Optimal. Leaf size=351 \[ \frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}-\frac {\sqrt {2} (a+b) \left (2 a^2+b^2 (2+n)^2\right ) F_1\left (\frac {1}{2};\frac {1}{2},-1-n;\frac {3}{2};\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n}}{b^3 d (2+n) (3+n) \sqrt {1+\sin (c+d x)}}+\frac {\sqrt {2} a \left (2 a^2+b^2 \left (4+5 n+n^2\right )\right ) F_1\left (\frac {1}{2};\frac {1}{2},-n;\frac {3}{2};\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n}}{b^3 d (2+n) (3+n) \sqrt {1+\sin (c+d x)}} \]
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Rubi [A]
time = 0.33, antiderivative size = 351, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2872, 3102,
2835, 2744, 144, 143} \begin {gather*} \frac {\sqrt {2} a \left (2 a^2+b^2 \left (n^2+5 n+4\right )\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac {1}{2};\frac {1}{2},-n;\frac {3}{2};\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right )}{b^3 d (n+2) (n+3) \sqrt {\sin (c+d x)+1}}-\frac {\sqrt {2} (a+b) \left (2 a^2+b^2 (n+2)^2\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac {1}{2};\frac {1}{2},-n-1;\frac {3}{2};\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right )}{b^3 d (n+2) (n+3) \sqrt {\sin (c+d x)+1}}+\frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{n+1}}{b^2 d (n+2) (n+3)}-\frac {\sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^{n+1}}{b d (n+3)} \end {gather*}
Antiderivative was successfully verified.
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Rule 143
Rule 144
Rule 2744
Rule 2835
Rule 2872
Rule 3102
Rubi steps
\begin {align*} \int \sin ^3(c+d x) (a+b \sin (c+d x))^n \, dx &=-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}+\frac {\int (a+b \sin (c+d x))^n \left (a+b (2+n) \sin (c+d x)-2 a \sin ^2(c+d x)\right ) \, dx}{b (3+n)}\\ &=\frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}+\frac {\int (a+b \sin (c+d x))^n \left (-a b n+\left (2 a^2+b^2 (2+n)^2\right ) \sin (c+d x)\right ) \, dx}{b^2 (2+n) (3+n)}\\ &=\frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}+\frac {\left (2 a^2+b^2 (2+n)^2\right ) \int (a+b \sin (c+d x))^{1+n} \, dx}{b^3 (2+n) (3+n)}-\frac {\left (a \left (2 a^2+b^2 \left (4+5 n+n^2\right )\right )\right ) \int (a+b \sin (c+d x))^n \, dx}{b^3 (2+n) (3+n)}\\ &=\frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}+\frac {\left (\left (2 a^2+b^2 (2+n)^2\right ) \cos (c+d x)\right ) \text {Subst}\left (\int \frac {(a+b x)^{1+n}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (c+d x)\right )}{b^3 d (2+n) (3+n) \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}}-\frac {\left (a \left (2 a^2+b^2 \left (4+5 n+n^2\right )\right ) \cos (c+d x)\right ) \text {Subst}\left (\int \frac {(a+b x)^n}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (c+d x)\right )}{b^3 d (2+n) (3+n) \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}}\\ &=\frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}-\frac {\left ((-a-b) \left (2 a^2+b^2 (2+n)^2\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (-\frac {a+b \sin (c+d x)}{-a-b}\right )^{-n}\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{1+n}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (c+d x)\right )}{b^3 d (2+n) (3+n) \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}}-\frac {\left (a \left (2 a^2+b^2 \left (4+5 n+n^2\right )\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (-\frac {a+b \sin (c+d x)}{-a-b}\right )^{-n}\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^n}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (c+d x)\right )}{b^3 d (2+n) (3+n) \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}}\\ &=\frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}-\frac {\sqrt {2} (a+b) \left (2 a^2+b^2 (2+n)^2\right ) F_1\left (\frac {1}{2};\frac {1}{2},-1-n;\frac {3}{2};\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n}}{b^3 d (2+n) (3+n) \sqrt {1+\sin (c+d x)}}+\frac {\sqrt {2} a \left (2 a^2+b^2 \left (4+5 n+n^2\right )\right ) F_1\left (\frac {1}{2};\frac {1}{2},-n;\frac {3}{2};\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n}}{b^3 d (2+n) (3+n) \sqrt {1+\sin (c+d x)}}\\ \end {align*}
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Mathematica [F]
time = 2.66, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^3(c+d x) (a+b \sin (c+d x))^n \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.49, size = 0, normalized size = 0.00 \[\int \left (\sin ^{3}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{n}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\sin \left (c+d\,x\right )}^3\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^n \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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