Optimal. Leaf size=174 \[ -\frac {\cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}-\frac {\sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {(a+b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}} \]
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Rubi [A]
time = 0.11, antiderivative size = 174, normalized size of antiderivative = 1.00, 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 = {3267, 486, 21,
434, 437, 435, 432, 430} \begin {gather*} \frac {(a+b) \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 \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {\cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f} \end {gather*}
Antiderivative was successfully verified.
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Rule 21
Rule 430
Rule 432
Rule 434
Rule 435
Rule 437
Rule 486
Rule 3267
Rubi steps
\begin {align*} \int \csc ^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 {\sqrt {a+b x^2}}{x^2 \sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac {\cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{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 )}{f}\\ &=-\frac {\cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{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 )}{f}\\ &=-\frac {\cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{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 )}{f}+\frac {\left ((a+b) \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 {\cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{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 )}{f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left ((a+b) \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 {\cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}-\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)}}{f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {(a+b) \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)}}\\ \end {align*}
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Mathematica [A]
time = 0.41, size = 137, normalized size = 0.79 \begin {gather*} \frac {-\sqrt {2} (2 a+b-b \cos (2 (e+f x))) \cot (e+f x)-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 )}{2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 6.20, size = 156, normalized size = 0.90
method | result | size |
default | \(\frac {\sin \left (f x +e \right ) \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \left (\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -\EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \right )+b \left (\cos ^{4}\left (f x +e \right )\right )+\left (-a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{\sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(156\) |
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 = 0.13, size = 626, normalized size = 3.60 \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}} 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}}) \sin \left (f x + e\right ) + 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}} 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}}) \sin \left (f x + e\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right ) + {\left (2 i \, \sqrt {-b} b \sqrt {\frac {a^{2} + a b}{b^{2}}} \sin \left (f x + e\right ) + {\left (2 i \, a + i \, b\right )} \sqrt {-b} \sin \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}}} \sin \left (f x + e\right ) + {\left (-2 i \, a - i \, b\right )} \sqrt {-b} \sin \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 \, b f \sin \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 \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \csc ^{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 {\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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