Optimal. Leaf size=305 \[ -\frac{\left (3 a^2-3 a b (2-m)+b^2 \left (m^2-4 m+3\right )\right ) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a-b}\right )}{16 d (m+1) (a-b)^3}+\frac{\left (3 a^2+3 a b (2-m)+b^2 \left (m^2-4 m+3\right )\right ) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a+b}\right )}{16 d (m+1) (a+b)^3}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{m+1}}{4 d \left (a^2-b^2\right )}+\frac{\sec ^2(c+d x) \left (a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)+b \left (b^2 (3-m)-a^2 (m+1)\right )\right ) (a+b \sin (c+d x))^{m+1}}{8 d \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 0.420683, antiderivative size = 305, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2668, 741, 823, 831, 68} \[ -\frac{\left (3 a^2-3 a b (2-m)+b^2 \left (m^2-4 m+3\right )\right ) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a-b}\right )}{16 d (m+1) (a-b)^3}+\frac{\left (3 a^2+3 a b (2-m)+b^2 \left (m^2-4 m+3\right )\right ) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a+b}\right )}{16 d (m+1) (a+b)^3}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{m+1}}{4 d \left (a^2-b^2\right )}+\frac{\sec ^2(c+d x) \left (a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)+b \left (b^2 (3-m)-a^2 (m+1)\right )\right ) (a+b \sin (c+d x))^{m+1}}{8 d \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 2668
Rule 741
Rule 823
Rule 831
Rule 68
Rubi steps
\begin{align*} \int \sec ^5(c+d x) (a+b \sin (c+d x))^m \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{(a+x)^m}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{4 \left (a^2-b^2\right ) d}+\frac{b^3 \operatorname{Subst}\left (\int \frac{(a+x)^m \left (3 a^2-b^2 (3-m)+a (2-m) x\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac{\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{4 \left (a^2-b^2\right ) d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^{1+m} \left (b \left (b^2 (3-m)-a^2 (1+m)\right )+a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^m \left (-3 a^4+a^2 b^2 \left (6-2 m-m^2\right )-b^4 \left (3-4 m+m^2\right )+a \left (3 a^2-b^2 (5-2 m)\right ) m x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac{\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{4 \left (a^2-b^2\right ) d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^{1+m} \left (b \left (b^2 (3-m)-a^2 (1+m)\right )+a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac{b \operatorname{Subst}\left (\int \left (\frac{\left (a b^2 \left (3 a^2-b^2 (5-2 m)\right ) m+b \left (-3 a^4+a^2 b^2 \left (6-2 m-m^2\right )-b^4 \left (3-4 m+m^2\right )\right )\right ) (a+x)^m}{2 b^2 (b-x)}+\frac{\left (-a b^2 \left (3 a^2-b^2 (5-2 m)\right ) m+b \left (-3 a^4+a^2 b^2 \left (6-2 m-m^2\right )-b^4 \left (3-4 m+m^2\right )\right )\right ) (a+x)^m}{2 b^2 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac{\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{4 \left (a^2-b^2\right ) d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^{1+m} \left (b \left (b^2 (3-m)-a^2 (1+m)\right )+a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac{\left (3 a^2-3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \operatorname{Subst}\left (\int \frac{(a+x)^m}{b+x} \, dx,x,b \sin (c+d x)\right )}{16 (a-b)^2 d}+\frac{\left (3 a^2+3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \operatorname{Subst}\left (\int \frac{(a+x)^m}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 (a+b)^2 d}\\ &=-\frac{\left (3 a^2-3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \, _2F_1\left (1,1+m;2+m;\frac{a+b \sin (c+d x)}{a-b}\right ) (a+b \sin (c+d x))^{1+m}}{16 (a-b)^3 d (1+m)}+\frac{\left (3 a^2+3 a b (2-m)+b^2 \left (3-4 m+m^2\right )\right ) \, _2F_1\left (1,1+m;2+m;\frac{a+b \sin (c+d x)}{a+b}\right ) (a+b \sin (c+d x))^{1+m}}{16 (a+b)^3 d (1+m)}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{4 \left (a^2-b^2\right ) d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^{1+m} \left (b \left (b^2 (3-m)-a^2 (1+m)\right )+a \left (3 a^2-b^2 (5-2 m)\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ \end{align*}
Mathematica [A] time = 4.06833, size = 260, normalized size = 0.85 \[ \frac{(a+b \sin (c+d x))^{m+1} \left (\frac{(a+b)^3 \left (3 a^2+3 a b (m-2)+b^2 \left (m^2-4 m+3\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a-b}\right )-(a-b)^3 \left (3 a^2-3 a b (m-2)+b^2 \left (m^2-4 m+3\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a+b}\right )}{(m+1) (a-b) (a+b) \left (a^2-b^2\right )}+\frac{2 \sec ^2(c+d x) \left (-a \left (3 a^2+b^2 (2 m-5)\right ) \sin (c+d x)+a^2 b (m+1)+b^3 (m-3)\right )}{a^2-b^2}+4 \sec ^4(c+d x) (b-a \sin (c+d x))\right )}{16 d \left (b^2-a^2\right )} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.563, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{5} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{m} \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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{m} \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 )}^{m} \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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