Optimal. Leaf size=499 \[ -\frac {\cos ^2(c+d x) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}\right ) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{2 \sqrt {b} d \sqrt {a+b \sin ^4(c+d x)}}-\frac {\sqrt [4]{a} \left (\sqrt {a+b}+\sqrt {a}\right ) \cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 b d \sqrt [4]{a+b} \sqrt {a+b \sin ^4(c+d x)}}+\frac {\left (\sqrt {a+b}+\sqrt {a}\right )^2 \cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} \Pi \left (-\frac {\left (\sqrt {a}-\sqrt {a+b}\right )^2}{4 \sqrt {a} \sqrt {a+b}};2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{4 \sqrt [4]{a} b d \sqrt [4]{a+b} \sqrt {a+b \sin ^4(c+d x)}} \]
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Rubi [A] time = 0.66, antiderivative size = 499, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3219, 1319, 1103, 1706} \[ -\frac {\cos ^2(c+d x) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}\right ) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{2 \sqrt {b} d \sqrt {a+b \sin ^4(c+d x)}}-\frac {\sqrt [4]{a} \left (\sqrt {a+b}+\sqrt {a}\right ) \cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 b d \sqrt [4]{a+b} \sqrt {a+b \sin ^4(c+d x)}}+\frac {\left (\sqrt {a+b}+\sqrt {a}\right )^2 \cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} \Pi \left (-\frac {\left (\sqrt {a}-\sqrt {a+b}\right )^2}{4 \sqrt {a} \sqrt {a+b}};2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{4 \sqrt [4]{a} b d \sqrt [4]{a+b} \sqrt {a+b \sin ^4(c+d x)}} \]
Antiderivative was successfully verified.
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Rule 1103
Rule 1319
Rule 1706
Rule 3219
Rubi steps
\begin {align*} \int \frac {\sin ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx &=\frac {\left (\cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^2\right ) \sqrt {a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\\ &=-\frac {\left (a \left (1+\frac {\sqrt {a+b}}{\sqrt {a}}\right ) \cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{b d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\left (a \left (1+\frac {\sqrt {a+b}}{\sqrt {a}}\right ) \cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1+\frac {\sqrt {a+b} x^2}{\sqrt {a}}}{\left (1+x^2\right ) \sqrt {a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{b d \sqrt {a+b \sin ^4(c+d x)}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}}\right ) \cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}}{2 \sqrt {b} d \sqrt {a+b \sin ^4(c+d x)}}-\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {a+b}\right ) \cos ^2(c+d x) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{2 b \sqrt [4]{a+b} d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\left (\sqrt {a}+\sqrt {a+b}\right )^2 \cos ^2(c+d x) \Pi \left (-\frac {\left (\sqrt {a}-\sqrt {a+b}\right )^2}{4 \sqrt {a} \sqrt {a+b}};2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{4 \sqrt [4]{a} b \sqrt [4]{a+b} d \sqrt {a+b \sin ^4(c+d x)}}\\ \end {align*}
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Mathematica [C] time = 2.88, size = 287, normalized size = 0.58 \[ -\frac {2 i \cos ^2(c+d x) \sqrt {1+\left (1+\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \sqrt {2+\left (2-\frac {2 i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \left (F\left (i \sinh ^{-1}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right )|\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )-\Pi \left (\frac {\sqrt {a}}{\sqrt {a}-i \sqrt {b}};i \sinh ^{-1}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right )|\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )\right )}{d \sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \sqrt {8 a-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))+3 b}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\cos \left (d x + c\right )^{2} - 1}{\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 7.36, size = 881, normalized size = 1.77 \[ -\frac {\sqrt {\left (4 a +\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )\right ) \left (\sin ^{2}\left (2 d x +2 c \right )\right )}\, \sqrt {-a b}\, \sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \left (\cos \left (2 d x +2 c \right )+1\right )^{2} \sqrt {\frac {-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \sqrt {\frac {b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \left (\EllipticF \left (\sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}, \sqrt {\frac {b +\sqrt {-a b}}{-b +\sqrt {-a b}}}\right )-2 \EllipticPi \left (\sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}, \frac {\sqrt {-a b}}{-b +\sqrt {-a b}}, \sqrt {\frac {b +\sqrt {-a b}}{-b +\sqrt {-a b}}}\right )\right )}{2 \left (-b +\sqrt {-a b}\right ) \sqrt {\frac {\left (-1+\cos \left (2 d x +2 c \right )\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \left (-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b \right ) \left (b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b \right )}{b}}\, \sin \left (2 d x +2 c \right ) \sqrt {4 a +\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )}\, d}-\frac {\sqrt {\left (4 a +\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )\right ) \left (\sin ^{2}\left (2 d x +2 c \right )\right )}\, \sqrt {-a b}\, \sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \left (\cos \left (2 d x +2 c \right )+1\right )^{2} \sqrt {\frac {-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \sqrt {\frac {b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \EllipticF \left (\sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}, \sqrt {\frac {b +\sqrt {-a b}}{-b +\sqrt {-a b}}}\right )}{2 \left (-b +\sqrt {-a b}\right ) \sqrt {\frac {\left (-1+\cos \left (2 d x +2 c \right )\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \left (-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b \right ) \left (b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b \right )}{b}}\, \sin \left (2 d x +2 c \right ) \sqrt {4 a +\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )}\, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (c+d\,x\right )}^2}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]
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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