Integrand size = 29, antiderivative size = 403 \[ \int (g+h x)^m \sqrt {a+c x^2} \left (d+e x+f x^2\right ) \, dx=\frac {f (g+h x)^{1+m} \left (a+c x^2\right )^{3/2}}{c h (4+m)}-\frac {\left (a f h^2 (1+m)-c \left (3 f g^2-h (e g-d h) (4+m)\right )\right ) (g+h x)^{1+m} \sqrt {a+c x^2} \operatorname {AppellF1}\left (1+m,-\frac {1}{2},-\frac {1}{2},2+m,\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}},\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}\right )}{c h^3 (1+m) (4+m) \sqrt {1-\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}}} \sqrt {1-\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}}}-\frac {(3 f g-e h (4+m)) (g+h x)^{2+m} \sqrt {a+c x^2} \operatorname {AppellF1}\left (2+m,-\frac {1}{2},-\frac {1}{2},3+m,\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}},\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}\right )}{h^3 (2+m) (4+m) \sqrt {1-\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}}} \sqrt {1-\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}}} \]
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Time = 0.27 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1668, 858, 774, 138} \[ \int (g+h x)^m \sqrt {a+c x^2} \left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {a+c x^2} (g+h x)^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {1}{2},-\frac {1}{2},m+2,\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}},\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}\right ) \left (-a f h^2 (m+1)-c h (m+4) (e g-d h)+3 c f g^2\right )}{c h^3 (m+1) (m+4) \sqrt {1-\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}}} \sqrt {1-\frac {g+h x}{\frac {\sqrt {-a} h}{\sqrt {c}}+g}}}-\frac {\sqrt {a+c x^2} (g+h x)^{m+2} (3 f g-e h (m+4)) \operatorname {AppellF1}\left (m+2,-\frac {1}{2},-\frac {1}{2},m+3,\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}},\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}\right )}{h^3 (m+2) (m+4) \sqrt {1-\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}}} \sqrt {1-\frac {g+h x}{\frac {\sqrt {-a} h}{\sqrt {c}}+g}}}+\frac {f \left (a+c x^2\right )^{3/2} (g+h x)^{m+1}}{c h (m+4)} \]
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Rule 138
Rule 774
Rule 858
Rule 1668
Rubi steps \begin{align*} \text {integral}& = \frac {f (g+h x)^{1+m} \left (a+c x^2\right )^{3/2}}{c h (4+m)}+\frac {\int (g+h x)^m \left (-h^2 (a f (1+m)-c d (4+m))-c h (3 f g-e h (4+m)) x\right ) \sqrt {a+c x^2} \, dx}{c h^2 (4+m)} \\ & = \frac {f (g+h x)^{1+m} \left (a+c x^2\right )^{3/2}}{c h (4+m)}-\frac {(3 f g-e h (4+m)) \int (g+h x)^{1+m} \sqrt {a+c x^2} \, dx}{h^2 (4+m)}+\frac {\left (3 c f g^2-a f h^2 (1+m)-c h (e g-d h) (4+m)\right ) \int (g+h x)^m \sqrt {a+c x^2} \, dx}{c h^2 (4+m)} \\ & = \frac {f (g+h x)^{1+m} \left (a+c x^2\right )^{3/2}}{c h (4+m)}-\frac {\left ((3 f g-e h (4+m)) \sqrt {a+c x^2}\right ) \text {Subst}\left (\int x^{1+m} \sqrt {1-\frac {x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}}} \sqrt {1-\frac {x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}} \, dx,x,g+h x\right )}{h^3 (4+m) \sqrt {1-\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}}} \sqrt {1-\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}}}+\frac {\left (\left (3 c f g^2-a f h^2 (1+m)-c h (e g-d h) (4+m)\right ) \sqrt {a+c x^2}\right ) \text {Subst}\left (\int x^m \sqrt {1-\frac {x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}}} \sqrt {1-\frac {x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}} \, dx,x,g+h x\right )}{c h^3 (4+m) \sqrt {1-\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}}} \sqrt {1-\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}}} \\ & = \frac {f (g+h x)^{1+m} \left (a+c x^2\right )^{3/2}}{c h (4+m)}+\frac {\left (3 c f g^2-a f h^2 (1+m)-c h (e g-d h) (4+m)\right ) (g+h x)^{1+m} \sqrt {a+c x^2} F_1\left (1+m;-\frac {1}{2},-\frac {1}{2};2+m;\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}},\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}\right )}{c h^3 (1+m) (4+m) \sqrt {1-\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}}} \sqrt {1-\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}}}-\frac {(3 f g-e h (4+m)) (g+h x)^{2+m} \sqrt {a+c x^2} F_1\left (2+m;-\frac {1}{2},-\frac {1}{2};3+m;\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}},\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}\right )}{h^3 (2+m) (4+m) \sqrt {1-\frac {g+h x}{g-\frac {\sqrt {-a} h}{\sqrt {c}}}} \sqrt {1-\frac {g+h x}{g+\frac {\sqrt {-a} h}{\sqrt {c}}}}} \\ \end{align*}
\[ \int (g+h x)^m \sqrt {a+c x^2} \left (d+e x+f x^2\right ) \, dx=\int (g+h x)^m \sqrt {a+c x^2} \left (d+e x+f x^2\right ) \, dx \]
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\[\int \left (h x +g \right )^{m} \left (f \,x^{2}+e x +d \right ) \sqrt {c \,x^{2}+a}d x\]
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\[ \int (g+h x)^m \sqrt {a+c x^2} \left (d+e x+f x^2\right ) \, dx=\int { \sqrt {c x^{2} + a} {\left (f x^{2} + e x + d\right )} {\left (h x + g\right )}^{m} \,d x } \]
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\[ \int (g+h x)^m \sqrt {a+c x^2} \left (d+e x+f x^2\right ) \, dx=\int \sqrt {a + c x^{2}} \left (g + h x\right )^{m} \left (d + e x + f x^{2}\right )\, dx \]
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\[ \int (g+h x)^m \sqrt {a+c x^2} \left (d+e x+f x^2\right ) \, dx=\int { \sqrt {c x^{2} + a} {\left (f x^{2} + e x + d\right )} {\left (h x + g\right )}^{m} \,d x } \]
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\[ \int (g+h x)^m \sqrt {a+c x^2} \left (d+e x+f x^2\right ) \, dx=\int { \sqrt {c x^{2} + a} {\left (f x^{2} + e x + d\right )} {\left (h x + g\right )}^{m} \,d x } \]
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Timed out. \[ \int (g+h x)^m \sqrt {a+c x^2} \left (d+e x+f x^2\right ) \, dx=\int {\left (g+h\,x\right )}^m\,\sqrt {c\,x^2+a}\,\left (f\,x^2+e\,x+d\right ) \,d x \]
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