3.17.0 ∫ (b+2cx)(d+ex)5∕2 __________________________

Integrand size = 28, antiderivative size = 350 \[ \int \frac {(b+2 c x) (d+e x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {5 e^2 \sqrt {d+e x}}{c}-\frac {(d+e x)^{5/2}}{a+b x+c x^2}-\frac {5 e \left (2 c^2 d^2+b \left (b-\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (b d-\sqrt {b^2-4 a c} d+a e\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}}\right )}{\sqrt {2} c^{3/2} \sqrt {b^2-4 a c} \sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}}+\frac {5 e \left (2 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt {b^2-4 a c} d+a e\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}\right )}{\sqrt {2} c^{3/2} \sqrt {b^2-4 a c} \sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \]

Rubi [A] (veriﬁed)

Time = 0.77 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {782, 717, 840, 1180, 214} \[ \int \frac {(b+2 c x) (d+e x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {5 e \left (-2 c e \left (-d \sqrt {b^2-4 a c}+a e+b d\right )+b e^2 \left (b-\sqrt {b^2-4 a c}\right )+2 c^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}\right )}{\sqrt {2} c^{3/2} \sqrt {b^2-4 a c} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}+\frac {5 e \left (-2 c e \left (d \sqrt {b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt {b^2-4 a c}+b\right )+2 c^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\right )}{\sqrt {2} c^{3/2} \sqrt {b^2-4 a c} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}-\frac {(d+e x)^{5/2}}{a+b x+c x^2}+\frac {5 e^2 \sqrt {d+e x}}{c} \]

Rubi steps \begin{align*} \text {integral}& = -\frac {(d+e x)^{5/2}}{a+b x+c x^2}+\frac {1}{2} (5 e) \int \frac {(d+e x)^{3/2}}{a+b x+c x^2} \, dx \\ & = \frac {5 e^2 \sqrt {d+e x}}{c}-\frac {(d+e x)^{5/2}}{a+b x+c x^2}+\frac {(5 e) \int \frac {c d^2-a e^2+e (2 c d-b e) x}{\sqrt {d+e x} \left (a+b x+c x^2\right )} \, dx}{2 c} \\ & = \frac {5 e^2 \sqrt {d+e x}}{c}-\frac {(d+e x)^{5/2}}{a+b x+c x^2}+\frac {(5 e) \text {Subst}\left (\int \frac {-d e (2 c d-b e)+e \left (c d^2-a e^2\right )+e (2 c d-b e) x^2}{c d^2-b d e+a e^2+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt {d+e x}\right )}{c} \\ & = \frac {5 e^2 \sqrt {d+e x}}{c}-\frac {(d+e x)^{5/2}}{a+b x+c x^2}+\frac {\left (5 e \left (2 c^2 d^2+b \left (b-\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (b d-\sqrt {b^2-4 a c} d+a e\right )\right )\right ) \text {Subst}\left (\int \frac {1}{-\frac {1}{2} \sqrt {b^2-4 a c} e+\frac {1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt {d+e x}\right )}{2 c \sqrt {b^2-4 a c}}-\frac {\left (5 e \left (2 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt {b^2-4 a c} d+a e\right )\right )\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{2} \sqrt {b^2-4 a c} e+\frac {1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt {d+e x}\right )}{2 c \sqrt {b^2-4 a c}} \\ & = \frac {5 e^2 \sqrt {d+e x}}{c}-\frac {(d+e x)^{5/2}}{a+b x+c x^2}-\frac {5 e \left (2 c^2 d^2+b \left (b-\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (b d-\sqrt {b^2-4 a c} d+a e\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}}\right )}{\sqrt {2} c^{3/2} \sqrt {b^2-4 a c} \sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}}+\frac {5 e \left (2 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt {b^2-4 a c} d+a e\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}\right )}{\sqrt {2} c^{3/2} \sqrt {b^2-4 a c} \sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \\ \end{align*}

Mathematica [C] (veriﬁed)

