Optimal. Leaf size=254 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (3 a e^2+5 c d^2\right )+a \left (a e^2 (B e+3 C d)+c d^2 (3 B e+C d)\right )\right )}{16 a^{7/2} c^{5/2}}-\frac{(d+e x) \left (a e (3 a B e+a C d+5 A c d)-x \left (3 c d (3 a B e+a C d+5 A c d)+4 a e^2 (2 a C+A c)\right )\right )}{48 a^3 c^2 \left (a+c x^2\right )}-\frac{(d+e x)^2 (2 a e (2 a C+A c)-c x (3 a B e+a C d+5 A c d))}{24 a^2 c^2 \left (a+c x^2\right )^2}-\frac{(d+e x)^3 (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.541702, antiderivative size = 288, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1645, 821, 778, 205} \[ -\frac{4 a e \left (A c \left (a e^2+5 c d^2\right )+a \left (2 a C e^2+c d (3 B e+C d)\right )\right )-c x \left (A c d \left (15 c d^2-a e^2\right )+a \left (a e^2 (7 C d-3 B e)+3 c d^2 (3 B e+C d)\right )\right )}{48 a^3 c^3 \left (a+c x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (3 a e^2+5 c d^2\right )+a \left (a e^2 (B e+3 C d)+c d^2 (3 B e+C d)\right )\right )}{16 a^{7/2} c^{5/2}}-\frac{(d+e x)^2 (2 a e (2 a C+A c)-c x (3 a B e+a C d+5 A c d))}{24 a^2 c^2 \left (a+c x^2\right )^2}-\frac{(d+e x)^3 (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 1645
Rule 821
Rule 778
Rule 205
Rubi steps
\begin{align*} \int \frac{(d+e x)^3 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx &=-\frac{(a B-(A c-a C) x) (d+e x)^3}{6 a c \left (a+c x^2\right )^3}-\frac{\int \frac{(d+e x)^2 (-5 A c d-a C d-3 a B e-2 (A c+2 a C) e x)}{\left (a+c x^2\right )^3} \, dx}{6 a c}\\ &=-\frac{(a B-(A c-a C) x) (d+e x)^3}{6 a c \left (a+c x^2\right )^3}-\frac{(d+e x)^2 (2 a (A c+2 a C) e-c (5 A c d+a C d+3 a B e) x)}{24 a^2 c^2 \left (a+c x^2\right )^2}-\frac{\int \frac{(d+e x) \left (-4 a (A c+2 a C) e^2-3 c d (5 A c d+a C d+3 a B e)-c e (5 A c d+a C d+3 a B e) x\right )}{\left (a+c x^2\right )^2} \, dx}{24 a^2 c^2}\\ &=-\frac{(a B-(A c-a C) x) (d+e x)^3}{6 a c \left (a+c x^2\right )^3}-\frac{(d+e x)^2 (2 a (A c+2 a C) e-c (5 A c d+a C d+3 a B e) x)}{24 a^2 c^2 \left (a+c x^2\right )^2}-\frac{4 a e \left (A c \left (5 c d^2+a e^2\right )+a \left (2 a C e^2+c d (C d+3 B e)\right )\right )-c \left (A c d \left (15 c d^2-a e^2\right )+a \left (a e^2 (7 C d-3 B e)+3 c d^2 (C d+3 B e)\right )\right ) x}{48 a^3 c^3 \left (a+c x^2\right )}+\frac{\left (A c d \left (5 c d^2+3 a e^2\right )+a \left (a e^2 (3 C d+B e)+c d^2 (C d+3 B e)\right )\right ) \int \frac{1}{a+c x^2} \, dx}{16 a^3 c^2}\\ &=-\frac{(a B-(A c-a C) x) (d+e x)^3}{6 a c \left (a+c x^2\right )^3}-\frac{(d+e x)^2 (2 a (A c+2 a C) e-c (5 A c d+a C d+3 a B e) x)}{24 a^2 c^2 \left (a+c x^2\right )^2}-\frac{4 a e \left (A c \left (5 c d^2+a e^2\right )+a \left (2 a C e^2+c d (C d+3 B e)\right )\right )-c \left (A c d \left (15 c d^2-a e^2\right )+a \left (a e^2 (7 C d-3 B e)+3 c d^2 (C d+3 B e)\right )\right ) x}{48 a^3 c^3 \left (a+c x^2\right )}+\frac{\left (A c d \left (5 c d^2+3 a e^2\right )+a \left (a e^2 (3 C d+B e)+c d^2 (C d+3 B e)\right )\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{16 a^{7/2} c^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.321771, size = 350, normalized size = 1.38 \[ \frac{-\frac{8 a^{5/2} \left (-a^2 c e (e (A e+3 B d+B e x)+3 C d (d+e x))+a^3 C e^3+a c^2 d \left (3 A e (d+e