\(\int \frac {(d+e x)^m}{a+c x^2} \, dx\) [724]

Optimal result

Integrand size = 17, antiderivative size = 167 \[ \int \frac {(d+e x)^m}{a+c x^2} \, dx=\frac {(d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 \sqrt {-a} \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+m)}-\frac {(d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {-a} \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+m)} \]

[Out]

1/2*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],(e*x+d)*c^(1/2)/(-e*(-a)^(1/2)+d*c^(1/2)))/(1+m)/(-a)^(1/2)/(-e*(-a
)^(1/2)+d*c^(1/2))-1/2*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],(e*x+d)*c^(1/2)/(e*(-a)^(1/2)+d*c^(1/2)))/(1+m)/
(-a)^(1/2)/(e*(-a)^(1/2)+d*c^(1/2))

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {726, 70} \[ \int \frac {(d+e x)^m}{a+c x^2} \, dx=\frac {(d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 \sqrt {-a} (m+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}-\frac {(d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {-a} (m+1) \left (\sqrt {-a} e+\sqrt {c} d\right )} \]

[In]

Int[(d + e*x)^m/(a + c*x^2),x]

[Out]

((d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(2*Sqrt[-
a]*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + m)) - ((d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x
))/(Sqrt[c]*d + Sqrt[-a]*e)])/(2*Sqrt[-a]*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + m))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 726

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\sqrt {-a} (d+e x)^m}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\sqrt {-a} (d+e x)^m}{2 a \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx \\ & = -\frac {\int \frac {(d+e x)^m}{\sqrt {-a}-\sqrt {c} x} \, dx}{2 \sqrt {-a}}-\frac {\int \frac {(d+e x)^m}{\sqrt {-a}+\sqrt {c} x} \, dx}{2 \sqrt {-a}} \\ & = \frac {(d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 \sqrt {-a} \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+m)}-\frac {(d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {-a} \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.87 \[ \int \frac {(d+e x)^m}{a+c x^2} \, dx=\frac {(d+e x)^{1+m} \left (\frac {\operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {c} d-\sqrt {-a} e}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {-a} (1+m)} \]

[In]

Integrate[(d + e*x)^m/(a + c*x^2),x]

[Out]

((d + e*x)^(1 + m)*(Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)]/(Sqrt[c]*
d - Sqrt[-a]*e) - Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]/(Sqrt[c]*d
+ Sqrt[-a]*e)))/(2*Sqrt[-a]*(1 + m))

Maple [F]

\[\int \frac {\left (e x +d \right )^{m}}{c \,x^{2}+a}d x\]

[In]

int((e*x+d)^m/(c*x^2+a),x)

[Out]

int((e*x+d)^m/(c*x^2+a),x)

Fricas [F]

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

[In]

integrate((e*x+d)^m/(c*x^2+a),x, algorithm="fricas")

[Out]

integral((e*x + d)^m/(c*x^2 + a), x)

Sympy [F]

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

[In]

integrate((e*x+d)**m/(c*x**2+a),x)

[Out]

Integral((d + e*x)**m/(a + c*x**2), x)

Maxima [F]

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

[In]

integrate((e*x+d)^m/(c*x^2+a),x, algorithm="maxima")

[Out]

integrate((e*x + d)^m/(c*x^2 + a), x)

Giac [F]

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

[In]

integrate((e*x+d)^m/(c*x^2+a),x, algorithm="giac")

[Out]

integrate((e*x + d)^m/(c*x^2 + a), x)

Mupad [F(-1)]

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

[In]

int((d + e*x)^m/(a + c*x^2),x)

[Out]

int((d + e*x)^m/(a + c*x^2), x)