Optimal. Leaf size=425 \[ \frac {g^{5/2} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a b^{3/2} f}-\frac {g^{5/2} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a b^{3/2} f}-\frac {2 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {\cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \]
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Rubi [A]
time = 0.76, antiderivative size = 425, normalized size of antiderivative = 1.00, number of steps
used = 21, number of rules used = 16, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.516, Rules used = {2977,
2645, 327, 335, 304, 209, 212, 2774, 2946, 2721, 2719, 2780, 2886, 2884, 211, 214}
\begin {gather*} -\frac {g^{5/2} \left (b^2-a^2\right )^{3/4} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a b^{3/2} f}+\frac {g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}+\frac {g^{5/2} \left (b^2-a^2\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a b^{3/2} f}+\frac {g^{5/2} \text {ArcTan}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {g^{5/2} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {2 g^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 211
Rule 212
Rule 214
Rule 304
Rule 327
Rule 335
Rule 2645
Rule 2719
Rule 2721
Rule 2774
Rule 2780
Rule 2884
Rule 2886
Rule 2946
Rule 2977
Rubi steps
\begin {align*} \int \frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx &=\int \left (\frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a}-\frac {b (g \cos (e+f x))^{5/2}}{a (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac {\int (g \cos (e+f x))^{5/2} \csc (e+f x) \, dx}{a}-\frac {b \int \frac {(g \cos (e+f x))^{5/2}}{a+b \sin (e+f x)} \, dx}{a}\\ &=-\frac {2 g (g \cos (e+f x))^{3/2}}{3 a f}-\frac {\text {Subst}\left (\int \frac {x^{5/2}}{1-\frac {x^2}{g^2}} \, dx,x,g \cos (e+f x)\right )}{a f g}-\frac {g^2 \int \frac {\sqrt {g \cos (e+f x)} (b+a \sin (e+f x))}{a+b \sin (e+f x)} \, dx}{a}\\ &=-\frac {g \text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{g^2}} \, dx,x,g \cos (e+f x)\right )}{a f}-\frac {g^2 \int \sqrt {g \cos (e+f x)} \, dx}{b}-\frac {\left (\left (-a^2+b^2\right ) g^2\right ) \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx}{a b}\\ &=-\frac {(2 g) \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{g^2}} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}-\frac {\left (\left (a^2-b^2\right ) g^3\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b^2}+\frac {\left (\left (a^2-b^2\right ) g^3\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b^2}+\frac {\left (\left (a^2-b^2\right ) g^3\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) g^2+b^2 x^2} \, dx,x,g \cos (e+f x)\right )}{a f}-\frac {\left (g^2 \sqrt {g \cos (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \, dx}{b \sqrt {\cos (e+f x)}}\\ &=-\frac {2 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {\cos (e+f x)}}-\frac {g^3 \text {Subst}\left (\int \frac {1}{g-x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}+\frac {g^3 \text {Subst}\left (\int \frac {1}{g+x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}+\frac {\left (2 \left (a^2-b^2\right ) g^3\right ) \text {Subst}\left (\int \frac {x^2}{\left (a^2-b^2\right ) g^2+b^2 x^4} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}-\frac {\left (\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b^2 \sqrt {g \cos (e+f x)}}+\frac {\left (\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b^2 \sqrt {g \cos (e+f x)}}\\ &=\frac {g^{5/2} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {g^{5/2} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {2 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {\cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {\left (\left (a^2-b^2\right ) g^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g-b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a b f}+\frac {\left (\left (a^2-b^2\right ) g^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g+b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a b f}\\ &=\frac {g^{5/2} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a b^{3/2} f}-\frac {g^{5/2} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a b^{3/2} f}-\frac {2 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {\cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in
optimal.
time = 43.37, size = 484, normalized size = 1.14 \begin {gather*} \frac {(g \cos (e+f x))^{5/2} \csc (e+f x) \left (8 a b^{5/2} F_1\left (\frac {7}{4};\frac {1}{2},1;\frac {11}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {7}{2}}(e+f x)+7 \left (a^2-b^2\right ) \left (-2 \sqrt {2} \left (a^2-b^2\right )^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \sqrt {2} \left (a^2-b^2\right )^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+4 b^{3/2} \tan ^{-1}\left (\sqrt {\cos (e+f x)}\right )+2 b^{3/2} \log \left (1-\sqrt {\cos (e+f x)}\right )-2 b^{3/2} \log \left (1+\sqrt {\cos (e+f x)}\right )+\sqrt {2} \left (a^2-b^2\right )^{3/4} \log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )-\sqrt {2} \left (a^2-b^2\right )^{3/4} \log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{28 a b^{3/2} \left (a^2-b^2\right ) f \cos ^{\frac {5}{2}}(e+f x) (b+a \csc (e+f x))} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [A]
time = 20.60, size = 257, normalized size = 0.60
method | result | size |
default | \(-\frac {3 g^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}+3 g^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}+4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right ) \sqrt {-g}+8 \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}\, \sqrt {-g}\, g^{2} \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+6 g^{3} \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )-4 g^{2} \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}\, \sqrt {-g}}{6 a \sqrt {-g}\, f}\) | \(257\) |
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 [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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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 [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 = -0.00 \begin {gather*} \int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}}{\sin \left (e+f\,x\right )\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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