Optimal. Leaf size=75 \[ \frac {a b \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{3 d}+\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d} \]
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Rubi [A]
time = 0.07, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2770, 2748,
3852, 8} \begin {gather*} \frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{3 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 2770
Rule 3852
Rubi steps
\begin {align*} \int \sec ^4(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{3 d}-\frac {1}{3} \int \sec ^2(c+d x) \left (-2 a^2+b^2-a b \sin (c+d x)\right ) \, dx\\ &=\frac {a b \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{3 d}-\frac {1}{3} \left (-2 a^2+b^2\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a b \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{3 d}-\frac {\left (2 a^2-b^2\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {a b \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{3 d}+\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A]
time = 0.34, size = 105, normalized size = 1.40 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (8 a b+3 \left (2 a^2+b^2\right ) \sin (c+d x)+\left (2 a^2-b^2\right ) \sin (3 (c+d x))\right )}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (-1+\sin (c+d x))^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.34, size = 62, normalized size = 0.83
method | result | size |
derivativedivides | \(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {2 a b}{3 \cos \left (d x +c \right )^{3}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}}{d}\) | \(62\) |
default | \(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {2 a b}{3 \cos \left (d x +c \right )^{3}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}}{d}\) | \(62\) |
risch | \(-\frac {2 i \left (8 i a b \,{\mathrm e}^{3 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 a^{2}+b^{2}\right )}{3 d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{3}}\) | \(71\) |
norman | \(\frac {-\frac {4 a b}{3 d}-\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 \left (a^{2}+b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 \left (a^{2}+b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 \left (a^{2}+4 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\) | \(219\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 51, normalized size = 0.68 \begin {gather*} \frac {b^{2} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + \frac {2 \, a b}{\cos \left (d x + c\right )^{3}}}{3 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.32, size = 52, normalized size = 0.69 \begin {gather*} \frac {2 \, a b + {\left ({\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \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 \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sec ^{4}{\left (c + d x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 4.61, size = 102, normalized size = 1.36 \begin {gather*} -\frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a b\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.26, size = 71, normalized size = 0.95 \begin {gather*} \frac {\frac {2\,a\,b}{3}+\frac {a^2\,\sin \left (c+d\,x\right )}{3}+\frac {b^2\,\sin \left (c+d\,x\right )}{3}+{\cos \left (c+d\,x\right )}^2\,\left (\frac {2\,a^2\,\sin \left (c+d\,x\right )}{3}-\frac {b^2\,\sin \left (c+d\,x\right )}{3}\right )}{d\,{\cos \left (c+d\,x\right )}^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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