Optimal. Leaf size=135 \[ \frac {5 a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}+\frac {5 a^2 \sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{64 d}+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{7/2}}{6 d}+\frac {5 a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{48 d} \]
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Rubi [A] time = 0.23, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2675, 2667, 63, 206} \[ \frac {5 a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}+\frac {5 a^2 \sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{64 d}+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{7/2}}{6 d}+\frac {5 a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{48 d} \]
Antiderivative was successfully verified.
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Rule 63
Rule 206
Rule 2667
Rule 2675
Rubi steps
\begin {align*} \int \sec ^7(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}+\frac {1}{12} (5 a) \int \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\\ &=\frac {5 a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{48 d}+\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}+\frac {1}{32} \left (5 a^2\right ) \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=\frac {5 a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{64 d}+\frac {5 a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{48 d}+\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}+\frac {1}{128} \left (5 a^3\right ) \int \sec (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=\frac {5 a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{64 d}+\frac {5 a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{48 d}+\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}+\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{128 d}\\ &=\frac {5 a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{64 d}+\frac {5 a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{48 d}+\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}+\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{64 d}\\ &=\frac {5 a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}+\frac {5 a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{64 d}+\frac {5 a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{48 d}+\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}\\ \end {align*}
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Mathematica [A] time = 0.56, size = 120, normalized size = 0.89 \[ -\frac {15 \sqrt {2} a^{7/2} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^6 \tanh ^{-1}\left (\frac {\sqrt {a (\sin (c+d x)+1)}}{\sqrt {2} \sqrt {a}}\right )+2 a^3 \left (15 \sin ^2(c+d x)-50 \sin (c+d x)+67\right ) \sqrt {a (\sin (c+d x)+1)}}{384 d (\sin (c+d x)-1)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 193, normalized size = 1.43 \[ \frac {15 \, {\left (3 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} - 4 \, \sqrt {2} a^{3} - {\left (\sqrt {2} a^{3} \cos \left (d x + c\right )^{2} - 4 \, \sqrt {2} a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, {\left (15 \, a^{3} \cos \left (d x + c\right )^{2} + 50 \, a^{3} \sin \left (d x + c\right ) - 82 \, a^{3}\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{768 \, {\left (3 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [A] time = 0.35, size = 144, normalized size = 1.07 \[ \frac {2 a^{7} \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{12 a \left (a \sin \left (d x +c \right )-a \right )^{3}}-\frac {5 \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{8 a \left (a \sin \left (d x +c \right )-a \right )^{2}}-\frac {3 \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{4 a \left (a \sin \left (d x +c \right )-a \right )}+\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{8 a}\right )}{12 a}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.46, size = 168, normalized size = 1.24 \[ -\frac {15 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{5} - 80 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{6} + 132 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{7}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3} - 6 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a + 12 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{2} - 8 \, a^{3}}}{768 \, a d} \]
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 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\cos \left (c+d\,x\right )}^7} \,d x \]
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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