Optimal. Leaf size=487 \[ \frac{3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,1;n+2;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 d (n+1)}+\frac{3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,1;n+2;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 d (n+1)}+\frac{3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,2;n+2;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 d (n+1)}+\frac{3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,2;n+2;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 d (n+1)}+\frac{\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,3;n+2;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right )}{8 d (n+1)}+\frac{\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,3;n+2;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right )}{8 d (n+1)} \]
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Rubi [A] time = 0.566236, antiderivative size = 487, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2837, 961, 135, 133, 912} \[ \frac{3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,1;n+2;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 d (n+1)}+\frac{3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,1;n+2;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 d (n+1)}+\frac{3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,2;n+2;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 d (n+1)}+\frac{3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,2;n+2;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 d (n+1)}+\frac{\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,3;n+2;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right )}{8 d (n+1)}+\frac{\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac{b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,3;n+2;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right )}{8 d (n+1)} \]
Antiderivative was successfully verified.
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Rule 2837
Rule 961
Rule 135
Rule 133
Rule 912
Rubi steps
\begin{align*} \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n (a+x)^p}{\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 (a+x)^p}{8 b^3 (b-x)^3}+\frac{3 \left (\frac{x}{b}\right )^n (a+x)^p}{16 b^4 (b-x)^2}+\frac{\left (\frac{x}{b}\right )^n (a+x)^p}{8 b^3 (b+x)^3}+\frac{3 \left (\frac{x}{b}\right )^n (a+x)^p}{16 b^4 (b+x)^2}+\frac{3 \left (\frac{x}{b}\right )^n (a+x)^p}{8 b^4 \left (b^2-x^2\right )}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n (a+x)^p}{(b-x)^2} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n (a+x)^p}{(b+x)^2} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n (a+x)^p}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n (a+x)^p}{(b-x)^3} \, dx,x,b \sin (c+d x)\right )}{8 d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n (a+x)^p}{(b+x)^3} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac{(3 b) \operatorname{Subst}\left (\int \left (\frac{\left (\frac{x}{b}\right )^n (a+x)^p}{2 b (b-x)}+\frac{\left (\frac{x}{b}\right )^n (a+x)^p}{2 b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d}+\frac{\left (3 b (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n \left (1+\frac{x}{a}\right )^p}{(b-x)^2} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{\left (3 b (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n \left (1+\frac{x}{a}\right )^p}{(b+x)^2} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{\left (b^2 (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n \left (1+\frac{x}{a}\right )^p}{(b-x)^3} \, dx,x,b \sin (c+d x)\right )}{8 d}+\frac{\left (b^2 (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n \left (1+\frac{x}{a}\right )^p}{(b+x)^3} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac{3 F_1\left (1+n;-p,2;2+n;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac{3 F_1\left (1+n;-p,2;2+n;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac{F_1\left (1+n;-p,3;2+n;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac{F_1\left (1+n;-p,3;2+n;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac{3 \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n (a+x)^p}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{3 \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n (a+x)^p}{b+x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=\frac{3 F_1\left (1+n;-p,2;2+n;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac{3 F_1\left (1+n;-p,2;2+n;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac{F_1\left (1+n;-p,3;2+n;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac{F_1\left (1+n;-p,3;2+n;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac{\left (3 (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n \left (1+\frac{x}{a}\right )^p}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{\left (3 (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{b}\right )^n \left (1+\frac{x}{a}\right )^p}{b+x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=\frac{3 F_1\left (1+n;-p,1;2+n;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac{3 F_1\left (1+n;-p,1;2+n;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac{3 F_1\left (1+n;-p,2;2+n;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac{3 F_1\left (1+n;-p,2;2+n;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac{F_1\left (1+n;-p,3;2+n;-\frac{b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac{F_1\left (1+n;-p,3;2+n;-\frac{b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac{b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}\\ \end{align*}
Mathematica [F] time = 11.9553, size = 0, normalized size = 0. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 1.485, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{n} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{p}\, 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 ({\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}, 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{\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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