Optimal. Leaf size=206 \[ \frac {5 b (2 a+b) \sin (c+d x) \cos (c+d x)}{24 a^2 d (a+b)^2 \left (a+b \sin ^2(c+d x)\right )^2}+\frac {(2 a+b) \left (8 a^2+8 a b+5 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{16 a^{7/2} d (a+b)^{7/2}}+\frac {b \left (44 a^2+44 a b+15 b^2\right ) \sin (c+d x) \cos (c+d x)}{48 a^3 d (a+b)^3 \left (a+b \sin ^2(c+d x)\right )}+\frac {b \sin (c+d x) \cos (c+d x)}{6 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^3} \]
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Rubi [A] time = 0.30, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3184, 3173, 12, 3181, 205} \[ \frac {(2 a+b) \left (8 a^2+8 a b+5 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{16 a^{7/2} d (a+b)^{7/2}}+\frac {b \left (44 a^2+44 a b+15 b^2\right ) \sin (c+d x) \cos (c+d x)}{48 a^3 d (a+b)^3 \left (a+b \sin ^2(c+d x)\right )}+\frac {5 b (2 a+b) \sin (c+d x) \cos (c+d x)}{24 a^2 d (a+b)^2 \left (a+b \sin ^2(c+d x)\right )^2}+\frac {b \sin (c+d x) \cos (c+d x)}{6 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 3173
Rule 3181
Rule 3184
Rubi steps
\begin {align*} \int \frac {1}{\left (a+b \sin ^2(c+d x)\right )^4} \, dx &=\frac {b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}-\frac {\int \frac {-6 a-5 b+4 b \sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx}{6 a (a+b)}\\ &=\frac {b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}+\frac {5 b (2 a+b) \cos (c+d x) \sin (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )^2}-\frac {\int \frac {-24 a^2-34 a b-15 b^2+10 b (2 a+b) \sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx}{24 a^2 (a+b)^2}\\ &=\frac {b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}+\frac {5 b (2 a+b) \cos (c+d x) \sin (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )^2}+\frac {b \left (44 a^2+44 a b+15 b^2\right ) \cos (c+d x) \sin (c+d x)}{48 a^3 (a+b)^3 d \left (a+b \sin ^2(c+d x)\right )}-\frac {\int -\frac {3 (2 a+b) \left (8 a^2+8 a b+5 b^2\right )}{a+b \sin ^2(c+d x)} \, dx}{48 a^3 (a+b)^3}\\ &=\frac {b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}+\frac {5 b (2 a+b) \cos (c+d x) \sin (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )^2}+\frac {b \left (44 a^2+44 a b+15 b^2\right ) \cos (c+d x) \sin (c+d x)}{48 a^3 (a+b)^3 d \left (a+b \sin ^2(c+d x)\right )}+\frac {\left ((2 a+b) \left (8 a^2+8 a b+5 b^2\right )\right ) \int \frac {1}{a+b \sin ^2(c+d x)} \, dx}{16 a^3 (a+b)^3}\\ &=\frac {b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}+\frac {5 b (2 a+b) \cos (c+d x) \sin (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )^2}+\frac {b \left (44 a^2+44 a b+15 b^2\right ) \cos (c+d x) \sin (c+d x)}{48 a^3 (a+b)^3 d \left (a+b \sin ^2(c+d x)\right )}+\frac {\left ((2 a+b) \left (8 a^2+8 a b+5 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{16 a^3 (a+b)^3 d}\\ &=\frac {(2 a+b) \left (8 a^2+8 a b+5 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{16 a^{7/2} (a+b)^{7/2} d}+\frac {b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}+\frac {5 b (2 a+b) \cos (c+d x) \sin (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )^2}+\frac {b \left (44 a^2+44 a b+15 b^2\right ) \cos (c+d x) \sin (c+d x)}{48 a^3 (a+b)^3 d \left (a+b \sin ^2(c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 1.47, size = 201, normalized size = 0.98 \[ \frac {\frac {32 a^{5/2} b \sin (2 (c+d x))}{(a+b) (2 a-b \cos (2 (c+d x))+b)^3}+\frac {20 a^{3/2} b (2 a+b) \sin (2 (c+d x))}{(a+b)^2 (2 a-b \cos (2 (c+d x))+b)^2}+\frac {\sqrt {a} b \left (44 a^2+44 a b+15 b^2\right ) \sin (2 (c+d x))}{(a+b)^3 (2 a-b \cos (2 (c+d x))+b)}+\frac {3 \left (16 a^3+24 a^2 b+18 a b^2+5 b^3\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{7/2}}}{48 a^{7/2} d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.54, size = 1361, normalized size = 6.61 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 344, normalized size = 1.67 \[ \frac {\frac {3 \, {\left (16 \, a^{3} + 24 \, a^{2} b + 18 \, a b^{2} + 5 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )}}{{\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \sqrt {a^{2} + a b}} + \frac {72 \, a^{4} b \tan \left (d x + c\right )^{5} + 198 \, a^{3} b^{2} \tan \left (d x + c\right )^{5} + 195 \, a^{2} b^{3} \tan \left (d x + c\right )^{5} + 84 \, a b^{4} \tan \left (d x + c\right )^{5} + 15 \, b^{5} \tan \left (d x + c\right )^{5} + 144 \, a^{4} b \tan \left (d x + c\right )^{3} + 288 \, a^{3} b^{2} \tan \left (d x + c\right )^{3} + 184 \, a^{2} b^{3} \tan \left (d x + c\right )^{3} + 