Optimal. Leaf size=158 \[ \frac {2}{3} a x^{3/2}-\frac {4 b \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 b \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 i b \sqrt {x} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b \sqrt {x} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d} \]
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Rubi [A] time = 0.13, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {14, 4204, 4181, 2531, 2282, 6589} \[ \frac {4 i b \sqrt {x} \text {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b \sqrt {x} \text {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 b \text {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 b \text {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2}{3} a x^{3/2}-\frac {4 i b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 14
Rule 2282
Rule 2531
Rule 4181
Rule 4204
Rule 6589
Rubi steps
\begin {align*} \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx &=\int \left (a \sqrt {x}+b \sqrt {x} \sec \left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {2}{3} a x^{3/2}+b \int \sqrt {x} \sec \left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {2}{3} a x^{3/2}+(2 b) \operatorname {Subst}\left (\int x^2 \sec (c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{3} a x^{3/2}-\frac {4 i b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {(4 b) \operatorname {Subst}\left (\int x \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(4 b) \operatorname {Subst}\left (\int x \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {2}{3} a x^{3/2}-\frac {4 i b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 i b \sqrt {x} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b \sqrt {x} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(4 i b) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {(4 i b) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {2}{3} a x^{3/2}-\frac {4 i b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 i b \sqrt {x} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b \sqrt {x} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}\\ &=\frac {2}{3} a x^{3/2}-\frac {4 i b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 i b \sqrt {x} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b \sqrt {x} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 b \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 b \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}\\ \end {align*}
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Mathematica [A] time = 0.14, size = 155, normalized size = 0.98 \[ \frac {2 \left (a d^3 x^{3/2}-6 i b d^2 x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )+6 i b d \sqrt {x} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )-6 i b d \sqrt {x} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-6 b \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+6 b \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )\right )}{3 d^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b \sqrt {x} \sec \left (d \sqrt {x} + c\right ) + a \sqrt {x}, x\right ) \]
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 \[ \int {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )} \sqrt {x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.18, size = 0, normalized size = 0.00 \[ \int \left (a +b \sec \left (c +d \sqrt {x}\right )\right ) \sqrt {x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.98, size = 374, normalized size = 2.37 \[ \frac {2 \, {\left (d \sqrt {x} + c\right )}^{3} a - 6 \, {\left (d \sqrt {x} + c\right )}^{2} a c + 6 \, {\left (d \sqrt {x} + c\right )} a c^{2} + 6 \, b c^{2} \log \left (\sec \left (d \sqrt {x} + c\right ) + \tan \left (d \sqrt {x} + c\right )\right ) + 3 \, {\left (-2 i \, {\left (d \sqrt {x} + c\right )}^{2} b + 4 i \, {\left (d \sqrt {x} + c\right )} b c\right )} \arctan \left (\cos \left (d \sqrt {x} + c\right ), \sin \left (d \sqrt {x} + c\right ) + 1\right ) + 3 \, {\left (-2 i \, {\left (d \sqrt {x} + c\right )}^{2} b + 4 i \, {\left (d \sqrt {x} + c\right )} b c\right )} \arctan \left (\cos \left (d \sqrt {x} + c\right ), -\sin \left (d \sqrt {x} + c\right ) + 1\right ) + 3 \, {\left (-4 i \, {\left (d \sqrt {x} + c\right )} b + 4 i \, b c\right )} {\rm Li}_2\left (i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) + 3 \, {\left (4 i \, {\left (d \sqrt {x} + c\right )} b - 4 i \, b c\right )} {\rm Li}_2\left (-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) + 3 \, {\left ({\left (d \sqrt {x} + c\right )}^{2} b - 2 \, {\left (d \sqrt {x} + c\right )} b c\right )} \log \left (\cos \left (d \sqrt {x} + c\right )^{2} + \sin \left (d \sqrt {x} + c\right )^{2} + 2 \, \sin \left (d \sqrt {x} + c\right ) + 1\right ) - 3 \, {\left ({\left (d \sqrt {x} + c\right )}^{2} b - 2 \, {\left (d \sqrt {x} + c\right )} b c\right )} \log \left (\cos \left (d \sqrt {x} + c\right )^{2} + \sin \left (d \sqrt {x} + c\right )^{2} - 2 \, \sin \left (d \sqrt {x} + c\right ) + 1\right ) + 12 \, b {\rm Li}_{3}(i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) - 12 \, b {\rm Li}_{3}(-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )})}{3 \, d^{3}} \]
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 \[ \int \sqrt {x}\,\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right ) \,d x \]
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 \[ \int \sqrt {x} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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