Optimal. Leaf size=97 \[ -\frac {a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{7/2} d}+\frac {a^2 \tan (c+d x)}{(a+b)^3 d}-\frac {a \tan ^3(c+d x)}{3 (a+b)^2 d}+\frac {\tan ^5(c+d x)}{5 (a+b) d} \]
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Rubi [A]
time = 0.07, antiderivative size = 97, 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 = {3274, 308, 211}
\begin {gather*} -\frac {a^{5/2} \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{d (a+b)^{7/2}}+\frac {a^2 \tan (c+d x)}{d (a+b)^3}+\frac {\tan ^5(c+d x)}{5 d (a+b)}-\frac {a \tan ^3(c+d x)}{3 d (a+b)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 308
Rule 3274
Rubi steps
\begin {align*} \int \frac {\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac {\text {Subst}\left (\int \frac {x^6}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {a^2}{(a+b)^3}-\frac {a x^2}{(a+b)^2}+\frac {x^4}{a+b}-\frac {a^3}{(a+b)^3 \left (a+(a+b) x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {a^2 \tan (c+d x)}{(a+b)^3 d}-\frac {a \tan ^3(c+d x)}{3 (a+b)^2 d}+\frac {\tan ^5(c+d x)}{5 (a+b) d}-\frac {a^3 \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{(a+b)^3 d}\\ &=-\frac {a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{7/2} d}+\frac {a^2 \tan (c+d x)}{(a+b)^3 d}-\frac {a \tan ^3(c+d x)}{3 (a+b)^2 d}+\frac {\tan ^5(c+d x)}{5 (a+b) d}\\ \end {align*}
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Mathematica [A]
time = 0.54, size = 111, normalized size = 1.14 \begin {gather*} \frac {-15 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )+\sqrt {a+b} \left (23 a^2+11 a b+3 b^2-\left (11 a^2+17 a b+6 b^2\right ) \sec ^2(c+d x)+3 (a+b)^2 \sec ^4(c+d x)\right ) \tan (c+d x)}{15 (a+b)^{7/2} d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.55, size = 121, normalized size = 1.25
method | result | size |
derivativedivides | \(\frac {\frac {\frac {a^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {2 a b \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {b^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {a b \left (\tan ^{3}\left (d x +c \right )\right )}{3}+a^{2} \tan \left (d x +c \right )}{\left (a +b \right )^{3}}-\frac {a^{3} \arctan \left (\frac {\tan \left (d x +c \right ) \left (a +b \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{3} \sqrt {a \left (a +b \right )}}}{d}\) | \(121\) |
default | \(\frac {\frac {\frac {a^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {2 a b \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {b^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {a b \left (\tan ^{3}\left (d x +c \right )\right )}{3}+a^{2} \tan \left (d x +c \right )}{\left (a +b \right )^{3}}-\frac {a^{3} \arctan \left (\frac {\tan \left (d x +c \right ) \left (a +b \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{3} \sqrt {a \left (a +b \right )}}}{d}\) | \(121\) |
risch | \(\frac {2 i \left (45 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+45 b \,{\mathrm e}^{8 i \left (d x +c \right )} a +15 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+90 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+30 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+140 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+80 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+30 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+70 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+10 b \,{\mathrm e}^{2 i \left (d x +c \right )} a +23 a^{2}+11 a b +3 b^{2}\right )}{15 d \left (a +b \right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {\sqrt {-a \left (a +b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right )^{4} d}+\frac {\sqrt {-a \left (a +b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right )^{4} d}\) | \(284\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 130, normalized size = 1.34 \begin {gather*} -\frac {\frac {15 \, a^{3} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{5} - 5 \, {\left (a^{2} + a b\right )} \tan \left (d x + c\right )^{3} + 15 \, a^{2} \tan \left (d x + c\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{15 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 181 vs.
\(2 (85) = 170\).
time = 0.43, size = 472, normalized size = 4.87 \begin {gather*} \left [\frac {15 \, a^{2} \sqrt {-\frac {a}{a + b}} \cos \left (d x + c\right )^{5} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) + 4 \, {\left ({\left (23 \, a^{2} + 11 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - {\left (11 \, a^{2} + 17 \, a b + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cos \left (d x + c\right )^{5}}, \frac {15 \, a^{2} \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{5} + 2 \, {\left ({\left (23 \, a^{2} + 11 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - {\left (11 \, a^{2} + 17 \, a b + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cos \left (d x + c\right )^{5}}\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 {\tan ^{6}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 296 vs.
\(2 (85) = 170\).
time = 2.19, size = 296, normalized size = 3.05 \begin {gather*} -\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} a^{3}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a^{2} + a b}} - \frac {3 \, a^{4} \tan \left (d x + c\right )^{5} + 12 \, a^{3} b \tan \left (d x + c\right )^{5} + 18 \, a^{2} b^{2} \tan \left (d x + c\right )^{5} + 12 \, a b^{3} \tan \left (d x + c\right )^{5} + 3 \, b^{4} \tan \left (d x + c\right )^{5} - 5 \, a^{4} \tan \left (d x + c\right )^{3} - 15 \, a^{3} b \tan \left (d x + c\right )^{3} - 15 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} - 5 \, a b^{3} \tan \left (d x + c\right )^{3} + 15 \, a^{4} \tan \left (d x + c\right ) + 30 \, a^{3} b \tan \left (d x + c\right ) + 15 \, a^{2} b^{2} \tan \left (d x + c\right )}{a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}}}{15 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 15.11, size = 112, normalized size = 1.15 \begin {gather*} \frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,d\,\left (a+b\right )}-\frac {a^{5/2}\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{7/2}}\right )}{d\,{\left (a+b\right )}^{7/2}}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d\,{\left (a+b\right )}^2}+\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )}{d\,{\left (a+b\right )}^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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