Time = 5.35 (sec) , antiderivative size = 715, normalized size of antiderivative = 2.04 \[ \int \frac {(b+2 c x) (d+e x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {e \left (\frac {2 \sqrt {c} \sqrt {d+e x} \left (5 e^2 (a+b x)-c \left (d^2+2 d e x-4 e^2 x^2\right )\right )}{e (a+x (b+c x))}+\frac {2 \sqrt {2} \left (12 c^2 d^2+b \left (5 b-3 \sqrt {b^2-4 a c}\right ) e^2-2 c e \left (6 b d-3 \sqrt {b^2-4 a c} d+4 a e\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {-2 c d+b e-\sqrt {b^2-4 a c} e}}\right )}{\sqrt {b^2-4 a c} \sqrt {-2 c d+\left (b-\sqrt {b^2-4 a c}\right ) e}}+\frac {\left (14 i c^2 d^2+b \left (5 i b+\sqrt {-b^2+4 a c}\right ) e^2-2 c e \left (7 i b d+\sqrt {-b^2+4 a c} d+3 i a e\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {-2 c d+b e-i \sqrt {-b^2+4 a c} e}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {-2 c d+\left (b-i \sqrt {-b^2+4 a c}\right ) e}}+\frac {\left (-14 i c^2 d^2+b \left (-5 i b+\sqrt {-b^2+4 a c}\right ) e^2+2 i c e \left (7 b d+i \sqrt {-b^2+4 a c} d+3 a e\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {-2 c d+b e+i \sqrt {-b^2+4 a c} e}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {-2 c d+\left (b+i \sqrt {-b^2+4 a c}\right ) e}}-\frac {2 \sqrt {2} \left (12 c^2 d^2+b \left (5 b+3 \sqrt {b^2-4 a c}\right ) e^2-2 c e \left (6 b d+3 \sqrt {b^2-4 a c} d+4 a e\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}}\right )}{\sqrt {b^2-4 a c} \sqrt {-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}}\right )}{2 c^{3/2}} \]

Maple [A] (veriﬁed)

