Optimal. Leaf size=144 \[ \frac {\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} (a+b)^{5/2} d}+\frac {b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}+\frac {3 b (2 a+b) \cos (c+d x) \sin (c+d x)}{8 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )} \]
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Rubi [A]
time = 0.10, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3263, 3252, 12,
3260, 211} \begin {gather*} \frac {3 b (2 a+b) \sin (c+d x) \cos (c+d x)}{8 a^2 d (a+b)^2 \left (a+b \sin ^2(c+d x)\right )}+\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} d (a+b)^{5/2}}+\frac {b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 211
Rule 3252
Rule 3260
Rule 3263
Rubi steps
\begin {align*} \int \frac {1}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx &=\frac {b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}-\frac {\int \frac {-4 a-3 b+2 b \sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx}{4 a (a+b)}\\ &=\frac {b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}+\frac {3 b (2 a+b) \cos (c+d x) \sin (c+d x)}{8 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}-\frac {\int \frac {-8 a^2-8 a b-3 b^2}{a+b \sin ^2(c+d x)} \, dx}{8 a^2 (a+b)^2}\\ &=\frac {b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}+\frac {3 b (2 a+b) \cos (c+d x) \sin (c+d x)}{8 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}+\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{a+b \sin ^2(c+d x)} \, dx}{8 a^2 (a+b)^2}\\ &=\frac {b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}+\frac {3 b (2 a+b) \cos (c+d x) \sin (c+d x)}{8 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}+\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{8 a^2 (a+b)^2 d}\\ &=\frac {\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} (a+b)^{5/2} d}+\frac {b \cos (c+d x) \sin (c+d x)}{4 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}+\frac {3 b (2 a+b) \cos (c+d x) \sin (c+d x)}{8 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.84, size = 125, normalized size = 0.87 \begin {gather*} \frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{5/2}}+\frac {\sqrt {a} b \left (16 a^2+16 a b+3 b^2-3 b (2 a+b) \cos (2 (c+d x))\right ) \sin (2 (c+d x))}{(a+b)^2 (2 a+b-b \cos (2 (c+d x)))^2}}{8 a^{5/2} d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.38, size = 148, normalized size = 1.03
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\left (8 a +3 b \right ) b \left (\tan ^{3}\left (d x +c \right )\right )}{8 a^{2} \left (a +b \right )}+\frac {b \left (8 a +5 b \right ) \tan \left (d x +c \right )}{8 a \left (a^{2}+2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+b \left (\tan ^{2}\left (d x +c \right )\right )+a \right )^{2}}+\frac {\left (8 a^{2}+8 a b +3 b^{2}\right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 a^{2} \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}}{d}\) | \(148\) |
default | \(\frac {\frac {\frac {\left (8 a +3 b \right ) b \left (\tan ^{3}\left (d x +c \right )\right )}{8 a^{2} \left (a +b \right )}+\frac {b \left (8 a +5 b \right ) \tan \left (d x +c \right )}{8 a \left (a^{2}+2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+b \left (\tan ^{2}\left (d x +c \right )\right )+a \right )^{2}}+\frac {\left (8 a^{2}+8 a b +3 b^{2}\right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 a^{2} \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}}{d}\) | \(148\) |
risch | \(-\frac {i \left (-8 b \,{\mathrm e}^{6 i \left (d x +c \right )} a^{2}-8 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-3 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+48 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+72 b \,{\mathrm e}^{4 i \left (d x +c \right )} a^{2}+42 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-40 b \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}-40 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-9 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+6 a \,b^{2}+3 b^{3}\right )}{4 a^{2} \left (a +b \right )^{2} d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d a}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b^{2}}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d a}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b^{2}}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d \,a^{2}}\) | \(782\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.55, size = 211, normalized size = 1.47 \begin {gather*} \frac {\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {{\left (8 \, a^{2} b + 11 \, a b^{2} + 3 \, b^{3}\right )} \tan \left (d x + c\right )^{3} + {\left (8 \, a^{2} b + 5 \, a b^{2}\right )} \tan \left (d x + c\right )}{a^{6} + 2 \, a^{5} b + a^{4} b^{2} + {\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \tan \left (d x + c\right )^{2}}}{8 \, 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. 371 vs.
\(2 (130) = 260\).
time = 0.43, size = 843, normalized size = 5.85 \begin {gather*} \left [-\frac {{\left ({\left (8 \, a^{2} b^{2} + 8 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} + 24 \, a^{3} b + 27 \, a^{2} b^{2} + 14 \, a b^{3} + 3 \, b^{4} - 2 \, {\left (8 \, a^{3} b + 16 \, a^{2} b^{2} + 11 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a^{2} - a b} \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 + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - 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 (3 \, {\left (2 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (8 \, a^{4} b + 19 \, a^{3} b^{2} + 14 \, a^{2} b^{3} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{32 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{7} b + 4 \, a^{6} b^{2} + 6 \, a^{5} b^{3} + 4 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} + 5 \, a^{7} b + 10 \, a^{6} b^{2} + 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} + a^{3} b^{5}\right )} d\right )}}, -\frac {{\left ({\left (8 \, a^{2} b^{2} + 8 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} + 24 \, a^{3} b + 27 \, a^{2} b^{2} + 14 \, a b^{3} + 3 \, b^{4} - 2 \, {\left (8 \, a^{3} b + 16 \, a^{2} b^{2} + 11 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 2 \, {\left (3 \, {\left (2 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (8 \, a^{4} b + 19 \, a^{3} b^{2} + 14 \, a^{2} b^{3} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{16 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{7} b + 4 \, a^{6} b^{2} + 6 \, a^{5} b^{3} + 4 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} + 5 \, a^{7} b + 10 \, a^{6} b^{2} + 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} + a^{3} b^{5}\right )} d\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.45, size = 211, normalized size = 1.47 \begin {gather*} \frac {\frac {{\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 (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )}}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {a^{2} + a b}} + \frac {8 \, a^{2} b \tan \left (d x + c\right )^{3} + 11 \, a b^{2} \tan \left (d x + c\right )^{3} + 3 \, b^{3} \tan \left (d x + c\right )^{3} + 8 \, a^{2} b \tan \left (d x + c\right ) + 5 \, a b^{2} \tan \left (d x + c\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{2}}}{8 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 13.82, size = 176, normalized size = 1.22 \begin {gather*} \frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,b^2+8\,a\,b\right )}{8\,a^2\,\left (a+b\right )}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (5\,b^2+8\,a\,b\right )}{8\,a\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^2+2\,a\,b+b^2\right )+a^2+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a^2+2\,b\,a\right )\right )}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{5/2}}\right )\,\left (8\,a^2+8\,a\,b+3\,b^2\right )}{8\,a^{5/2}\,d\,{\left (a+b\right )}^{5/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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