Optimal. Leaf size=123 \[ -\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right )}+\frac {b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac {(a+2 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac {(a-2 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]
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Rubi [A] time = 0.16, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2668, 741, 801} \[ \frac {b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right )}-\frac {(a+2 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac {(a-2 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]
Antiderivative was successfully verified.
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Rule 741
Rule 801
Rule 2668
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d}+\frac {b \operatorname {Subst}\left (\int \frac {a^2-2 b^2+a x}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d}+\frac {b \operatorname {Subst}\left (\int \left (\frac {(a-b) (a+2 b)}{2 b (a+b) (b-x)}+\frac {2 b^2}{(a-b) (a+b) (a+x)}+\frac {(a-2 b) (a+b)}{2 (a-b) b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\frac {(a+2 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {(a-2 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac {b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d}\\ \end {align*}
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Mathematica [A] time = 0.58, size = 170, normalized size = 1.38 \[ \frac {\frac {4 b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {1}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {1}{(a-b) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {2 (a+2 b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{(a+b)^2}+\frac {2 (a-2 b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{(a-b)^2}}{4 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 153, normalized size = 1.24 \[ \frac {4 \, b^{3} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.43, size = 177, normalized size = 1.44 \[ \frac {\frac {4 \, b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} + \frac {{\left (a - 2 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a + 2 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (b^{3} \sin \left (d x + c\right )^{2} - a^{3} \sin \left (d x + c\right ) + a b^{2} \sin \left (d x + c\right ) + a^{2} b - 2 \, b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 164, normalized size = 1.33 \[ -\frac {1}{d \left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a}{4 d \left (a +b \right )^{2}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) b}{2 d \left (a +b \right )^{2}}+\frac {b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{d \left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) a}{4 d \left (a -b \right )^{2}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right ) b}{2 d \left (a -b \right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 139, normalized size = 1.13 \[ \frac {\frac {4 \, b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a - 2 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a + 2 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (a \sin \left (d x + c\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.39, size = 148, normalized size = 1.20 \[ \frac {\frac {b}{2\,\left (a^2-b^2\right )}-\frac {a\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {1}{4\,\left (a+b\right )}\right )}{d}+\frac {b^3\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-2\,b\right )}{4\,d\,{\left (a-b\right )}^2} \]
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 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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