Time = 0.66 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.29


method	result	size

derivativedivides	\(2 e^{2} \left (\frac {2 \sqrt {e x +d}}{c}-\frac {\frac {\left (-\frac {b e}{2}+c d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {1}{2} e^{2} a +\frac {1}{2} b d e -\frac {1}{2} c \,d^{2}\right ) \sqrt {e x +d}}{c \left (e x +d \right )^{2}+b e \left (e x +d \right )-2 c d \left (e x +d \right )+e^{2} a -b d e +c \,d^{2}}+10 c \left (\frac {\left (-2 a c \,e^{2}+b^{2} e^{2}-2 b c d e +2 c^{2} d^{2}+\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, b e -2 \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, c d \right ) \sqrt {2}\, \arctan \left (\frac {c \sqrt {e x +d}\, \sqrt {2}}{\sqrt {\left (b e -2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{8 c \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (b e -2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}-\frac {\left (2 a c \,e^{2}-b^{2} e^{2}+2 b c d e -2 c^{2} d^{2}+\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, b e -2 \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, c d \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \sqrt {e x +d}\, \sqrt {2}}{\sqrt {\left (-b e +2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{8 c \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (-b e +2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{c}\right )\)	\(452\)
default	\(2 e^{2} \left (\frac {2 \sqrt {e x +d}}{c}-\frac {\frac {\left (-\frac {b e}{2}+c d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {1}{2} e^{2} a +\frac {1}{2} b d e -\frac {1}{2} c \,d^{2}\right ) \sqrt {e x +d}}{c \left (e x +d \right )^{2}+b e \left (e x +d \right )-2 c d \left (e x +d \right )+e^{2} a -b d e +c \,d^{2}}+10 c \left (\frac {\left (-2 a c \,e^{2}+b^{2} e^{2}-2 b c d e +2 c^{2} d^{2}+\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, b e -2 \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, c d \right ) \sqrt {2}\, \arctan \left (\frac {c \sqrt {e x +d}\, \sqrt {2}}{\sqrt {\left (b e -2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{8 c \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (b e -2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}-\frac {\left (2 a c \,e^{2}-b^{2} e^{2}+2 b c d e -2 c^{2} d^{2}+\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, b e -2 \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, c d \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \sqrt {e x +d}\, \sqrt {2}}{\sqrt {\left (-b e +2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{8 c \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (-b e +2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{c}\right )\)	\(452\)
risch	\(\frac {4 e^{2} \sqrt {e x +d}}{c}-\frac {2 e^{2} \left (\frac {\left (-\frac {b e}{2}+c d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {1}{2} e^{2} a +\frac {1}{2} b d e -\frac {1}{2} c \,d^{2}\right ) \sqrt {e x +d}}{c \left (e x +d \right )^{2}+b e \left (e x +d \right )-2 c d \left (e x +d \right )+e^{2} a -b d e +c \,d^{2}}+10 c \left (\frac {\left (-2 a c \,e^{2}+b^{2} e^{2}-2 b c d e +2 c^{2} d^{2}+\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, b e -2 \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, c d \right ) \sqrt {2}\, \arctan \left (\frac {c \sqrt {e x +d}\, \sqrt {2}}{\sqrt {\left (b e -2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{8 c \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (b e -2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}-\frac {\left (2 a c \,e^{2}-b^{2} e^{2}+2 b c d e -2 c^{2} d^{2}+\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, b e -2 \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, c d \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \sqrt {e x +d}\, \sqrt {2}}{\sqrt {\left (-b e +2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )}{8 c \sqrt {-e^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (-b e +2 c d +\sqrt {-e^{2} \left (4 a c -b^{2}\right )}\right ) c}}\right )\right )}{c}\)	\(453\)
pseudoelliptic	\(\frac {5 \sqrt {\left (b e -2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}\, \sqrt {2}\, \left (\left (\frac {b e}{2}-c d \right ) \sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}+\left (a c -\frac {b^{2}}{2}\right ) e^{2}+b c d e -c^{2} d^{2}\right ) e^{2} \left (c \,x^{2}+b x +a \right ) \operatorname {arctanh}\left (\frac {c \sqrt {e x +d}\, \sqrt {2}}{\sqrt {\left (-b e +2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}}\right )+5 \sqrt {\left (-b e +2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}\, \left (\left (\left (-\frac {b e}{2}+c d \right ) \sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}+\left (a c -\frac {b^{2}}{2}\right ) e^{2}+b c d e -c^{2} d^{2}\right ) \sqrt {2}\, e^{2} \left (c \,x^{2}+b x +a \right ) \arctan \left (\frac {c \sqrt {e x +d}\, \sqrt {2}}{\sqrt {\left (b e -2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}}\right )+\sqrt {\left (b e -2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}\, \left (\left (a +\frac {4}{5} c \,x^{2}+b x \right ) e^{2}-\frac {2 c d e x}{5}-\frac {c \,d^{2}}{5}\right ) \sqrt {e x +d}\, \sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right )}{\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\, \sqrt {\left (b e -2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}\, \sqrt {\left (-b e +2 c d +\sqrt {-4 e^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) c}\, c \left (c \,x^{2}+b x +a \right )}\)	\(453\)

Fricas [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2928 vs. \(2 (298) = 596\).

Time = 0.39 (sec) , antiderivative size = 2928, normalized size of antiderivative = 8.37 \[ \int \frac {(b+2 c x) (d+e x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \]

Sympy [F(-1)]

Timed out. \[ \int \frac {(b+2 c x) (d+e x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \]

Maxima [F]

\[ \int \frac {(b+2 c x) (d+e x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\int { \frac {{\left (2 \, c x + b\right )} {\left (e x + d\right )}^{\frac {5}{2}}}{{\left (c x^{2} + b x + a\right )}^{2}} \,d x } \]

Giac [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 930 vs. \(2 (298) = 596\).

