\(\int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx\) [635]

Optimal result

Integrand size = 21, antiderivative size = 183 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx=-\frac {(a-b (1-m)) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a-b}\right ) (a+b \sin (c+d x))^{1+m}}{4 (a-b)^2 d (1+m)}+\frac {(a+b-b m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a+b}\right ) (a+b \sin (c+d x))^{1+m}}{4 (a+b)^2 d (1+m)}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{2 \left (a^2-b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2747, 755, 845, 70} \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx=-\frac {\sec ^2(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{m+1}}{2 d \left (a^2-b^2\right )}-\frac {(a-b (1-m)) (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \sin (c+d x)}{a-b}\right )}{4 d (m+1) (a-b)^2}+\frac {(a-b m+b) (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \sin (c+d x)}{a+b}\right )}{4 d (m+1) (a+b)^2} \]

[In]

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

[Out]

-1/4*((a - b*(1 - m))*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Sin[c + d*x])/(a - b)]*(a + b*Sin[c + d*x])^(1
+ m))/((a - b)^2*d*(1 + m)) + ((a + b - b*m)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Sin[c + d*x])/(a + b)]
*(a + b*Sin[c + d*x])^(1 + m))/(4*(a + b)^2*d*(1 + m)) - (Sec[c + d*x]^2*(b - a*Sin[c + d*x])*(a + b*Sin[c + d
*x])^(1 + m))/(2*(a^2 - b^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 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^3 \text {Subst}\left (\int \frac {(a+x)^m}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {\sec ^2(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{2 \left (a^2-b^2\right ) d}+\frac {b \text {Subst}\left (\int \frac {(a+x)^m \left (a^2-b^2 (1-m)-a m x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d} \\ & = -\frac {\sec ^2(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{2 \left (a^2-b^2\right ) d}+\frac {b \text {Subst}\left (\int \left (\frac {\left (b \left (a^2-b^2 (1-m)\right )-a b^2 m\right ) (a+x)^m}{2 b^2 (b-x)}+\frac {\left (b \left (a^2-b^2 (1-m)\right )+a b^2 m\right ) (a+x)^m}{2 b^2 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d} \\ & = -\frac {\sec ^2(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{2 \left (a^2-b^2\right ) d}+\frac {((a+b) (a-b (1-m))) \text {Subst}\left (\int \frac {(a+x)^m}{b+x} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d}+\frac {(a+b-b m) \text {Subst}\left (\int \frac {(a+x)^m}{b-x} \, dx,x,b \sin (c+d x)\right )}{4 (a+b) d} \\ & = -\frac {(a-b (1-m)) \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}}{4 (a-b)^2 d (1+m)}+\frac {(a+b-b m) \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}}{4 (a+b)^2 d (1+m)}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{2 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.86 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\frac {(a+b \sin (c+d x))^{1+m} \left (\frac {b \left ((a+b)^2 (a+b (-1+m)) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a-b}\right )-(a-b)^2 (a+b-b m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a+b}\right )\right )}{(a-b) (a+b) (1+m)}+2 b \sec ^2(c+d x) (b-a \sin (c+d x))\right )}{4 b \left (-a^2+b^2\right ) d} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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