Optimal. Leaf size=370 \[ -\frac {b^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 d^2 \sqrt {a^2-b^2}}+\frac {b^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 d^2 \sqrt {a^2-b^2}}-\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 d \sqrt {a^2-b^2}}+\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a^2 d \sqrt {a^2-b^2}}-\frac {i b f \text {Li}_2\left (-e^{i (c+d x)}\right )}{a^2 d^2}+\frac {i b f \text {Li}_2\left (e^{i (c+d x)}\right )}{a^2 d^2}+\frac {2 b (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}+\frac {f \log (\sin (c+d x))}{a d^2}-\frac {(e+f x) \cot (c+d x)}{a d} \]
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Rubi [A] time = 0.62, antiderivative size = 370, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {4535, 4184, 3475, 4183, 2279, 2391, 3323, 2264, 2190} \[ -\frac {b^2 f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 d^2 \sqrt {a^2-b^2}}+\frac {b^2 f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a^2 d^2 \sqrt {a^2-b^2}}-\frac {i b f \text {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a^2 d^2}+\frac {i b f \text {PolyLog}\left (2,e^{i (c+d x)}\right )}{a^2 d^2}-\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 d \sqrt {a^2-b^2}}+\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a^2 d \sqrt {a^2-b^2}}+\frac {2 b (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}+\frac {f \log (\sin (c+d x))}{a d^2}-\frac {(e+f x) \cot (c+d x)}{a d} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2279
Rule 2391
Rule 3323
Rule 3475
Rule 4183
Rule 4184
Rule 4535
Rubi steps
\begin {align*} \int \frac {(e+f x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x) \csc ^2(c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=-\frac {(e+f x) \cot (c+d x)}{a d}-\frac {b \int (e+f x) \csc (c+d x) \, dx}{a^2}+\frac {b^2 \int \frac {e+f x}{a+b \sin (c+d x)} \, dx}{a^2}+\frac {f \int \cot (c+d x) \, dx}{a d}\\ &=\frac {2 b (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x) \cot (c+d x)}{a d}+\frac {f \log (\sin (c+d x))}{a d^2}+\frac {\left (2 b^2\right ) \int \frac {e^{i (c+d x)} (e+f x)}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{a^2}+\frac {(b f) \int \log \left (1-e^{i (c+d x)}\right ) \, dx}{a^2 d}-\frac {(b f) \int \log \left (1+e^{i (c+d x)}\right ) \, dx}{a^2 d}\\ &=\frac {2 b (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x) \cot (c+d x)}{a d}+\frac {f \log (\sin (c+d x))}{a d^2}-\frac {\left (2 i b^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{a^2 \sqrt {a^2-b^2}}+\frac {\left (2 i b^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{a^2 \sqrt {a^2-b^2}}-\frac {(i b f) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a^2 d^2}+\frac {(i b f) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a^2 d^2}\\ &=\frac {2 b (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x) \cot (c+d x)}{a d}-\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {f \log (\sin (c+d x))}{a d^2}-\frac {i b f \text {Li}_2\left (-e^{i (c+d x)}\right )}{a^2 d^2}+\frac {i b f \text {Li}_2\left (e^{i (c+d x)}\right )}{a^2 d^2}+\frac {\left (i b^2 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{a^2 \sqrt {a^2-b^2} d}-\frac {\left (i b^2 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{a^2 \sqrt {a^2-b^2} d}\\ &=\frac {2 b (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x) \cot (c+d x)}{a d}-\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {f \log (\sin (c+d x))}{a d^2}-\frac {i b f \text {Li}_2\left (-e^{i (c+d x)}\right )}{a^2 d^2}+\frac {i b f \text {Li}_2\left (e^{i (c+d x)}\right )}{a^2 d^2}+\frac {\left (b^2 f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a^2 \sqrt {a^2-b^2} d^2}-\frac {\left (b^2 f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a^2 \sqrt {a^2-b^2} d^2}\\ &=\frac {2 b (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x) \cot (c+d x)}{a d}-\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {f \log (\sin (c+d x))}{a d^2}-\frac {i b f \text {Li}_2\left (-e^{i (c+d x)}\right )}{a^2 d^2}+\frac {i b f \text {Li}_2\left (e^{i (c+d x)}\right )}{a^2 d^2}-\frac {b^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^2}+\frac {b^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^2}\\ \end {align*}