Time = 0.54 (sec) , antiderivative size = 930, normalized size of antiderivative = 2.66 \[ \int \frac {(b+2 c x) (d+e x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {4 \, \sqrt {e x + d} e^{2}}{c} - \frac {2 \, {\left (e x + d\right )}^{\frac {3}{2}} c d e^{2} - \sqrt {e x + d} c d^{2} e^{2} - {\left (e x + d\right )}^{\frac {3}{2}} b e^{3} + \sqrt {e x + d} b d e^{3} - \sqrt {e x + d} a e^{4}}{{\left ({\left (e x + d\right )}^{2} c - 2 \, {\left (e x + d\right )} c d + c d^{2} + {\left (e x + d\right )} b e - b d e + a e^{2}\right )} c} + \frac {5 \, {\left ({\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d e^{2} - {\left (b^{3} - 4 \, a b c\right )} e^{3}\right )} \sqrt {-4 \, c^{2} d + 2 \, {\left (b c - \sqrt {b^{2} - 4 \, a c} c\right )} e} c^{2} e^{2} - 2 \, {\left (\sqrt {b^{2} - 4 \, a c} c^{3} d^{2} e^{2} - \sqrt {b^{2} - 4 \, a c} b c^{2} d e^{3} + \sqrt {b^{2} - 4 \, a c} a c^{2} e^{4}\right )} \sqrt {-4 \, c^{2} d + 2 \, {\left (b c - \sqrt {b^{2} - 4 \, a c} c\right )} e} {\left | c \right |} {\left | e \right |} - {\left (4 \, c^{5} d^{3} e^{2} - 6 \, b c^{4} d^{2} e^{3} + 4 \, {\left (b^{2} c^{3} - a c^{4}\right )} d e^{4} - {\left (b^{3} c^{2} - 2 \, a b c^{3}\right )} e^{5}\right )} \sqrt {-4 \, c^{2} d + 2 \, {\left (b c - \sqrt {b^{2} - 4 \, a c} c\right )} e}\right )} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {e x + d}}{\sqrt {-\frac {2 \, c^{2} d - b c e + \sqrt {-4 \, {\left (c^{2} d^{2} - b c d e + a c e^{2}\right )} c^{2} + {\left (2 \, c^{2} d - b c e\right )}^{2}}}{c^{2}}}}\right )}{8 \, {\left (\sqrt {b^{2} - 4 \, a c} c^{4} d^{2} - \sqrt {b^{2} - 4 \, a c} b c^{3} d e + \sqrt {b^{2} - 4 \, a c} a c^{3} e^{2}\right )} c^{2} {\left | e \right |}} - \frac {5 \, {\left ({\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d e^{2} - {\left (b^{3} - 4 \, a b c\right )} e^{3}\right )} \sqrt {-4 \, c^{2} d + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} e} c^{2} e^{2} + 2 \, {\left (\sqrt {b^{2} - 4 \, a c} c^{3} d^{2} e^{2} - \sqrt {b^{2} - 4 \, a c} b c^{2} d e^{3} + \sqrt {b^{2} - 4 \, a c} a c^{2} e^{4}\right )} \sqrt {-4 \, c^{2} d + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} e} {\left | c \right |} {\left | e \right |} - {\left (4 \, c^{5} d^{3} e^{2} - 6 \, b c^{4} d^{2} e^{3} + 4 \, {\left (b^{2} c^{3} - a c^{4}\right )} d e^{4} - {\left (b^{3} c^{2} - 2 \, a b c^{3}\right )} e^{5}\right )} \sqrt {-4 \, c^{2} d + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} e}\right )} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {e x + d}}{\sqrt {-\frac {2 \, c^{2} d - b c e - \sqrt {-4 \, {\left (c^{2} d^{2} - b c d e + a c e^{2}\right )} c^{2} + {\left (2 \, c^{2} d - b c e\right )}^{2}}}{c^{2}}}}\right )}{8 \, {\left (\sqrt {b^{2} - 4 \, a c} c^{4} d^{2} - \sqrt {b^{2} - 4 \, a c} b c^{3} d e + \sqrt {b^{2} - 4 \, a c} a c^{3} e^{2}\right )} c^{2} {\left | e \right |}} \]

Mupad [B] (veriﬁcation not implemented)

Time = 1.49 (sec) , antiderivative size = 8776, normalized size of antiderivative = 25.07 \[ \int \frac {(b+2 c x) (d+e x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \]

\(\int \frac {(b+2 c x) (d+e x)^{5/2}}{(a+b x+c x^2)^2} \, dx\) [1618]

Optimal result