Optimal. Leaf size=183 \[ -\frac {\sec ^2(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{m+1}}{2 d \left (a^2-b^2\right )}-\frac {(a-b (1-m)) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a-b}\right )}{4 d (m+1) (a-b)^2}+\frac {(a-b m+b) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a+b}\right )}{4 d (m+1) (a+b)^2} \]
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Rubi [A] time = 0.23, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2668, 741, 831, 68} \[ -\frac {\sec ^2(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^{m+1}}{2 d \left (a^2-b^2\right )}-\frac {(a-b (1-m)) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a-b}\right )}{4 d (m+1) (a-b)^2}+\frac {(a-b m+b) (a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a+b}\right )}{4 d (m+1) (a+b)^2} \]
Antiderivative was successfully verified.
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Rule 68
Rule 741
Rule 831
Rule 2668
Rubi steps
\begin {align*} \int \sec ^3(c+d x) (a+b \sin (c+d x))^m \, dx &=\frac {b^3 \operatorname {Subst}\left (\int \frac {(a+x)^m}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\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}+\frac {b \operatorname {Subst}\left (\int \frac {(a+x)^m \left (a^2-b^2 (1-m)-a m x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\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}+\frac {b \operatorname {Subst}\left (\int \left (\frac {\left (b \left (a^2-b^2 (1-m)\right )-a b^2 m\right ) (a+x)^m}{2 b^2 (b-x)}+\frac {\left (b \left (a^2-b^2 (1-m)\right )+a b^2 m\right ) (a+x)^m}{2 b^2 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\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}+\frac {((a+b) (a-b (1-m))) \operatorname {Subst}\left (\int \frac {(a+x)^m}{b+x} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d}+\frac {(a+b-b m) \operatorname {Subst}\left (\int \frac {(a+x)^m}{b-x} \, dx,x,b \sin (c+d x)\right )}{4 (a+b) d}\\ &=-\frac {(a-b (1-m)) \, _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}}{4 (a-b)^2 d (1+m)}+\frac {(a+b-b m) \, _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}}{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}\\ \end {align*}
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Mathematica [A] time = 0.62, size = 157, normalized size = 0.86 \[ \frac {(a+b \sin (c+d x))^{m+1} \left (\frac {b \left ((a+b)^2 (a+b (m-1)) \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a-b}\right )-(a-b)^2 (a-b m+b) \, _2F_1\left (1,m+1;m+2;\frac {a+b \sin (c+d x)}{a+b}\right )\right )}{(m+1) (a-b) (a+b)}+2 b \sec ^2(c+d x) (b-a \sin (c+d x))\right )}{4 b d \left (b^2-a^2\right )} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{3}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.52, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{3}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m}{{\cos \left (c+d\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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