Optimal. Leaf size=140 \[ -\frac {E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {F\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f} \]
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Rubi [A]
time = 0.11, antiderivative size = 180, normalized size of antiderivative = 1.29, number of steps
used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3271, 425, 21,
434, 437, 435, 432, 430} \begin {gather*} \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{f (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {\tan (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f (a+b)} \end {gather*}
Antiderivative was successfully verified.
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Rule 21
Rule 425
Rule 430
Rule 432
Rule 434
Rule 435
Rule 437
Rule 3271
Rubi steps
\begin {align*} \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx &=\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^{3/2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {b-b x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f}\\ &=\frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f}+\frac {\left (b \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {1-x^2}}{\sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f}\\ &=\frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f}\\ &=\frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a+b \sin ^2(e+f x)}}\\ &=-\frac {\sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f}\\ \end {align*}
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Mathematica [A]
time = 0.44, size = 141, normalized size = 1.01 \begin {gather*} \frac {-2 a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+2 (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )+\sqrt {2} (2 a+b-b \cos (2 (e+f x))) \tan (e+f x)}{2 (a+b) f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 13.32, size = 278, normalized size = 1.99
method | result | size |
default | \(\frac {\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )+b \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) b +a \sin \left (f x +e \right )+b \sin \left (f x +e \right )\right )}{\left (a +b \right ) \sqrt {-\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(278\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] Result contains complex when optimal does not.
time = 0.15, size = 632, normalized size = 4.51 \begin {gather*} -\frac {2 \, {\left (-2 i \, a - i \, b\right )} \sqrt {-b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} \cos \left (f x + e\right ) F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + 2 \, {\left (2 i \, a + i \, b\right )} \sqrt {-b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} \cos \left (f x + e\right ) F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (2 i \, \sqrt {-b} b \sqrt {\frac {a^{2} + a b}{b^{2}}} \cos \left (f x + e\right ) + {\left (2 i \, a + i \, b\right )} \sqrt {-b} \cos \left (f x + e\right )\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (-2 i \, \sqrt {-b} b \sqrt {\frac {a^{2} + a b}{b^{2}}} \cos \left (f x + e\right ) + {\left (-2 i \, a - i \, b\right )} \sqrt {-b} \cos \left (f x + e\right )\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \sin \left (f x + e\right )}{2 \, {\left (a b + b^{2}\right )} f \cos \left (f x + e\right )} \end {gather*}
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 \begin {gather*} \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\cos \left (e+f\,x\right )}^2\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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