} \, dx,x,b \sin (c+d x)\right )}{16 (a-b)^2 d}+\frac {\left (3 a^2+3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \text {Subst}\left (\int \frac {(a+x)^m}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 (a+b)^2 d} \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.58 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.85 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^m \, dx=\frac {(a+b \sin (c+d x))^{1+m} \left (\frac {(a+b)^3 \left (3 a^2+3 a b (-2+m)+b^2 \left (3-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a-b}\right )-(a-b)^3 \left (3 a^2-3 a b (-2+m)+b^2 \left (3-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a+b}\right )}{(a-b) (a+b) \left (a^2-b^2\right ) (1+m)}+4 \sec ^4(c+d x) (b-a \sin (c+d x))+\frac {2 \sec ^2(c+d x) \left (b^3 (-3+m)+a^2 b (1+m)-a \left (3 a^2+b^2 (-5+2 m)\right ) \sin (c+d x)\right )}{a^2-b^2}\right )}{16 \left (-a^2+b^2\right ) d} \]

[In]

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

[Out]

((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))*Hyperge
ometric2F1[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)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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