Optimal. Leaf size=77 \[ \frac {(a-b)^2 \text {ArcTan}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2} d}-\frac {(a-2 b) \tan (c+d x)}{b^2 d}+\frac {\tan ^3(c+d x)}{3 b d} \]
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Rubi [A]
time = 0.06, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3756, 398, 211}
\begin {gather*} \frac {(a-b)^2 \text {ArcTan}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2} d}-\frac {(a-2 b) \tan (c+d x)}{b^2 d}+\frac {\tan ^3(c+d x)}{3 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 398
Rule 3756
Rubi steps
\begin {align*} \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {a-2 b}{b^2}+\frac {x^2}{b}+\frac {a^2-2 a b+b^2}{b^2 \left (a+b x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {(a-2 b) \tan (c+d x)}{b^2 d}+\frac {\tan ^3(c+d x)}{3 b d}+\frac {(a-b)^2 \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{b^2 d}\\ &=\frac {(a-b)^2 \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2} d}-\frac {(a-2 b) \tan (c+d x)}{b^2 d}+\frac {\tan ^3(c+d x)}{3 b d}\\ \end {align*}
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Mathematica [A]
time = 0.23, size = 74, normalized size = 0.96 \begin {gather*} \frac {\frac {3 (a-b)^2 \text {ArcTan}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a}}+\sqrt {b} \left (-3 a+5 b+b \sec ^2(c+d x)\right ) \tan (c+d x)}{3 b^{5/2} d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.35, size = 74, normalized size = 0.96
method | result | size |
derivativedivides | \(\frac {-\frac {-\frac {b \left (\tan ^{3}\left (d x +c \right )\right )}{3}+a \tan \left (d x +c \right )-2 b \tan \left (d x +c \right )}{b^{2}}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}}{d}\) | \(74\) |
default | \(\frac {-\frac {-\frac {b \left (\tan ^{3}\left (d x +c \right )\right )}{3}+a \tan \left (d x +c \right )-2 b \tan \left (d x +c \right )}{b^{2}}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}}{d}\) | \(74\) |
risch | \(-\frac {2 i \left (3 a \,{\mathrm e}^{4 i \left (d x +c \right )}-3 b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 b \,{\mathrm e}^{2 i \left (d x +c \right )}+3 a -5 b \right )}{3 d \,b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a^{2}}{2 \sqrt {-a b}\, d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a}{\sqrt {-a b}\, d b}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\sqrt {-a b}\, \left (a -b \right )}\right ) a^{2}}{2 \sqrt {-a b}\, d \,b^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\sqrt {-a b}\, \left (a -b \right )}\right ) a}{\sqrt {-a b}\, d b}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\sqrt {-a b}\, \left (a -b \right )}\right )}{2 \sqrt {-a b}\, d}\) | \(446\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 69, normalized size = 0.90 \begin {gather*} \frac {\frac {3 \, {\left (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} b^{2}} + \frac {b \tan \left (d x + c\right )^{3} - 3 \, {\left (a - 2 \, b\right )} \tan \left (d x + c\right )}{b^{2}}}{3 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.93, size = 339, normalized size = 4.40 \begin {gather*} \left [-\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {-a b} \cos \left (d x + c\right )^{3} \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 (a b^{2} - {\left (3 \, a^{2} b - 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, a b^{3} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\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 ) \cos \left (d x + c\right )^{3} - 2 \, {\left (a b^{2} - {\left (3 \, a^{2} b - 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, a b^{3} d \cos \left (d x + c\right )^{3}}\right ] \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 \frac {\sec ^{6}{\left (c + d x \right )}}{a + b \tan ^{2}{\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 = 0.65, size = 96, normalized size = 1.25 \begin {gather*} \frac {\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} {\left (a^{2} - 2 \, a b + b^{2}\right )}}{\sqrt {a b} b^{2}} + \frac {b^{2} \tan \left (d x + c\right )^{3} - 3 \, a b \tan \left (d x + c\right ) + 6 \, b^{2} \tan \left (d x + c\right )}{b^{3}}}{3 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 12.23, size = 90, normalized size = 1.17 \begin {gather*} \frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,b\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {a}{b^2}-\frac {2}{b}\right )}{d}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )\,{\left (a-b\right )}^2}{\sqrt {a}\,\left (a^2-2\,a\,b+b^2\right )}\right )\,{\left (a-b\right )}^2}{\sqrt {a}\,b^{5/2}\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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