Integrand size = 29, antiderivative size = 186 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a \left (a^2 (3-n)-3 b^2 (1+n)\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{4 d (1+n)}+\frac {b \left (3 a^2 (2-n)-b^2 (2+n)\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{4 d (2+n)}+\frac {\sec ^4(c+d x) \sin ^{1+n}(c+d x) \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d} \]
[Out]
Time = 0.20 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2916, 1820, 822, 371} \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a \left (a^2 (3-n)-3 b^2 (n+1)\right ) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (2,\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{4 d (n+1)}+\frac {b \left (3 a^2 (2-n)-b^2 (n+2)\right ) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (2,\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{4 d (n+2)}+\frac {\sec ^4(c+d x) \sin ^{n+1}(c+d x) \left (b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )\right )}{4 d} \]
[In]
[Out]
Rule 371
Rule 822
Rule 1820
Rule 2916
Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)^3}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \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}-\frac {b^3 \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (-a \left (a^2 (3-n)-3 b^2 (1+n)\right )-\left (3 a^2 (2-n)-b^2 (2+n)\right ) x\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \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}+\frac {\left (a b^3 \left (a^2 (3-n)-3 b^2 (1+n)\right )\right ) \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}+\frac {\left (b^4 \left (3 a^2 (2-n)-b^2 (2+n)\right )\right ) \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^{1+n}}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \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} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.85 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {\left (6 a (a-b) (a+b) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )+3 (a-b)^2 (a+b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,-\sin (c+d x))+3 (a-b) (a+b)^2 \operatorname {Hypergeometric2F1}(2,1+n,2+n,\sin (c+d x))+2 (a-b)^3 \operatorname {Hypergeometric2F1}(3,1+n,2+n,-\sin (c+d x))+2 (a+b)^3 \operatorname {Hypergeometric2F1}(3,1+n,2+n,\sin (c+d x))\right ) \sin ^{1+n}(c+d x)}{16 d (1+n)} \]
[In]
[Out]
\[\int \left (\sec ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{3}d x\]
[In]
[Out]
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]
[In]
[Out]
Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]
[In]
[Out]
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]
[In]
[Out]
Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\int \frac {{\sin \left (c+d\,x\right )}^n\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^3}{{\cos \left (c+d\,x\right )}^5} \,d x \]
[In]
[Out]