Optimal. Leaf size=227 \[ -\frac {\sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}+\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} f}+\frac {\sqrt {d} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{8 \sqrt {2} f}-\frac {\sqrt {d} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{8 \sqrt {2} f}+\frac {\cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 d f} \]
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Rubi [A] time = 0.17, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2607, 290, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {\sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}+\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} f}+\frac {\sqrt {d} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{8 \sqrt {2} f}-\frac {\sqrt {d} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{8 \sqrt {2} f}+\frac {\cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 d f} \]
Antiderivative was successfully verified.
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Rule 204
Rule 290
Rule 297
Rule 329
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 2607
Rubi steps
\begin {align*} \int \cos ^2(e+f x) \sqrt {d \tan (e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {d x}}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 d f}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {d x}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {\cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 d f}+\frac {\operatorname {Subst}\left (\int \frac {x^2}{1+\frac {x^4}{d^2}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 d f}\\ &=\frac {\cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 d f}-\frac {\operatorname {Subst}\left (\int \frac {d-x^2}{1+\frac {x^4}{d^2}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 d f}+\frac {\operatorname {Subst}\left (\int \frac {d+x^2}{1+\frac {x^4}{d^2}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 d f}\\ &=\frac {\cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 d f}+\frac {\sqrt {d} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}+\frac {\sqrt {d} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}+\frac {d \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 f}+\frac {d \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 f}\\ &=\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}-\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}+\frac {\cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 d f}+\frac {\sqrt {d} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}-\frac {\sqrt {d} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}\\ &=-\frac {\sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}+\frac {\sqrt {d} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}+\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}-\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}+\frac {\cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 d f}\\ \end {align*}
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Mathematica [A] time = 0.19, size = 102, normalized size = 0.45 \[ -\frac {\sqrt {\sin (2 (e+f x))} \sqrt {d \tan (e+f x)} \left (-2 \sqrt {\sin (2 (e+f x))}+\csc (e+f x) \sin ^{-1}(\cos (e+f x)-\sin (e+f x))+\csc (e+f x) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{8 f} \]
Antiderivative was successfully verified.
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fricas [B] time = 110.91, size = 1897, normalized size = 8.36 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.56, size = 227, normalized size = 1.00 \[ \frac {\frac {8 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )}{{\left (d^{2} \tan \left (f x + e\right )^{2} + d^{2}\right )} f} + \frac {2 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{f} + \frac {2 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{f} - \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{f} + \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{f}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.56, size = 522, normalized size = 2.30 \[ \frac {\left (-1+\cos \left (f x +e \right )\right ) \left (i \EllipticPi \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}-i \EllipticPi \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}+\EllipticPi \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}+\EllipticPi \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}+2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}-2 \cos \left (f x +e \right ) \sqrt {2}\right ) \left (1+\cos \left (f x +e \right )\right )^{2} \sqrt {\frac {d \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {2}}{8 f \sin \left (f x +e \right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.75, size = 193, normalized size = 0.85 \[ \frac {d^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + \frac {8 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{2}}{d^{2} \tan \left (f x + e\right )^{2} + d^{2}}}{16 \, d f} \]
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 \[ \int {\cos \left (e+f\,x\right )}^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )} \,d x \]
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 \[ \int \sqrt {d \tan {\left (e + f x \right )}} \cos ^{2}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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