Optimal. Leaf size=360 \[ \frac{\left (3 a^2-9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;-\sin (c+d x))}{16 d (n+1) (a-b)^3}+\frac{\left (3 a^2+9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;\sin (c+d x))}{16 d (n+1) (a+b)^3}-\frac{b^6 \sin ^{n+1}(c+d x) \, _2F_1\left (1,n+1;n+2;-\frac{b \sin (c+d x)}{a}\right )}{a d (n+1) \left (a^2-b^2\right )^3}+\frac{(3 a-5 b) \sin ^{n+1}(c+d x) \, _2F_1(2,n+1;n+2;-\sin (c+d x))}{16 d (n+1) (a-b)^2}+\frac{(3 a+5 b) \sin ^{n+1}(c+d x) \, _2F_1(2,n+1;n+2;\sin (c+d x))}{16 d (n+1) (a+b)^2}+\frac{\sin ^{n+1}(c+d x) \, _2F_1(3,n+1;n+2;-\sin (c+d x))}{8 d (n+1) (a-b)}+\frac{\sin ^{n+1}(c+d x) \, _2F_1(3,n+1;n+2;\sin (c+d x))}{8 d (n+1) (a+b)} \]
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Rubi [A] time = 0.541221, antiderivative size = 360, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2837, 961, 64} \[ \frac{\left (3 a^2-9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;-\sin (c+d x))}{16 d (n+1) (a-b)^3}+\frac{\left (3 a^2+9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;\sin (c+d x))}{16 d (n+1) (a+b)^3}-\frac{b^6 \sin ^{n+1}(c+d x) \, _2F_1\left (1,n+1;n+2;-\frac{b \sin (c+d x)}{a}\right )}{a d (n+1) \left (a^2-b^2\right )^3}+\frac{(3 a-5 b) \sin ^{n+1}(c+d x) \, _2F_1(2,n+1;n+2;-\sin (c+d x))}{16 d (n+1) (a-b)^2}+\frac{(3 a+5 b) \sin ^{n+1}(c+d x) \, _2F_1(2,n+1;n+2;\sin (c+d x))}{16 d (n+1) (a+b)^2}+\frac{\sin ^{n+1}(c+d x) \, _2F_1(3,n+1;n+2;-\sin (c+d x))}{8 d (n+1) (a-b)}+\frac{\sin ^{n+1}(c+d x) \, _2F_1(3,n+1;n+2;\sin (c+d x))}{8 d (n+1) (a+b)} \]
Antiderivative was successfully verified.
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Rule 2837
Rule 961
Rule 64
Rubi steps
\begin{align*} \int \frac{\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^5 \operatorname{Subst}\left (\int \left (\frac{\left (\frac{x}{b}\right )^n}{8 b^3 (a+b) (b-x)^3}+\frac{(3 a+5 b) \left (\frac{x}{b}\right )^n}{16 b^4 (a+b)^2 (b-x)^2}+\frac{\left (3 a^2+9 a b+8 b^2\right ) \left (\frac{x}{b}\right )^n}{16 b^5 (a+b)^3 (b-x)}-\frac{\left (\frac{x}{b}\right )^n}{(a-b)^3 (a+b)^3 (a+x)}-\frac{\left (\frac{x}{b}\right )^n}{8 b^3 (-a+b) (b+x)^3}+\frac{(3 a-5 b) \left (\frac{x}{b}\right )^n}{16 (a-b)^2 b^4 (b+x)^2}+\frac{\left (3 a^2-9 a b+8 b^2\right ) \left (\frac{x}{b}\right )^n}{16 (a-b)^3 b^5 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{((3 a-5 b) b) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n}{(b+x)^2} \, dx,x,b \sin (c+d x)\right )}{16 (a-b)^2 d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n}{(b+x)^3} \, dx,x,b \sin (c+d x)\right )}{8 (a-b) d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n}{(b-x)^3} \, dx,x,b \sin (c+d x)\right )}{8 (a+b) d}+\frac{(b (3 a+5 b)) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n}{(b-x)^2} \, dx,x,b \sin (c+d x)\right )}{16 (a+b)^2 d}-\frac{b^5 \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n}{a+x} \, dx,x,b \sin (c+d x)\right )}{\left (a^2-b^2\right )^3 d}+\frac{\left (3 a^2-9 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n}{b+x} \, dx,x,b \sin (c+d x)\right )}{16 (a-b)^3 d}+\frac{\left (3 a^2+9 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 (a+b)^3 d}\\ &=\frac{\left (3 a^2-9 a b+8 b^2\right ) \, _2F_1(1,1+n;2+n;-\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a-b)^3 d (1+n)}+\frac{\left (3 a^2+9 a b+8 b^2\right ) \, _2F_1(1,1+n;2+n;\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a+b)^3 d (1+n)}-\frac{b^6 \, _2F_1\left (1,1+n;2+n;-\frac{b \sin (c+d x)}{a}\right ) \sin ^{1+n}(c+d x)}{a \left (a^2-b^2\right )^3 d (1+n)}+\frac{(3 a-5 b) \, _2F_1(2,1+n;2+n;-\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a-b)^2 d (1+n)}+\frac{(3 a+5 b) \, _2F_1(2,1+n;2+n;\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a+b)^2 d (1+n)}+\frac{\, _2F_1(3,1+n;2+n;-\sin (c+d x)) \sin ^{1+n}(c+d x)}{8 (a-b) d (1+n)}+\frac{\, _2F_1(3,1+n;2+n;\sin (c+d x)) \sin ^{1+n}(c+d x)}{8 (a+b) d (1+n)}\\ \end{align*}
Mathematica [A] time = 0.451814, size = 241, normalized size = 0.67 \[ \frac{\sin ^{n+1}(c+d x) \left (\frac{\left (3 a^2-9 a b+8 b^2\right ) \, _2F_1(1,n+1;n+2;-\sin (c+d x))}{(a-b)^3}+\frac{\left (3 a^2+9 a b+8 b^2\right ) \, _2F_1(1,n+1;n+2;\sin (c+d x))}{(a+b)^3}-\frac{16 b^6 \, _2F_1\left (1,n+1;n+2;-\frac{b \sin (c+d x)}{a}\right )}{a (a-b)^3 (a+b)^3}+\frac{(3 a-5 b) \, _2F_1(2,n+1;n+2;-\sin (c+d x))}{(a-b)^2}+\frac{(3 a+5 b) \, _2F_1(2,n+1;n+2;\sin (c+d x))}{(a+b)^2}+\frac{2 \, _2F_1(3,n+1;n+2;-\sin (c+d x))}{a-b}+\frac{2 \, _2F_1(3,n+1;n+2;\sin (c+d x))}{a+b}\right )}{16 d (n+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.071, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{n}}{a+b\sin \left ( dx+c \right ) }}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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