Optimal. Leaf size=1092 \[ \text{result too large to display} \]
[Out]
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Rubi [A] time = 18.8666, antiderivative size = 1092, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {1019, 1066, 1076, 621, 206, 1032, 724} \[ \frac{B \left (c x^2+b x+a\right )^{3/2}}{3 f}-\frac{\left (2 A c f (4 c e-5 b f)-B \left (8 \left (e^2-d f\right ) c^2-2 f (5 b e-4 a f) c+b^2 f^2\right )+2 c f (2 B c e-b B f-2 A c f) x\right ) \sqrt{c x^2+b x+a}}{8 c f^3}+\frac{\left (2 A c f \left (8 \left (e^2-d f\right ) c^2-12 f (b e-a f) c+3 b^2 f^2\right )-B \left (16 \left (e^3-2 d e f\right ) c^3-24 f \left (b e^2-a f e-b d f\right ) c^2+6 b f^2 (b e-2 a f) c+b^3 f^3\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{c x^2+b x+a}}\right )}{16 c^{3/2} f^4}-\frac{\left (2 c f \left (B d (c e-b f) \left (c e^2-b f e+2 a f^2-2 c d f\right )+A f \left (-d \left (e^2-d f\right ) c^2+2 d f (b e-a f) c-f^2 \left (b^2 d-a^2 f\right )\right )\right )-c \left (e-\sqrt{e^2-4 d f}\right ) \left (A f (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+B \left (\left (e^4-3 d f e^2+d^2 f^2\right ) c^2+2 f \left (a f \left (e^2-d f\right )-b \left (e^3-2 d e f\right )\right ) c-f^2 \left (-\left (e^2-d f\right ) b^2+2 a e f b-a^2 f^2\right )\right )\right )\right ) \tanh ^{-1}\left (\frac{4 a f-b \left (e-\sqrt{e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right ) x}{2 \sqrt{2} \sqrt{c e^2-b f e+2 a f^2-2 c d f-(c e-b f) \sqrt{e^2-4 d f}} \sqrt{c x^2+b x+a}}\right )}{\sqrt{2} c f^4 \sqrt{e^2-4 d f} \sqrt{c e^2-b f e+2 a f^2-2 c d f-(c e-b f) \sqrt{e^2-4 d f}}}+\frac{\left (2 f \left (B d (c e-b f) \left (c e^2-b f e+2 a f^2-2 c d f\right )+A f \left (-d \left (e^2-d f\right ) c^2+2 d f (b e-a f) c-f^2 \left (b^2 d-a^2 f\right )\right )\right )-\left (e+\sqrt{e^2-4 d f}\right ) \left (A f (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+B \left (\left (e^4-3 d f e^2+d^2 f^2\right ) c^2+2 f \left (a f \left (e^2-d f\right )-b \left (e^3-2 d e f\right )\right ) c-f^2 \left (-\left (e^2-d f\right ) b^2+2 a e f b-a^2 f^2\right )\right )\right )\right ) \tanh ^{-1}\left (\frac{4 a f-b \left (e+\sqrt{e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt{e^2-4 d f}\right )\right ) x}{2 \sqrt{2} \sqrt{c e^2-b f e+2 a f^2-2 c d f+(c e-b f) \sqrt{e^2-4 d f}} \sqrt{c x^2+b x+a}}\right )}{\sqrt{2} f^4 \sqrt{e^2-4 d f} \sqrt{c e^2-b f e+2 a f^2-2 c d f+(c e-b f) \sqrt{e^2-4 d f}}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 1019
Rule 1066
Rule 1076
Rule 621
Rule 206
Rule 1032
Rule 724
Rubi steps
\begin{align*} \int \frac{(A+B x) \left (a+b x+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx &=\frac{B \left (a+b x+c x^2\right )^{3/2}}{3 f}-\frac{\int \frac{\sqrt{a+b x+c x^2} \left (\frac{3}{2} (b B d-2 a A f)-\frac{3}{2} (2 A b f-B (2 c d+b e-2 a f)) x+\frac{3}{2} (2 B c e-b B f-2 A c f) x^2\right )}{d+e x+f x^2} \, dx}{3 f}\\ &=-\frac{\left (2 A c f (4 c e-5 b f)-B \left (b^2 f^2-2 c f (5 b e-4 a f)+8 c^2 \left (e^2-d f\right )\right )+2 c f (2 B c e-b B f-2 A c f) x\right ) \sqrt{a+b x+c x^2}}{8 c f^3}+\frac{B \left (a+b x+c x^2\right )^{3/2}}{3 f}+\frac{\int \frac{-\frac{3}{8} \left (b^3 B d f^2-10 b^2 c d f (B e-A f)-8 a c f (B c d e-A f (c d-2 a f))-4 b c d \left (2 A c e f-B \left (2 c e^2-2 c d f+5 a f^2\right )\right )\right )+\frac{3}{8} \left (2 A c f \left (8 c^2 d e-4 a c e f-b f (5 b e-16 a f)+4 b c \left (e^2-4 d f\right )\right )-B \left (b^3 e f^2+16 c^3 d \left (e^2-d f\right )+2 c f \left (10 a b e f-8 a^2 f^2-b^2 \left (5 e^2-8 d f\right )\right )-8 