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Mathematica [B] time = 11.37, size = 933, normalized size = 2.52 \[ \frac {(d e+d f x) \left (\frac {2 (d e-c f) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {i f \left (\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt {b^2-a^2}}{-i a+b+\sqrt {b^2-a^2}}\right )+\text {Li}_2\left (\frac {a \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a+i \left (b+\sqrt {b^2-a^2}\right )}\right )\right )}{\sqrt {b^2-a^2}}+\frac {i f \left (\log \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right ) \log \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt {b^2-a^2}}{i a+b+\sqrt {b^2-a^2}}\right )+\text {Li}_2\left (\frac {a \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right )}{a-i \left (b+\sqrt {b^2-a^2}\right )}\right )\right )}{\sqrt {b^2-a^2}}+\frac {i f \left (\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (-\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )-\sqrt {b^2-a^2}}{i a-b+\sqrt {b^2-a^2}}\right )+\text {Li}_2\left (\frac {a \left (\tan \left (\frac {1}{2} (c+d x)\right )+i\right )}{i a-b+\sqrt {b^2-a^2}}\right )\right )}{\sqrt {b^2-a^2}}-\frac {i f \left (\log \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right ) \log \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )-\sqrt {b^2-a^2}}{i a+b-\sqrt {b^2-a^2}}\right )+\text {Li}_2\left (\frac {i \tan \left (\frac {1}{2} (c+d x)\right ) a+a}{a+i \left (\sqrt {b^2-a^2}-b\right )}\right )\right )}{\sqrt {b^2-a^2}}\right ) b^2}{a^2 d^2 \left (d e-c f+i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )-i f \log \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}-\frac {e \log \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ) b}{a^2 d}+\frac {c f \log \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ) b}{a^2 d^2}-\frac {f \left ((c+d x) \left (\log \left (1-e^{i (c+d x)}\right )-\log \left (1+e^{i (c+d x)}\right )\right )+i \left (\text {Li}_2\left (-e^{i (c+d x)}\right )-\text {Li}_2\left (e^{i (c+d x)}\right )\right )\right ) b}{a^2 d^2}+\frac {\left (-d e \cos \left (\frac {1}{2} (c+d x)\right )+c f \cos \left (\frac {1}{2} (c+d x)\right )-f (c+d x) \cos \left (\frac {1}{2} (c+d x)\right )\right ) \csc \left (\frac {1}{2} (c+d x)\right )}{2 a d^2}+\frac {f \log (\sin (c+d x))}{a d^2}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (d e \sin \left (\frac {1}{2} (c+d x)\right )-c f \sin \left (\frac {1}{2} (c+d x)\right )+f (c+d x) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a d^2} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.76, size = 1700, normalized size = 4.59 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [B] time = 0.29, size = 766, normalized size = 2.07 \[ -\frac {i b f \dilog \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a^{2} d^{2}}-\frac {2 i b^{2} f c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{2} d^{2} \sqrt {-a^{2}+b^{2}}}-\frac {i b^{2} f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{a^{2} d^{2} \sqrt {-a^{2}+b^{2}}}+\frac {b f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) x}{a^{2} d}-\frac {2 i \left (f x +e \right )}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {2 i b^{2} e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{2} d \sqrt {-a^{2}+b^{2}}}+\frac {b^{2} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{a^{2} d \sqrt {-a^{2}+b^{2}}}+\frac {b^{2} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) c}{a^{2} d^{2} \sqrt {-a^{2}+b^{2}}}+\frac {i b^{2} f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{a^{2} d^{2} \sqrt {-a^{2}+b^{2}}}+\frac {b e \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d}-\frac {b e \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {b f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{2} d^{2}}-\frac {2 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}-\frac {b^{2} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{a^{2} d \sqrt {-a^{2}+b^{2}}}-\frac {b^{2} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{a^{2} d^{2} \sqrt {-a^{2}+b^{2}}}-\frac {i b f \dilog \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d^{2}}+\frac {f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a \,d^{2}}+\frac {f \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a \,d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
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 \[ \int \frac {\left (e + f x\right ) \csc ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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