Optimal. Leaf size=474 \[ \frac {12 a^2 b^2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {3 a^5 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac {a^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {8 a^4 b^2 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {a^4 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}+\frac {4 a^3 b^3 \cos (c+d x)}{d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}-\frac {3 a \cos (c+d x)}{4 d (a+b)^4 (1-\sin (c+d x))}+\frac {3 a \cos (c+d x)}{4 d (a-b)^4 (\sin (c+d x)+1)}+\frac {\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)}+\frac {\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)^2} \]
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Rubi [A] time = 0.87, antiderivative size = 474, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2731, 2650, 2648, 2664, 2754, 12, 2660, 618, 204} \[ \frac {a^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {8 a^4 b^2 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {12 a^2 b^2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {3 a^5 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}+\frac {4 a^3 b^3 \cos (c+d x)}{d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}-\frac {3 a \cos (c+d x)}{4 d (a+b)^4 (1-\sin (c+d x))}+\frac {3 a \cos (c+d x)}{4 d (a-b)^4 (\sin (c+d x)+1)}+\frac {\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)}+\frac {\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 204
Rule 618
Rule 2648
Rule 2650
Rule 2660
Rule 2664
Rule 2731
Rule 2754
Rubi steps
\begin {align*} \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \left (\frac {1}{4 (a+b)^3 (-1+\sin (c+d x))^2}+\frac {3 a}{4 (a+b)^4 (-1+\sin (c+d x))}+\frac {1}{4 (a-b)^3 (1+\sin (c+d x))^2}-\frac {3 a}{4 (a-b)^4 (1+\sin (c+d x))}+\frac {a^4}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^3}+\frac {4 a^3 b^2}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}+\frac {6 a^2 b^2 \left (a^2+b^2\right )}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))}\right ) \, dx\\ &=-\frac {(3 a) \int \frac {1}{1+\sin (c+d x)} \, dx}{4 (a-b)^4}+\frac {\int \frac {1}{(1+\sin (c+d x))^2} \, dx}{4 (a-b)^3}+\frac {(3 a) \int \frac {1}{-1+\sin (c+d x)} \, dx}{4 (a+b)^4}+\frac {\int \frac {1}{(-1+\sin (c+d x))^2} \, dx}{4 (a+b)^3}+\frac {\left (4 a^3 b^2\right ) \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{\left (a^2-b^2\right )^3}+\frac {a^4 \int \frac {1}{(a+b \sin (c+d x))^3} \, dx}{\left (a^2-b^2\right )^2}+\frac {\left (6 a^2 b^2 \left (a^2+b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^4}\\ &=\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {\int \frac {1}{1+\sin (c+d x)} \, dx}{12 (a-b)^3}-\frac {\int \frac {1}{-1+\sin (c+d x)} \, dx}{12 (a+b)^3}+\frac {\left (4 a^3 b^2\right ) \int \frac {a}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^4}-\frac {a^4 \int \frac {-2 a+b \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )^3}+\frac {\left (12 a^2 b^2 \left (a^2+b^2\right )\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 )^4 d}\\ &=\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {a^4 \int \frac {2 a^2+b^2}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^4}+\frac {\left (4 a^4 b^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^4}-\frac {\left (24 a^2 b^2 \left (a^2+b^2\right )\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 )^4 d}\\ &=\frac {12 a^2 b^2 \left (a^2+b^2\right ) \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 )^{9/2} d}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {\left (a^4 \left (2 a^2+b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^4}+\frac {\left (8 a^4 b^2\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 )^4 d}\\ &=\frac {12 a^2 b^2 \left (a^2+b^2\right ) \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 )^{9/2} d}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}-\frac {\left (16 a^4 b^2\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 )^4 d}+\frac {\left (a^4 \left (2 a^2+b^2\right )\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 )^4 d}\\ &=\frac {8 a^4 b^2 \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 )^{9/2} d}+\frac {12 a^2 b^2 \left (a^2+b^2\right ) \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 )^{9/2} d}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}-\frac {\left (2 a^4 \left (2 a^2+b^2\right )\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 )^4 d}\\ &=\frac {8 a^4 b^2 \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 )^{9/2} d}+\frac {12 a^2 b^2 \left (a^2+b^2\right ) \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 )^{9/2} d}+\frac {a^4 \left (2 a^2+b^2\right ) \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 )^{9/2} d}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 1.04, size = 351, normalized size = 0.74 \[ \frac {\frac {96 a^2 \left (2 a^4+21 a^2 b^2+12 b^4\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2}}-\frac {\sec ^3(c+d x) \left (32 a^7 \sin (3 (c+d x))-264 a^6 b+22 a^5 b^2 \sin (c+d x)-91 a^5 b^2 \sin (3 (c+d x))-17 a^5 b^2 \sin (5 (c+d x))-358 a^4 b^3-264 a^3 b^4 \sin (c+d x)-244 a^3 b^4 \sin (3 (c+d x))-76 a^3 b^4 \sin (5 (c+d x))+8 a^2 b^5-2 \left (28 a^6 b+89 a^4 b^3-12 a^2 b^5\right ) \cos (4 (c+d x))-8 \left (44 a^6 b+55 a^4 b^3+8 a^2 b^5-2 b^7\right ) \cos (2 (c+d x))+32 a b^6 \sin (c+d x)-12 a b^6 \sin (3 (c+d x))-12 a b^6 \sin (5 (c+d x))-16 b^7\right )}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))^2}}{96 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 1249, normalized size = 2.64 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.42, size = 632, normalized size = 1.33 \[ \frac {\frac {3 \, {\left (2 \, a^{6} + 21 \, a^{4} b^{2} + 12 \, a^{2} b^{4}\right )} {\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 )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, {\left (5 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 14 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 11 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 22 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{6} b + 7 \, a^{4} b^{3}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}} + \frac {2 \, {\left (3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 56 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, a^{4} b - 20 \, a^{2} b^{3} - b^{5}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.32, size = 922, normalized size = 1.95 \[ \text {result too large to display} \]
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 = 11.22, size = 1099, normalized size = 2.32 \[ \frac {\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (2\,a^6\,b+21\,a^4\,b^3+12\,a^2\,b^5\right )}{a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (14\,a^7+23\,a^5\,b^2+242\,a^3\,b^4+36\,a\,b^6\right )}{3\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (2\,a^5+21\,a^3\,b^2+12\,a\,b^4\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (30\,a^4\,b+71\,a^2\,b^3+4\,b^5\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {a^2\,\left (42\,a^4\,b+61\,a^2\,b^3+2\,b^5\right )}{3\,\left (a^2-b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^6\,b+16\,a^4\,b^3+43\,a^2\,b^5\right )}{3\,\left (a^2-b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-2\,a^7+65\,a^5\,b^2+131\,a^3\,b^4+16\,a\,b^6\right )}{3\,\left (a^2-b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (22\,a^6\,b+233\,a^4\,b^3+153\,a^2\,b^5+12\,b^7\right )}{3\,\left (a^2-b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (2\,a^4+21\,a^2\,b^2+12\,b^4\right )}{a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-6\,a^6+105\,a^4\,b^2+208\,a^2\,b^4+8\,b^6\right )}{3\,\left (a^2-b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2+12\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (2\,a^2+12\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-a^2+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2-4\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (a^2-4\,b^2\right )+8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {\frac {a^2\,\left (2\,a^4+21\,a^2\,b^2+12\,b^4\right )\,\left (2\,a^8\,b-8\,a^6\,b^3+12\,a^4\,b^5-8\,a^2\,b^7+2\,b^9\right )}{2\,{\left (a+b\right )}^{9/2}\,{\left (a-b\right )}^{9/2}}+\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^4+21\,a^2\,b^2+12\,b^4\right )\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )}{{\left (a+b\right )}^{9/2}\,{\left (a-b\right )}^{9/2}}}{2\,a^6+21\,a^4\,b^2+12\,a^2\,b^4}\right )\,\left (2\,a^4+21\,a^2\,b^2+12\,b^4\right )}{d\,{\left (a+b\right )}^{9/2}\,{\left (a-b\right )}^{9/2}} \]
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 \[ \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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