c^2 \left (a f \left (e^2-4 d f\right )-b \left (e^3-5 d e f\right )\right )\right )\right ) x+\frac{3}{8} \left (2 A c f \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right )-B \left (b^3 f^3+6 b c f^2 (b e-2 a f)-24 c^2 f \left (b e^2-b d f-a e f\right )+16 c^3 \left (e^3-2 d e f\right )\right )\right ) x^2}{\sqrt{a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{6 c f^3}\\ &=-\frac{\left (2 A c f (4 c e-5 b f)-B \left (b^2 f^2-2 c f (5 b e-4 a f)+8 c^2 \left (e^2-d f\right )\right )+2 c f (2 B c e-b B f-2 A c f) x\right ) \sqrt{a+b x+c x^2}}{8 c f^3}+\frac{B \left (a+b x+c x^2\right )^{3/2}}{3 f}+\frac{\int \frac{-\frac{3}{8} d \left (2 A c f \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right )-B \left (b^3 f^3+6 b c f^2 (b e-2 a f)-24 c^2 f \left (b e^2-b d f-a e f\right )+16 c^3 \left (e^3-2 d e f\right )\right )\right )-\frac{3}{8} f \left (b^3 B d f^2-10 b^2 c d f (B e-A f)-8 a c f (B c d e-A f (c d-2 a f))-4 b c d \left (2 A c e f-B \left (2 c e^2-2 c d f+5 a f^2\right )\right )\right )+\left (-\frac{3}{8} e \left (2 A c f \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right )-B \left (b^3 f^3+6 b c f^2 (b e-2 a f)-24 c^2 f \left (b e^2-b d f-a e f\right )+16 c^3 \left (e^3-2 d e f\right )\right )\right )+\frac{3}{8} f \left (2 A c f \left (8 c^2 d e-4 a c e f-b f (5 b e-16 a f)+4 b c \left (e^2-4 d f\right )\right )-B \left (b^3 e f^2+16 c^3 d \left (e^2-d f\right )+2 c f \left (10 a b e f-8 a^2 f^2-b^2 \left (5 e^2-8 d f\right )\right )-8 c^2 \left (a f \left (e^2-4 d f\right )-b \left (e^3-5 d e f\right )\right )\right )\right )\right ) x}{\sqrt{a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{6 c f^4}+\frac{\left (2 A c f \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right )-B \left (b^3 f^3+6 b c f^2 (b e-2 a f)-24 c^2 f \left (b e^2-b d f-a e f\right )+16 c^3 \left (e^3-2 d e f\right )\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{16 c f^4}\\ &=-\frac{\left (2 A c f (4 c e-5 b f)-B \left (b^2 f^2-2 c f (5 b e-4 a f)+8 c^2 \left (e^2-d f\right )\right )+2 c f (2 B c e-b B f-2 A c f) x\right ) \sqrt{a+b x+c x^2}}{8 c f^3}+\frac{B \left (a+b x+c x^2\right )^{3/2}}{3 f}+\frac{\left (2 A c f \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right )-B \left (b^3 f^3+6 b c f^2 (b e-2 a f)-24 c^2 f \left (b e^2-b d f-a e f\right )+16 c^3 \left (e^3-2 d e f\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{8 c f^4}-\frac{\left (A f \left (c^2 \left (e^4-4 d e^2 f+2 d^2 f^2+e^3 \sqrt{e^2-4 d f}-2 d e f \sqrt{e^2-4 d f}\right )+f^2 \left (2 a^2 f^2-2 a b f \left (e+\sqrt{e^2-4 d f}\right )+b^2 \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )\right )+2 c f \left (a f \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )-b \left (e^3-3 d e f+e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}\right )\right )\right )-B \left (c^2 \left (e^5-5 d e^3 f+5 d^2 e f^2+e^4 \sqrt{e^2-4 d f}-3 d e^2 f \sqrt{e^2-4 d f}+d^2 f^2 \sqrt{e^2-4 d f}\right )+f^2 \left (a^2 f^2 \left (e+\sqrt{e^2-4 d f}\right )-2 a b f \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )+b^2 \left (e^3-3 d e f+e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}\right )\right )+2 c f \left (a f \left (e^3-3 d e f+e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}\right )-b \left (e^4-4 d e^2 f+2 d^2 f^2+e^3 \sqrt{e^2-4 d f}-2 d e f \sqrt{e^2-4 d f}\right )\right )\right )\right ) \int \frac{1}{\left (e+\sqrt{e^2-4 d f}+2 f x\right ) \sqrt{a+b x+c x^2}} \, dx}{f^4 \sqrt{e^2-4 d f}}--\frac{\left (2 f \left (-\frac{3}{8} d \left (2 A c f \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right )-B \left (b^3 f^3+6 b c f^2 (b e-2 a f)-24 c^2 f \left (b e^2-b d f-a e f\right )+16 c^3 \left (e^3-2 d e f\right )\right )\right )-\frac{3}{8} f \left (b^3 B d f^2-10 b^2 c d f (B e-A f)-8 a c f (B c d e-A f (c d-2 a f))-4 b c d \left (2 A c e f-B \left (2 c e^2-2 c d f+5 a f^2\right )\right )\right )\right )-\left (e-\sqrt{e^2-4 d f}\right ) \left (-\frac{3}{8} e \left (2 A c f \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right )-B \left (b^3 f^3+6 b c f^2 (b e-2 a f)-24 c^2 f \left (b e^2-b d f-a e f\right )+16 c^3 \left (e^3-2 d e f\right )\right )\right )+\frac{3}{8} f \left (2 A c f \left (8 c^2 d e-4 a c e f-b f (5 b e-16 a f)+4 b c \left (e^2-4 d f\right )\right )-B \left (b^3 e f^2+16 c^3 d \left (e^2-d f\right )+2 c f \left (10 a b e f-8 a^2 f^2-b^2 \left (5 e^2-8 d f\right )\right )-8 c^2 \left (a f \left (e^2-4 d f\right )-b \left (e^3-5 d e f\right )\right )\right )\right )\right )\right ) \int \frac{1}{\left (e-\sqrt{e^2-4 d f}+2 f x\right ) \sqrt{a+b x+c x^2}} \, dx}{6 c f^4 \sqrt{e^2-4 d f}}\\ &=-\frac{\left (2 A c f (4 c e-5 b f)-B \left (b^2 f^2-2 c f (5 b e-4 a f)+8 c^2 \left (e^2-d f\right )\right )+2 c f (2 B c e-b B f-2 A c f) x\right ) \sqrt{a+b x+c x^2}}{8 c f^3}+\frac{B \left (a+b x+c x^2\right )^{3/2}}{3 f}+\frac{\left (2 A c f \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right )-B \left (b^3 f^3+6 b c f^2 (b e-2 a f)-24 c^2 f \left (b e^2-b d f-a e f\right )+16 c^3 \left (e^3-2 d e f\right )\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{3/2} f^4}+\frac{\left (2 \left (A f \left (c^2 \left (e^4-4 d e^2 f+2 d^2 f^2+e^3 \sqrt{e^2-4 d f}-2 d e f \sqrt{e^2-4 d f}\right )+f^2 \left (2 a^2 f^2-2 a b f \left (e+\sqrt{e^2-4 d f}\right )+b^2 \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )\right )+2 c f \left (a f \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )-b \left (e^3-3 d e f+e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}\right )\right )\right )-B \left (c^2 \left (e^5-5 d e^3 f+5 d^2 e f^2+e^4 \sqrt{e^2-4 d f}-3 d e^2 f \sqrt{e^2-4 d f}+d^2 f^2 \sqrt{e^2-4 d f}\right )+f^2 \left (a^2 f^2 \left (e+\sqrt{e^2-4 d f}\right )-2 a b f \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )+b^2 \left (e^3-3 d e f+e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}\right )\right )+2 c f \left (a f \left (e^3-3 d e f+e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}\right )-b \left (e^4-4 d e^2 f+2 d^2 f^2+e^3 \sqrt{e^2-4 d f}-2 d e f \sqrt{e^2-4 d f}\right )\right )\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{16 a f^2-8 b f \left (e+\sqrt{e^2-4 d f}\right )+4 c \left (e+\sqrt{e^2-4 d f}\right )^2-x^2} \, dx,x,\frac{4 a f-b \left (e+\sqrt{e^2-4 d f}\right )-\left (-2 b f+2 c \left (e+\sqrt{e^2-4 d f}\right )\right ) x}{\sqrt{a+b x+c x^2}}\right )}{f^4 \sqrt{e^2-4 d f}}+-\frac{\left (2 f \left (-\frac{3}{8} d \left (2 A c f \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right )-B \left (b^3 f^3+6 b c f^2 (b e-2 a f)-24 c^2 f \left (b e^2-b d f-a e f\right )+16 c^3 \left (e^3-2 d e f\right )\right )\right )-\frac{3}{8} f \left (b^3 B d f^2-10 b^2 c d f (B e-A f)-8 a c f (B c d e-A f (c d-2 a f))-4 b c d \left (2 A c e f-B \left (2 c e^2-2 c d f+5 a f^2\right )\right )\right )\right )-\left (e-\sqrt{e^2-4 d f}\right ) \left (-\frac{3}{8} e \left (2 A c f \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right )-B \left (b^3 f^3+6 b c f^2 (b e-2 a f)-24 c^2 f \left (b e^2-b d f-a e f\right )+16 c^3 \left (e^3-2 d e f\right )\right )\right )+\frac{3}{8} f \left (2 A c f \left (8 c^2 d e-4 a c e f-b f (5 b e-16 a f)+4 b c \left (e^2-4 d f\right )\right )-B \left (b^3 e f^2+16 c^3 d \left (e^2-d f\right )+2 c f \left (10 a b e f-8 a^2 f^2-b^2 \left (5 e^2-8 d f\right )\right )-8 c^2 \left (a f \left (e^2-4 d f\right )-b \left (e^3-5 d e f\right )\right )\right )\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{16 a f^2-8 b f \left (e-\sqrt{e^2-4 d f}\right )+4 c \left (e-\sqrt{e^2-4 d f}\right )^2-x^2} \, dx,x,\frac{4 a f-b \left (e-\sqrt{e^2-4 d f}\right )-\left (-2 b f+2 c \left (e-\sqrt{e^2-4 d f}\right )\right ) x}{\sqrt{a+b x+c x^2}}\right )}{3 c f^4 \sqrt{e^2-4 d f}}\\ &=-\frac{\left (2 A c f (4 c e-5 b f)-B \left (b^2 f^2-2 c f (5 b e-4 a f)+8 c^2 \left (e^2-d f\right )\right )+2 c f (2 B c e-b B f-2 A c f) x\right ) \sqrt{a+b x+c x^2}}{8 c f^3}+\frac{B \left (a+b x+c x^2\right )^{3/2}}{3 f}+\frac{\left (2 A c f \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right )-B \left (b^3 f^3+6 b c f^2 (b e-2 a f)-24 c^2 f \left (b e^2-b d f-a e f\right )+16 c^3 \left (e^3-2 d e f\right )\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{3/2} f^4}+\frac{\left (2 c f \left (B d (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )-A f \left (2 c d f (b e-a f)-f^2 \left (b^2 d-a^2 f\right )-c^2 d \left (e^2-d f\right )\right )\right )+c \left (e-\sqrt{e^2-4 d f}\right ) \left (A f (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+B \left (c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )-f^2 \left (2 a b e f-a^2 f^2-b^2 \left (e^2-d f\right )\right )+2 c f \left (a f \left (e^2-d f\right )-b \left (e^3-2 d e f\right )\right )\right )\right )\right ) \tanh ^{-1}\left (\frac{4 a f-b \left (e-\sqrt{e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right ) x}{2 \sqrt{2} \sqrt{c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt{e^2-4 d f}} \sqrt{a+b x+c x^2}}\right )}{\sqrt{2} c f^4 \sqrt{e^2-4 d f} \sqrt{c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt{e^2-4 d f}}}+\frac{\left (A f \left (c^2 \left (e^4-4 d e^2 f+2 d^2 f^2+e^3 \sqrt{e^2-4 d f}-2 d e f \sqrt{e^2-4 d f}\right )+f^2 \left (2 a^2 f^2-2 a b f \left (e+\sqrt{e^2-4 d f}\right )+b^2 \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )\right )+2 c f \left (a f \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )-b \left (e^3-3 d e f+e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}\right )\right )\right )-B \left (c^2 \left (e^5-5 d e^3 f+5 d^2 e f^2+e^4 \sqrt{e^2-4 d f}-3 d e^2 f \sqrt{e^2-4 d f}+d^2 f^2 \sqrt{e^2-4 d f}\right )+f^2 \left (a^2 f^2 \left (e+\sqrt{e^2-4 d f}\right )-2 a b f \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )+b^2 \left (e^3-3 d e f+e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}\right )\right )+2 c f \left (a f \left (e^3-3 d e f+e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}\right )-b \left (e^4-4 d e^2 f+2 d^2 f^2+e^3 \sqrt{e^2-4 d f}-2 d e f \sqrt{e^2-4 d f}\right )\right )\right )\right ) \tanh ^{-1}\left (\frac{4 a f-b \left (e+\sqrt{e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt{e^2-4 d f}\right )\right ) x}{2 \sqrt{2} \sqrt{c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt{e^2-4 d f}} \sqrt{a+b x+c x^2}}\right )}{\sqrt{2} f^4 \sqrt{e^2-4 d f} \sqrt{c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt{e^2-4 d f}}}\\ \end{align*}
Mathematica [A] time = 6.53623, size = 1627, normalized size = 1.49 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.306, size = 59465, normalized size = 54.5 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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