Optimal. Leaf size=556 \[ -\frac {\left (2 a^2+b^2\right ) f x}{2 g}+\frac {\left (2 a^2+b^2\right ) (f+g x)^2}{4 g^2}-\frac {i a b e^{i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right )}{g^2 n}+\frac {i a b e^{-i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )}{g^2 n}-\frac {2^{-2-\frac {1}{n}} b^2 e^{2 i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )}{g^2 n}-\frac {2^{-2-\frac {1}{n}} b^2 e^{-2 i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )}{g^2 n}+\frac {i a b e^{i c} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i d (f+g x)^n\right )}{g^2 n}-\frac {i a b e^{-i c} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i d (f+g x)^n\right )}{g^2 n}+\frac {4^{-1-\frac {1}{n}} b^2 e^{2 i c} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-2 i d (f+g x)^n\right )}{g^2 n}+\frac {4^{-1-\frac {1}{n}} b^2 e^{-2 i c} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},2 i d (f+g x)^n\right )}{g^2 n} \]
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Rubi [A]
time = 0.30, antiderivative size = 556, normalized size of antiderivative = 1.00, number of steps
used = 19, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3514, 3448,
3447, 2239, 3446, 3506, 6, 3505, 2250, 3504} \begin {gather*} \frac {i a b e^{i c} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},-i d (f+g x)^n\right )}{g^2 n}-\frac {i a b e^{i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},-i d (f+g x)^n\right )}{g^2 n}+\frac {i a b e^{-i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},i d (f+g x)^n\right )}{g^2 n}-\frac {i a b e^{-i c} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},i d (f+g x)^n\right )}{g^2 n}+\frac {b^2 e^{2 i c} 4^{-\frac {1}{n}-1} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},-2 i d (f+g x)^n\right )}{g^2 n}-\frac {b^2 e^{2 i c} f 2^{-\frac {1}{n}-2} (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},-2 i d (f+g x)^n\right )}{g^2 n}-\frac {b^2 e^{-2 i c} f 2^{-\frac {1}{n}-2} (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},2 i d (f+g x)^n\right )}{g^2 n}+\frac {b^2 e^{-2 i c} 4^{-\frac {1}{n}-1} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},2 i d (f+g x)^n\right )}{g^2 n}+\frac {\left (2 a^2+b^2\right ) (f+g x)^2}{4 g^2}-\frac {f x \left (2 a^2+b^2\right )}{2 g} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 2239
Rule 2250
Rule 3446
Rule 3447
Rule 3448
Rule 3504
Rule 3505
Rule 3506
Rule 3514
Rubi steps
\begin {align*} \int x \left (a+b \sin \left (c+d (f+g x)^n\right )\right )^2 \, dx &=\frac {\text {Subst}\left (\int \left (-f \left (a+b \sin \left (c+d x^n\right )\right )^2+x \left (a+b \sin \left (c+d x^n\right )\right )^2\right ) \, dx,x,f+g x\right )}{g^2}\\ &=\frac {\text {Subst}\left (\int x \left (a+b \sin \left (c+d x^n\right )\right )^2 \, dx,x,f+g x\right )}{g^2}-\frac {f \text {Subst}\left (\int \left (a+b \sin \left (c+d x^n\right )\right )^2 \, dx,x,f+g x\right )}{g^2}\\ &=\frac {\text {Subst}\left (\int \left (a^2 x+\frac {b^2 x}{2}-\frac {1}{2} b^2 x \cos \left (2 c+2 d x^n\right )+2 a b x \sin \left (c+d x^n\right )\right ) \, dx,x,f+g x\right )}{g^2}-\frac {f \text {Subst}\left (\int \left (a^2+\frac {b^2}{2}-\frac {1}{2} b^2 \cos \left (2 c+2 d x^n\right )+2 a b \sin \left (c+d x^n\right )\right ) \, dx,x,f+g x\right )}{g^2}\\ &=-\frac {\left (2 a^2+b^2\right ) f x}{2 g}+\frac {\text {Subst}\left (\int \left (\left (a^2+\frac {b^2}{2}\right ) x-\frac {1}{2} b^2 x \cos \left (2 c+2 d x^n\right )+2 a b x \sin \left (c+d x^n\right )\right ) \, dx,x,f+g x\right )}{g^2}-\frac {(2 a b f) \text {Subst}\left (\int \sin \left (c+d x^n\right ) \, dx,x,f+g x\right )}{g^2}+\frac {\left (b^2 f\right ) \text {Subst}\left (\int \cos \left (2 c+2 d x^n\right ) \, dx,x,f+g x\right )}{2 g^2}\\ &=-\frac {\left (2 a^2+b^2\right ) f x}{2 g}+\frac {\left (2 a^2+b^2\right ) (f+g x)^2}{4 g^2}+\frac {(2 a b) \text {Subst}\left (\int x \sin \left (c+d x^n\right ) \, dx,x,f+g x\right )}{g^2}-\frac {b^2 \text {Subst}\left (\int x \cos \left (2 c+2 d x^n\right ) \, dx,x,f+g x\right )}{2 g^2}-\frac {(i a b f) \text {Subst}\left (\int e^{-i c-i d x^n} \, dx,x,f+g x\right )}{g^2}+\frac {(i a b f) \text {Subst}\left (\int e^{i c+i d x^n} \, dx,x,f+g x\right )}{g^2}+\frac {\left (b^2 f\right ) \text {Subst}\left (\int e^{-2 i c-2 i d x^n} \, dx,x,f+g x\right )}{4 g^2}+\frac {\left (b^2 f\right ) \text {Subst}\left (\int e^{2 i c+2 i d x^n} \, dx,x,f+g x\right )}{4 g^2}\\ &=-\frac {\left (2 a^2+b^2\right ) f x}{2 g}+\frac {\left (2 a^2+b^2\right ) (f+g x)^2}{4 g^2}-\frac {i a b e^{i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right )}{g^2 n}+\frac {i a b e^{-i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )}{g^2 n}-\frac {2^{-2-\frac {1}{n}} b^2 e^{2 i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )}{g^2 n}-\frac {2^{-2-\frac {1}{n}} b^2 e^{-2 i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )}{g^2 n}+\frac {(i a b) \text {Subst}\left (\int e^{-i c-i d x^n} x \, dx,x,f+g x\right )}{g^2}-\frac {(i a b) \text {Subst}\left (\int e^{i c+i d x^n} x \, dx,x,f+g x\right )}{g^2}-\frac {b^2 \text {Subst}\left (\int e^{-2 i c-2 i d x^n} x \, dx,x,f+g x\right )}{4 g^2}-\frac {b^2 \text {Subst}\left (\int e^{2 i c+2 i d x^n} x \, dx,x,f+g x\right )}{4 g^2}\\ &=-\frac {\left (2 a^2+b^2\right ) f x}{2 g}+\frac {\left (2 a^2+b^2\right ) (f+g x)^2}{4 g^2}-\frac {i a b e^{i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right )}{g^2 n}+\frac {i a b e^{-i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )}{g^2 n}-\frac {2^{-2-\frac {1}{n}} b^2 e^{2 i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )}{g^2 n}-\frac {2^{-2-\frac {1}{n}} b^2 e^{-2 i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )}{g^2 n}+\frac {i a b e^{i c} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i d (f+g x)^n\right )}{g^2 n}-\frac {i a b e^{-i c} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i d (f+g x)^n\right )}{g^2 n}+\frac {4^{-1-\frac {1}{n}} b^2 e^{2 i c} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-2 i d (f+g x)^n\right )}{g^2 n}+\frac {4^{-1-\frac {1}{n}} b^2 e^{-2 i c} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},2 i d (f+g x)^n\right )}{g^2 n}\\ \end {align*}
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Mathematica [A]
time = 27.06, size = 552, normalized size = 0.99 \begin {gather*} \frac {2 a^2 g^2 n x^2+b^2 g^2 n x^2-4 i a b (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i d (f+g x)^n\right ) (\cos (c)-i \sin (c))+4 i a b (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i d (f+g x)^n\right ) (\cos (c)+i \sin (c))+4 a b f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right ) (-i \cos (c)+\sin (c))+4 a b f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right ) (i \cos (c)+\sin (c))-b^2 f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right ) (\cos (c)-i \sin (c))^2 \left (\cosh \left (\frac {\log (2)}{n}\right )-\sinh \left (\frac {\log (2)}{n}\right )\right )-b^2 f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right ) (\cos (c)+i \sin (c))^2 \left (\cosh \left (\frac {\log (2)}{n}\right )-\sinh \left (\frac {\log (2)}{n}\right )\right )+b^2 (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},2 i d (f+g x)^n\right ) (\cos (c)-i \sin (c))^2 \left (\cosh \left (\frac {\log (4)}{n}\right )-\sinh \left (\frac {\log (4)}{n}\right )\right )+b^2 (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-2 i d (f+g x)^n\right ) (\cos (c)+i \sin (c))^2 \left (\cosh \left (\frac {\log (4)}{n}\right )-\sinh \left (\frac {\log (4)}{n}\right )\right )}{4 g^2 n} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int x \left (a +b \sin \left (c +d \left (g x +f \right )^{n}\right )\right )^{2}\, dx\]
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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (a + b \sin {\left (c + d \left (f + g x\right )^{n} \right )}\right )^{2}\, dx \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 x\,{\left (a+b\,\sin \left (c+d\,{\left (f+g\,x\right )}^n\right )\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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