Integrand size = 21, antiderivative size = 183 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx=-\frac {(a-b (1-m)) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a-b}\right ) (a+b \sin (c+d x))^{1+m}}{4 (a-b)^2 d (1+m)}+\frac {(a+b-b m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a+b}\right ) (a+b \sin (c+d x))^{1+m}}{4 (a+b)^2 d (1+m)}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{2 \left (a^2-b^2\right ) d} \]
-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
Time = 0.40 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.86 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\frac {(a+b \sin (c+d x))^{1+m} \left (\frac {b \left ((a+b)^2 (a+b (-1+m)) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a-b}\right )-(a-b)^2 (a+b-b m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a+b}\right )\right )}{(a-b) (a+b) (1+m)}+2 b \sec ^2(c+d x) (b-a \sin (c+d x))\right )}{4 b \left (-a^2+b^2\right ) d} \]
((a + b*Sin[c + d*x])^(1 + m)*((b*((a + b)^2*(a + b*(-1 + m))*Hypergeometr ic2F1[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)
Time = 0.41 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.26, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3147, 496, 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.
\(\displaystyle \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x))^m}{\cos (c+d x)^3}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {b^3 \int \frac {(a+b \sin (c+d x))^m}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 496 |
\(\displaystyle \frac {b^3 \left (\frac {\int \frac {(a+b \sin (c+d x))^m \left (a^2-b m \sin (c+d x) a-b^2 (1-m)\right )}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2 \left (a^2-b^2\right )}-\frac {\left (b^2-a b \sin (c+d x)\right ) (a+b \sin (c+d x))^{m+1}}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \frac {b^3 \left (\frac {\int \left (\frac {\left (b \left (a^2-b^2 (1-m)\right )-a b^2 m\right ) (a+b \sin (c+d x))^m}{2 b^2 (b-b \sin (c+d x))}+\frac {\left (a m b^2+\left (a^2-b^2 (1-m)\right ) b\right ) (a+b \sin (c+d x))^m}{2 b^2 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 b^2 \left (a^2-b^2\right )}-\frac {\left (b^2-a b \sin (c+d x)\right ) (a+b \sin (c+d x))^{m+1}}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^3 \left (\frac {\frac {(a-b) (a-b m+b) (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \sin (c+d x)}{a+b}\right )}{2 b (m+1) (a+b)}-\frac {(a+b) (a-b (1-m)) (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \sin (c+d x)}{a-b}\right )}{2 b (m+1) (a-b)}}{2 b^2 \left (a^2-b^2\right )}-\frac {\left (b^2-a b \sin (c+d x)\right ) (a+b \sin (c+d x))^{m+1}}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
(b^3*(-1/2*((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)) + (-1/2*((a + b)*(a - b*(1 - m))*Hyp ergeometric2F1[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)*(a + b - b*m)*Hypergeomet ric2F1[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))))/d
3.7.35.3.1 Defintions of rubi rules used
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]
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]
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]
\[\int \left (\sec ^{3}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{m}d x\]
\[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{3} \,d x } \]
\[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{m} \sec ^{3}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{3} \,d x } \]
\[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m}{{\cos \left (c+d\,x\right )}^3} \,d x \]