40 \, a b^{4} \tan \left (d x + c\right )^{3} + 72 \, a^{4} b \tan \left (d x + c\right ) + 90 \, a^{3} b^{2} \tan \left (d x + c\right ) + 33 \, a^{2} b^{3} \tan \left (d x + c\right )}{{\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{3}}}{48 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.40, size = 705, normalized size = 3.42 \[ \frac {3 b \left (\tan ^{5}\left (d x +c \right )\right )}{2 d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{3} a \left (a +b \right )}+\frac {9 b^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{8 d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{3} a^{2} \left (a +b \right )}+\frac {5 b^{3} \left (\tan ^{5}\left (d x +c \right )\right )}{16 d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{3} a^{3} \left (a +b \right )}+\frac {3 b \left (\tan ^{3}\left (d x +c \right )\right )}{d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{3} \left (a^{2}+2 a b +b^{2}\right )}+\frac {3 b^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{3} a \left (a^{2}+2 a b +b^{2}\right )}+\frac {5 b^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{6 d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{3} a^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {3 b a \tan \left (d x +c \right )}{2 d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{3} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {15 b^{2} \tan \left (d x +c \right )}{8 d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{3} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {11 b^{3} \tan \left (d x +c \right )}{16 d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{3} a \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {\arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \sqrt {a \left (a +b \right )}}+\frac {3 \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right ) b}{2 d a \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \sqrt {a \left (a +b \right )}}+\frac {9 \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right ) b^{2}}{8 d \,a^{2} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \sqrt {a \left (a +b \right )}}+\frac {5 \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right ) b^{3}}{16 d \,a^{3} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \sqrt {a \left (a +b \right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.67, size = 378, normalized size = 1.83 \[ \frac {\frac {3 \, {\left (16 \, a^{3} + 24 \, a^{2} b + 18 \, a b^{2} + 5 \, b^{3}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {3 \, {\left (24 \, a^{4} b + 66 \, a^{3} b^{2} + 65 \, a^{2} b^{3} + 28 \, a b^{4} + 5 \, b^{5}\right )} \tan \left (d x + c\right )^{5} + 8 \, {\left (18 \, a^{4} b + 36 \, a^{3} b^{2} + 23 \, a^{2} b^{3} + 5 \, a b^{4}\right )} \tan \left (d x + c\right )^{3} + 3 \, {\left (24 \, a^{4} b + 30 \, a^{3} b^{2} + 11 \, a^{2} b^{3}\right )} \tan \left (d x + c\right )}{a^{9} + 3 \, a^{8} b + 3 \, a^{7} b^{2} + a^{6} b^{3} + {\left (a^{9} + 6 \, a^{8} b + 15 \, a^{7} b^{2} + 20 \, a^{6} b^{3} + 15 \, a^{5} b^{4} + 6 \, a^{4} b^{5} + a^{3} b^{6}\right )} \tan \left (d x + c\right )^{6} + 3 \, {\left (a^{9} + 5 \, a^{8} b + 10 \, a^{7} b^{2} + 10 \, a^{6} b^{3} + 5 \, a^{5} b^{4} + a^{4} b^{5}\right )} \tan \left (d x + c\right )^{4} + 3 \, {\left (a^{9} + 4 \, a^{8} b + 6 \, a^{7} b^{2} + 4 \, a^{6} b^{3} + a^{5} b^{4}\right )} \tan \left (d x + c\right )^{2}}}{48 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 15.44, size = 339, normalized size = 1.65 \[ \frac {\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (24\,a^2\,b+30\,a\,b^2+11\,b^3\right )}{16\,a\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (18\,a^2\,b+18\,a\,b^2+5\,b^3\right )}{6\,a^2\,\left (a^2+2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (24\,a^2\,b+18\,a\,b^2+5\,b^3\right )}{16\,a^3\,\left (a+b\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^6\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (3\,a^3+3\,b\,a^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (3\,a^3+6\,a^2\,b+3\,a\,b^2\right )+a^3\right )}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+b\right )\,\left (2\,a+2\,b\right )\,\left (8\,a^2+8\,a\,b+5\,b^2\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{7/2}\,\left (16\,a^3+24\,a^2\,b+18\,a\,b^2+5\,b^3\right )}\right )\,\left (2\,a+b\right )\,\left (8\,a^2+8\,a\,b+5\,b^2\right )}{16\,a^{7/2}\,d\,{\left (a+b\right )}^{7/2}} \]
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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