Optimal. Leaf size=177 \[ \frac {a \tan ^3(c+d x)}{3 d \left (a^2-b^2\right )}-\frac {b \sec ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac {a^2 b \sec (c+d x)}{d \left (a^2-b^2\right )^2}+\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )}+\frac {2 a^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac {a^3 \tan (c+d x)}{d \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 0.23, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2727, 2607, 30, 2606, 3767, 8, 2660, 618, 204} \[ \frac {2 a^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\frac {a \tan ^3(c+d x)}{3 d \left (a^2-b^2\right )}-\frac {a^3 \tan (c+d x)}{d \left (a^2-b^2\right )^2}-\frac {b \sec ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac {a^2 b \sec (c+d x)}{d \left (a^2-b^2\right )^2}+\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 204
Rule 618
Rule 2606
Rule 2607
Rule 2660
Rule 2727
Rule 3767
Rubi steps
\begin {align*} \int \frac {\tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {a \int \sec ^2(c+d x) \tan ^2(c+d x) \, dx}{a^2-b^2}-\frac {a^2 \int \frac {\tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{a^2-b^2}-\frac {b \int \sec (c+d x) \tan ^3(c+d x) \, dx}{a^2-b^2}\\ &=-\frac {a^3 \int \sec ^2(c+d x) \, dx}{\left (a^2-b^2\right )^2}+\frac {a^4 \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}+\frac {\left (a^2 b\right ) \int \sec (c+d x) \tan (c+d x) \, dx}{\left (a^2-b^2\right )^2}+\frac {a \operatorname {Subst}\left (\int x^2 \, dx,x,\tan (c+d x)\right )}{\left (a^2-b^2\right ) d}-\frac {b \operatorname {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{\left (a^2-b^2\right ) d}\\ &=\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {a^3 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {\left (2 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}+\frac {\left (a^2 b\right ) \operatorname {Subst}(\int 1 \, dx,x,\sec (c+d x))}{\left (a^2-b^2\right )^2 d}\\ &=\frac {a^2 b \sec (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {a^3 \tan (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {\left (4 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=\frac {2 a^4 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}+\frac {a^2 b \sec (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {a^3 \tan (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}\\ \end {align*}
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Mathematica [A] time = 1.42, size = 195, normalized size = 1.10 \[ \frac {\frac {48 a^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\sec ^3(c+d x) \left (8 a^3 \sin (3 (c+d x))+3 b \left (11 a^2-5 b^2\right ) \cos (c+d x)+12 b \left (b^2-2 a^2\right ) \cos (2 (c+d x))+11 a^2 b \cos (3 (c+d x))-16 a^2 b+6 a b^2 \sin (c+d x)-2 a b^2 \sin (3 (c+d x))-5 b^3 \cos (3 (c+d x))+4 b^3\right )}{(a-b)^2 (a+b)^2}}{24 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.74, size = 476, normalized size = 2.69 \[ \left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} a^{4} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - 6 \, {\left (2 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (4 \, a^{5} - 5 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, \sqrt {a^{2} - b^{2}} a^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + a^{4} b - 2 \, a^{2} b^{3} + b^{5} - 3 \, {\left (2 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (4 \, a^{5} - 5 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.27, size = 241, normalized size = 1.36 \[ \frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{4}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{2} b + 2 \, b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.39, size = 269, normalized size = 1.52 \[ -\frac {32}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (32 a +32 b \right )}-\frac {16}{d \left (32 a +32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b}{2 d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {32}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (32 a -32 b \right )}+\frac {16}{d \left (32 a -32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {a}{d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b}{2 d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}} \]
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 [B] time = 17.43, size = 372, normalized size = 2.10 \[ \frac {\frac {2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,\left (5\,a^2\,b-2\,b^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a\,b^2-5\,a^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2\,b-b^3\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {2\,a^4\,\mathrm {atan}\left (\frac {\frac {a^4\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {2\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{2\,a^4}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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