Optimal. Leaf size=255 \[ \frac {2}{3} a^2 x^{3/2}-\frac {8 a b \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 a b \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 i a b \sqrt {x} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 i b^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x}{d} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.32, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4204, 4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391} \[ \frac {8 i a b \sqrt {x} \text {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \text {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 a b \text {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 a b \text {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {2 i b^2 \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x}{d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 2190
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 3719
Rule 4181
Rule 4184
Rule 4190
Rule 4204
Rule 6589
Rubi steps
\begin {align*} \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \operatorname {Subst}\left (\int x^2 (a+b \sec (c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (a^2 x^2+2 a b x^2 \sec (c+d x)+b^2 x^2 \sec ^2(c+d x)\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{3} a^2 x^{3/2}+(4 a b) \operatorname {Subst}\left (\int x^2 \sec (c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \operatorname {Subst}\left (\int x^2 \sec ^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(8 a b) \operatorname {Subst}\left (\int x \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(8 a b) \operatorname {Subst}\left (\int x \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}-\frac {\left (4 b^2\right ) \operatorname {Subst}\left (\int x \tan (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x}{d}+\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i a b \sqrt {x} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(8 i a b) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {(8 i a b) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {\left (8 i b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x}{d}+\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {8 i a b \sqrt {x} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(8 a 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 {(8 a 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 {\left (4 b^2\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 i b^2 x}{d}+\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {8 i a b \sqrt {x} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 a b \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 a b \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}\\ &=-\frac {2 i b^2 x}{d}+\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {8 i a b \sqrt {x} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 i b^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {8 a b \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 a b \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.82, size = 247, normalized size = 0.97 \[ \frac {2 \left (a^2 d^3 x^{3/2}-12 i a b d^2 x \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )+12 i a b d \sqrt {x} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )-12 i a b d \sqrt {x} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-12 a b \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+12 a b \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )+3 b^2 d^2 x \tan \left (c+d \sqrt {x}\right )-3 i b^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )+6 b^2 d \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )-3 i b^2 d^2 x\right )}{3 d^3} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} \sqrt {x} \sec \left (d \sqrt {x} + c\right )^{2} + 2 \, a b \sqrt {x} \sec \left (d \sqrt {x} + c\right ) + a^{2} \sqrt {x}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2} \sqrt {x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 1.29, size = 0, normalized size = 0.00 \[ \int \left (a +b \sec \left (c +d \sqrt {x}\right )\right )^{2} \sqrt {x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 1.00, size = 1279, normalized size = 5.02 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {x}\,{\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________