x)+B d (d+3 e x)+C d^2 x\right )-A c^3 d^3 x\right )}{\left (a+c x^2\right )^3}+\frac{2 a^{3/2} \left (-a^2 c e (e (6 A e+18 B d+7 B e x)+3 C d (6 d+7 e x))+12 a^3 C e^3+a c^2 d x \left (3 e (A e+B d)+C d^2\right )+5 A c^3 d^3 x\right )}{\left (a+c x^2\right )^2}-\frac{3 \sqrt{a} \left (-a^2 c e^2 x (B e+3 C d)+8 a^3 C e^3-a c^2 d x \left (3 e (A e+B d)+C d^2\right )-5 A c^3 d^3 x\right )}{a+c x^2}+3 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (3 a e^2+5 c d^2\right )+a \left (a e^2 (B e+3 C d)+c d^2 (3 B e+C d)\right )\right )}{48 a^{7/2} c^3} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A] time = 0.053, size = 464, normalized size = 1.8 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{3}} \left ({\frac{ \left ( 3\,Acd{e}^{2}a+5\,A{d}^{3}{c}^{2}+{a}^{2}B{e}^{3}+3\,Bc{d}^{2}ae+3\,C{a}^{2}d{e}^{2}+Cac{d}^{3} \right ){x}^{5}}{16\,{a}^{3}}}-{\frac{C{e}^{3}{x}^{4}}{2\,c}}+{\frac{ \left ( 3\,Acd{e}^{2}a+5\,A{d}^{3}{c}^{2}-{a}^{2}B{e}^{3}+3\,Bc{d}^{2}ae-3\,C{a}^{2}d{e}^{2}+Cac{d}^{3} \right ){x}^{3}}{6\,{a}^{2}c}}-{\frac{e \left ( Ac{e}^{2}+3\,Bcde+2\,aC{e}^{2}+3\,Cc{d}^{2} \right ){x}^{2}}{4\,{c}^{2}}}-{\frac{ \left ( 3\,Acd{e}^{2}a-11\,A{d}^{3}{c}^{2}+{a}^{2}B{e}^{3}+3\,Bc{d}^{2}ae+3\,C{a}^{2}d{e}^{2}+Cac{d}^{3} \right ) x}{16\,a{c}^{2}}}-{\frac{aA{e}^{3}c+6\,A{c}^{2}{d}^{2}e+3\,aBd{e}^{2}c+2\,B{c}^{2}{d}^{3}+2\,{a}^{2}C{e}^{3}+3\,Cac{d}^{2}e}{12\,{c}^{3}}} \right ) }+{\frac{3\,Ad{e}^{2}}{16\,{a}^{2}c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{5\,A{d}^{3}}{16\,{a}^{3}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{B{e}^{3}}{16\,a{c}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{3\,B{d}^{2}e}{16\,{a}^{2}c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{3\,Cd{e}^{2}}{16\,a{c}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{C{d}^{3}}{16\,{a}^{2}c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Fricas [B] time = 1.6375, size = 2807, normalized size = 11.05 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Giac [B] time = 1.18846, size = 641, normalized size = 2.52 \begin{align*} \frac{{\left (C a c d^{3} + 5 \, A c^{2} d^{3} + 3 \, B a c d^{2} e + 3 \, C a^{2} d e^{2} + 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{16 \, \sqrt{a c} a^{3} c^{2}} + \frac{3 \, C a c^{4} d^{3} x^{5} + 15 \, A c^{5} d^{3} x^{5} + 9 \, B a c^{4} d^{2} x^{5} e + 9 \, C a^{2} c^{3} d x^{5} e^{2} + 9 \, A a c^{4} d x^{5} e^{2} + 8 \, C a^{2} c^{3} d^{3} x^{3} + 40 \, A a c^{4} d^{3} x^{3} + 3 \, B a^{2} c^{3} x^{5} e^{3} + 24 \, B a^{2} c^{3} d^{2} x^{3} e - 24 \, C a^{3} c^{2} x^{4} e^{3} - 24 \, C a^{3} c^{2} d x^{3} e^{2} + 24 \, A a^{2} c^{3} d x^{3} e^{2} - 36 \, C a^{3} c^{2} d^{2} x^{2} e - 3 \, C a^{3} c^{2} d^{3} x + 33 \, A a^{2} c^{3} d^{3} x - 8 \, B a^{3} c^{2} x^{3} e^{3} - 36 \, B a^{3} c^{2} d x^{2} e^{2} - 9 \, B a^{3} c^{2} d^{2} x e - 8 \, B a^{3} c^{2} d^{3} - 24 \, C a^{4} c x^{2} e^{3} - 12 \, A a^{3} c^{2} x^{2} e^{3} - 9 \, C a^{4} c d x e^{2} - 9 \, A a^{3} c^{2} d x e^{2} - 12 \, C a^{4} c d^{2} e - 24 \, A a^{3} c^{2} d^{2} e - 3 \, B a^{4} c x e^{3} - 12 \, B a^{4} c d e^{2} - 8 \, C a^{5} e^{3} - 4 \, A a^{4} c e^{3}}{48 \,{\left (c x^{2} + a\right )}^{3} a^{3} c^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]