Optimal. Leaf size=104 \[ -\frac {\left (3 a^2-2 a b-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2} d}+\frac {(a-b)^2 \tan (c+d x)}{2 a b^2 d \left (a+b \tan ^2(c+d x)\right )}+\frac {\tan (c+d x)}{b^2 d} \]
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Rubi [A] time = 0.14, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3675, 390, 385, 205} \[ -\frac {\left (3 a^2-2 a b-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2} d}+\frac {(a-b)^2 \tan (c+d x)}{2 a b^2 d \left (a+b \tan ^2(c+d x)\right )}+\frac {\tan (c+d x)}{b^2 d} \]
Antiderivative was successfully verified.
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Rule 205
Rule 385
Rule 390
Rule 3675
Rubi steps
\begin {align*} \int \frac {\sec ^6(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{\left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{b^2}-\frac {a^2-b^2+2 (a-b) b x^2}{b^2 \left (a+b x^2\right )^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\tan (c+d x)}{b^2 d}-\frac {\operatorname {Subst}\left (\int \frac {a^2-b^2+2 (a-b) b x^2}{\left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{b^2 d}\\ &=\frac {\tan (c+d x)}{b^2 d}+\frac {(a-b)^2 \tan (c+d x)}{2 a b^2 d \left (a+b \tan ^2(c+d x)\right )}-\frac {((a-b) (3 a+b)) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{2 a b^2 d}\\ &=-\frac {(a-b) (3 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2} d}+\frac {\tan (c+d x)}{b^2 d}+\frac {(a-b)^2 \tan (c+d x)}{2 a b^2 d \left (a+b \tan ^2(c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.66, size = 104, normalized size = 1.00 \[ \frac {-\frac {(3 a+b) (a-b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2}}+\frac {\sqrt {b} (a-b)^2 \sin (2 (c+d x))}{a ((a-b) \cos (2 (c+d x))+a+b)}+2 \sqrt {b} \tan (c+d x)}{2 b^{5/2} d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.54, size = 479, normalized size = 4.61 \[ \left [\frac {{\left ({\left (3 \, a^{3} - 5 \, a^{2} b + a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} b - 2 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a b} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt {-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 4 \, {\left (2 \, a^{2} b^{2} + {\left (3 \, a^{3} b - 4 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{2} b^{4} d \cos \left (d x + c\right ) + {\left (a^{3} b^{3} - a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{3}\right )}}, \frac {{\left ({\left (3 \, a^{3} - 5 \, a^{2} b + a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} b - 2 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a b} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 2 \, {\left (2 \, a^{2} b^{2} + {\left (3 \, a^{3} b - 4 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} b^{4} d \cos \left (d x + c\right ) + {\left (a^{3} b^{3} - a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{3}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.44, size = 128, normalized size = 1.23 \[ \frac {\frac {2 \, \tan \left (d x + c\right )}{b^{2}} - \frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} {\left (3 \, a^{2} - 2 \, a b - b^{2}\right )}}{\sqrt {a b} a b^{2}} + \frac {a^{2} \tan \left (d x + c\right ) - 2 \, a b \tan \left (d x + c\right ) + b^{2} \tan \left (d x + c\right )}{{\left (b \tan \left (d x + c\right )^{2} + a\right )} a b^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.65, size = 181, normalized size = 1.74 \[ \frac {\tan \left (d x +c \right )}{b^{2} d}+\frac {a \tan \left (d x +c \right )}{2 d \,b^{2} \left (a +b \left (\tan ^{2}\left (d x +c \right )\right )\right )}-\frac {\tan \left (d x +c \right )}{d b \left (a +b \left (\tan ^{2}\left (d x +c \right )\right )\right )}+\frac {\tan \left (d x +c \right )}{2 a d \left (a +b \left (\tan ^{2}\left (d x +c \right )\right )\right )}-\frac {3 a \arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d \,b^{2} \sqrt {a b}}+\frac {\arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{d b \sqrt {a b}}+\frac {\arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d a \sqrt {a b}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.94, size = 100, normalized size = 0.96 \[ \frac {\frac {{\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )}{a b^{3} \tan \left (d x + c\right )^{2} + a^{2} b^{2}} + \frac {2 \, \tan \left (d x + c\right )}{b^{2}} - \frac {{\left (3 \, a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a b^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 12.41, size = 119, normalized size = 1.14 \[ \frac {\mathrm {tan}\left (c+d\,x\right )}{b^2\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,a\,d\,\left (b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\,b^2\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )\,\left (a-b\right )\,\left (3\,a+b\right )}{\sqrt {a}\,\left (-3\,a^2+2\,a\,b+b^2\right )}\right )\,\left (a-b\right )\,\left (3\,a+b\right )}{2\,a^{3/2}\,b^{5/2}\,d} \]
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 \[ \int \frac {\sec ^{6}{\left (c + d x \right )}}{\left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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