Optimal. Leaf size=220 \[ \frac {a x^2}{2}-\frac {12 i b \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {12 i b \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {12 b \sqrt {x} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b \sqrt {x} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d} \]
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Rubi [A] time = 0.19, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {14, 4204, 4181, 2531, 6609, 2282, 6589} \[ \frac {6 i b x \text {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b x \text {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 b \sqrt {x} \text {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b \sqrt {x} \text {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {12 i b \text {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {12 i b \text {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {a x^2}{2}-\frac {4 i b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 14
Rule 2282
Rule 2531
Rule 4181
Rule 4204
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx &=\int \left (a x+b x \sec \left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {a x^2}{2}+b \int x \sec \left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {a x^2}{2}+(2 b) \operatorname {Subst}\left (\int x^3 \sec (c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a x^2}{2}-\frac {4 i b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {(6 b) \operatorname {Subst}\left (\int x^2 \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(6 b) \operatorname {Subst}\left (\int x^2 \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {a x^2}{2}-\frac {4 i b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 i b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(12 i b) \operatorname {Subst}\left (\int x \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {(12 i b) \operatorname {Subst}\left (\int x \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {a x^2}{2}-\frac {4 i b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 i b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 b \sqrt {x} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b \sqrt {x} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(12 b) \operatorname {Subst}\left (\int \text {Li}_3\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {(12 b) \operatorname {Subst}\left (\int \text {Li}_3\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {a x^2}{2}-\frac {4 i b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 i b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 b \sqrt {x} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b \sqrt {x} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {(12 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {(12 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}\\ &=\frac {a x^2}{2}-\frac {4 i b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 i b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 b \sqrt {x} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b \sqrt {x} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {12 i b \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {12 i b \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 223, normalized size = 1.01 \[ \frac {a x^2}{2}-\frac {12 i b \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {12 i b \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {12 b \sqrt {x} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b \sqrt {x} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b x^{3/2} \tan ^{-1}\left (e^{i c+i d \sqrt {x}}\right )}{d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x \sec \left (d \sqrt {x} + c\right ) + a x, x\right ) \]
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 \[ \int {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )} x\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.15, size = 0, normalized size = 0.00 \[ \int x \left (a +b \sec \left (c +d \sqrt {x}\right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.89, size = 540, normalized size = 2.45 \[ \frac {{\left (d \sqrt {x} + c\right )}^{4} a - 4 \, {\left (d \sqrt {x} + c\right )}^{3} a c + 6 \, {\left (d \sqrt {x} + c\right )}^{2} a c^{2} - 4 \, {\left (d \sqrt {x} + c\right )} a c^{3} - 4 \, b c^{3} \log \left (\sec \left (d \sqrt {x} + c\right ) + \tan \left (d \sqrt {x} + c\right )\right ) + 2 \, {\left (-2 i \, {\left (d \sqrt {x} + c\right )}^{3} b + 6 i \, {\left (d \sqrt {x} + c\right )}^{2} b c - 6 i \, {\left (d \sqrt {x} + c\right )} b c^{2}\right )} \arctan \left (\cos \left (d \sqrt {x} + c\right ), \sin \left (d \sqrt {x} + c\right ) + 1\right ) + 2 \, {\left (-2 i \, {\left (d \sqrt {x} + c\right )}^{3} b + 6 i \, {\left (d \sqrt {x} + c\right )}^{2} b c - 6 i \, {\left (d \sqrt {x} + c\right )} b c^{2}\right )} \arctan \left (\cos \left (d \sqrt {x} + c\right ), -\sin \left (d \sqrt {x} + c\right ) + 1\right ) + 2 \, {\left (-6 i \, {\left (d \sqrt {x} + c\right )}^{2} b + 12 i \, {\left (d \sqrt {x} + c\right )} b c - 6 i \, b c^{2}\right )} {\rm Li}_2\left (i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) + 2 \, {\left (6 i \, {\left (d \sqrt {x} + c\right )}^{2} b - 12 i \, {\left (d \sqrt {x} + c\right )} b c + 6 i \, b c^{2}\right )} {\rm Li}_2\left (-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) + 2 \, {\left ({\left (d \sqrt {x} + c\right )}^{3} b - 3 \, {\left (d \sqrt {x} + c\right )}^{2} b c + 3 \, {\left (d \sqrt {x} + c\right )} b c^{2}\right )} \log \left (\cos \left (d \sqrt {x} + c\right )^{2} + \sin \left (d \sqrt {x} + c\right )^{2} + 2 \, \sin \left (d \sqrt {x} + c\right ) + 1\right ) - 2 \, {\left ({\left (d \sqrt {x} + c\right )}^{3} b - 3 \, {\left (d \sqrt {x} + c\right )}^{2} b c + 3 \, {\left (d \sqrt {x} + c\right )} b c^{2}\right )} \log \left (\cos \left (d \sqrt {x} + c\right )^{2} + \sin \left (d \sqrt {x} + c\right )^{2} - 2 \, \sin \left (d \sqrt {x} + c\right ) + 1\right ) + 24 i \, b {\rm Li}_{4}(i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) - 24 i \, b {\rm Li}_{4}(-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) + 24 \, {\left ({\left (d \sqrt {x} + c\right )} b - b c\right )} {\rm Li}_{3}(i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) - 24 \, {\left ({\left (d \sqrt {x} + c\right )} b - b c\right )} {\rm Li}_{3}(-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )})}{2 \, d^{4}} \]
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 \[ \int x\,\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right ) \,d x \]
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 \[ \int x \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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