Optimal. Leaf size=339 \[ -\frac {\left (12 a^2-54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 d (a-b)^{9/2}}+\frac {\left (12 a^2+54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 d (a+b)^{9/2}}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}+\frac {\sec ^2(c+d x) \left (2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)+b \left (3 a^2+11 b^2\right )\right )}{16 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (3 a^4-16 a^2 b^2-127 b^4\right )}{8 d \left (a^2-b^2\right )^4 \sqrt {a+b \sin (c+d x)}}-\frac {b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^{3/2}} \]
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Rubi [A] time = 0.64, antiderivative size = 339, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2668, 741, 823, 829, 827, 1166, 206} \[ -\frac {a b \left (-16 a^2 b^2+3 a^4-127 b^4\right )}{8 d \left (a^2-b^2\right )^4 \sqrt {a+b \sin (c+d x)}}-\frac {b \left (-81 a^2 b^2+18 a^4-77 b^4\right )}{48 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^{3/2}}-\frac {\left (12 a^2-54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 d (a-b)^{9/2}}+\frac {\left (12 a^2+54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 d (a+b)^{9/2}}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}+\frac {\sec ^2(c+d x) \left (2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)+b \left (3 a^2+11 b^2\right )\right )}{16 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 206
Rule 741
Rule 823
Rule 827
Rule 829
Rule 1166
Rule 2668
Rubi steps
\begin {align*} \int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {1}{(a+x)^{5/2} \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac {b^3 \operatorname {Subst}\left (\int \frac {\frac {1}{2} \left (6 a^2-11 b^2\right )+\frac {9 a x}{2}}{(a+x)^{5/2} \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac {\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac {b \operatorname {Subst}\left (\int \frac {\frac {1}{4} \left (-12 a^4+19 a^2 b^2-77 b^4\right )-\frac {5}{2} a \left (3 a^2-10 b^2\right ) x}{(a+x)^{5/2} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac {b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac {\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {b \operatorname {Subst}\left (\int \frac {\frac {1}{4} a \left (12 a^4-49 a^2 b^2+177 b^4\right )+\frac {1}{4} \left (18 a^4-81 a^2 b^2-77 b^4\right ) x}{(a+x)^{3/2} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^3 d}\\ &=-\frac {b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (3 a^4-16 a^2 b^2-127 b^4\right )}{8 \left (a^2-b^2\right )^4 d \sqrt {a+b \sin (c+d x)}}+\frac {\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac {b \operatorname {Subst}\left (\int \frac {\frac {1}{4} \left (-12 a^6+67 a^4 b^2-258 a^2 b^4-77 b^6\right )-\frac {1}{2} a \left (3 a^4-16 a^2 b^2-127 b^4\right ) x}{\sqrt {a+x} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^4 d}\\ &=-\frac {b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (3 a^4-16 a^2 b^2-127 b^4\right )}{8 \left (a^2-b^2\right )^4 d \sqrt {a+b \sin (c+d x)}}+\frac {\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac {b \operatorname {Subst}\left (\int \frac {\frac {1}{2} a^2 \left (3 a^4-16 a^2 b^2-127 b^4\right )+\frac {1}{4} \left (-12 a^6+67 a^4 b^2-258 a^2 b^4-77 b^6\right )-\frac {1}{2} a \left (3 a^4-16 a^2 b^2-127 b^4\right ) x^2}{-a^2+b^2+2 a x^2-x^4} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{4 \left (a^2-b^2\right )^4 d}\\ &=-\frac {b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (3 a^4-16 a^2 b^2-127 b^4\right )}{8 \left (a^2-b^2\right )^4 d \sqrt {a+b \sin (c+d x)}}+\frac {\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac {\left (12 a^2-54 a b+77 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{32 (a-b)^4 d}+\frac {\left (12 a^2+54 a b+77 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{32 (a+b)^4 d}\\ &=-\frac {\left (12 a^2-54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 (a-b)^{9/2} d}+\frac {\left (12 a^2+54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 (a+b)^{9/2} d}-\frac {b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (3 a^4-16 a^2 b^2-127 b^4\right )}{8 \left (a^2-b^2\right )^4 d \sqrt {a+b \sin (c+d x)}}+\frac {\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}\\ \end {align*}
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Mathematica [C] time = 3.43, size = 296, normalized size = 0.87 \[ \frac {-15 a \left (3 a^2-10 b^2\right ) (a+b \sin (c+d x)) \left ((a+b) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \sin (c+d x)}{a-b}\right )+(b-a) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \sin (c+d x)}{a+b}\right )\right )+\frac {1}{2} \left (18 a^4-81 a^2 b^2-77 b^4\right ) \left ((a+b) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {a+b \sin (c+d x)}{a-b}\right )+(b-a) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {a+b \sin (c+d x)}{a+b}\right )\right )-3 (a-b) (a+b) \sec ^2(c+d x) \left (\left (6 a^3-20 a b^2\right ) \sin (c+d x)+3 a^2 b+11 b^3\right )-12 (a-b)^2 (a+b)^2 \sec ^4(c+d x) (a \sin (c+d x)-b)}{48 d \left (a^2-b^2\right )^2 \left (b^2-a^2\right ) (a+b \sin (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 2.36, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{5}}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \]
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 \[ \int \frac {\sec \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.05, size = 682, normalized size = 2.01 \[ \frac {2 b^{5}}{3 d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {12 b^{5} a}{d \left (a -b \right )^{4} \left (a +b \right )^{4} \sqrt {a +b \sin \left (d x +c \right )}}-\frac {3 b \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a}{16 d \left (a -b \right )^{4} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {17 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{32 d \left (a -b \right )^{4} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {3 b \sqrt {a +b \sin \left (d x +c \right )}\, a^{2}}{16 d \left (a -b \right )^{4} \left (b \sin \left (d x +c \right )+b \right )^{2}}-\frac {25 b^{2} \sqrt {a +b \sin \left (d x +c \right )}\, a}{32 d \left (a -b \right )^{4} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {19 b^{3} \sqrt {a +b \sin \left (d x +c \right )}}{32 d \left (a -b \right )^{4} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a^{2}}{8 d \left (a -b \right )^{4} \sqrt {-a +b}}-\frac {27 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{16 d \left (a -b \right )^{4} \sqrt {-a +b}}+\frac {77 b^{2} \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{32 d \left (a -b \right )^{4} \sqrt {-a +b}}-\frac {3 b \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a}{16 d \left (a +b \right )^{4} \left (b \sin \left (d x +c \right )-b \right )^{2}}-\frac {17 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{32 d \left (a +b \right )^{4} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {3 b \sqrt {a +b \sin \left (d x +c \right )}\, a^{2}}{16 d \left (a +b \right )^{4} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {25 b^{2} \sqrt {a +b \sin \left (d x +c \right )}\, a}{32 d \left (a +b \right )^{4} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {19 b^{3} \sqrt {a +b \sin \left (d x +c \right )}}{32 d \left (a +b \right )^{4} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {3 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a^{2}}{8 d \left (a +b \right )^{\frac {9}{2}}}+\frac {27 b \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{16 d \left (a +b \right )^{\frac {9}{2}}}+\frac {77 b^{2} \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{32 d \left (a +b \right )^{\frac {9}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
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 \[ \int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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