Optimal. Leaf size=110 \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 \sqrt {a} (a+b)^{5/2} d}-\frac {\tan ^3(c+d x)}{4 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}-\frac {3 \tan (c+d x)}{8 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )} \]
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Rubi [A]
time = 0.07, antiderivative size = 110, 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 = {3266, 294, 211}
\begin {gather*} \frac {3 \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 \sqrt {a} d (a+b)^{5/2}}-\frac {3 \tan (c+d x)}{8 d (a+b)^2 \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {\tan ^3(c+d x)}{4 d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 294
Rule 3266
Rubi steps
\begin {align*} \int \frac {\sin ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx &=\frac {\text {Subst}\left (\int \frac {x^4}{\left (a+(a+b) x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {\tan ^3(c+d x)}{4 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}+\frac {3 \text {Subst}\left (\int \frac {x^2}{\left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 (a+b) d}\\ &=-\frac {\tan ^3(c+d x)}{4 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}-\frac {3 \tan (c+d x)}{8 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )}+\frac {3 \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{8 (a+b)^2 d}\\ &=\frac {3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 \sqrt {a} (a+b)^{5/2} d}-\frac {\tan ^3(c+d x)}{4 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}-\frac {3 \tan (c+d x)}{8 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.89, size = 97, normalized size = 0.88 \begin {gather*} \frac {\frac {3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{5/2}}+\frac {(-8 a-5 b+(2 a+5 b) \cos (2 (c+d x))) \sin (2 (c+d x))}{(a+b)^2 (2 a+b-b \cos (2 (c+d x)))^2}}{8 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.24, size = 109, normalized size = 0.99
method | result | size |
derivativedivides | \(\frac {\frac {-\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{8 \left (a +b \right )}-\frac {3 a \tan \left (d x +c \right )}{8 \left (a^{2}+2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+b \left (\tan ^{2}\left (d x +c \right )\right )+a \right )^{2}}+\frac {3 \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}}{d}\) | \(109\) |
default | \(\frac {\frac {-\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{8 \left (a +b \right )}-\frac {3 a \tan \left (d x +c \right )}{8 \left (a^{2}+2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+b \left (\tan ^{2}\left (d x +c \right )\right )+a \right )^{2}}+\frac {3 \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}}{d}\) | \(109\) |
risch | \(\frac {i \left (-8 b \,{\mathrm e}^{6 i \left (d x +c \right )} a^{2}-16 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-5 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+16 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+56 b \,{\mathrm e}^{4 i \left (d x +c \right )} a^{2}+46 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+15 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-8 b \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}-32 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-15 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 a \,b^{2}+5 b^{3}\right )}{4 b^{2} \left (a +b \right )^{2} d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}\) | \(398\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.53, size = 158, normalized size = 1.44 \begin {gather*} -\frac {\frac {5 \, {\left (a + b\right )} \tan \left (d x + c\right )^{3} + 3 \, a \tan \left (d x + c\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} + 2 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (d x + c\right )^{2}} - \frac {3 \, \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a^{2} + 2 \, a b + b^{2}\right )}}}{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. 291 vs.
\(2 (96) = 192\).
time = 0.43, size = 683, normalized size = 6.21 \begin {gather*} \left [-\frac {3 \, {\left (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 )} \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 ({\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 5 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{32 \, {\left ({\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} b + 4 \, a^{4} b^{2} + 6 \, a^{3} b^{3} + 4 \, a^{2} b^{4} + a b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} + 5 \, a^{5} b + 10 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 5 \, a^{2} b^{4} + a b^{5}\right )} d\right )}}, -\frac {3 \, {\left (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 )} \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 ({\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 5 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{16 \, {\left ({\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} b + 4 \, a^{4} b^{2} + 6 \, a^{3} b^{3} + 4 \, a^{2} b^{4} + a b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} + 5 \, a^{5} b + 10 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 5 \, a^{2} b^{4} + a 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 = 0.48, size = 152, normalized size = 1.38 \begin {gather*} \frac {\frac {3 \, {\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^{2} + 2 \, a b + b^{2}\right )} \sqrt {a^{2} + a b}} - \frac {5 \, a \tan \left (d x + c\right )^{3} + 5 \, b \tan \left (d x + c\right )^{3} + 3 \, a \tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{2} {\left (a^{2} + 2 \, a b + b^{2}\right )}}}{8 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 13.69, size = 149, normalized size = 1.35 \begin {gather*} \frac {3\,\mathrm {atan}\left (\frac {3\,\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (\frac {8\,a^2}{3}+\frac {16\,a\,b}{3}+\frac {8\,b^2}{3}\right )}{16\,\sqrt {a}\,{\left (a+b\right )}^{5/2}}\right )}{8\,\sqrt {a}\,d\,{\left (a+b\right )}^{5/2}}-\frac {\frac {5\,{\mathrm {tan}\left (c+d\,x\right )}^3}{8\,\left (a+b\right )}+\frac {3\,a\,\mathrm {tan}\left (c+d\,x\right )}{8\,\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 )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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