3.8.29 \(\int \frac {(d+e x)^m}{\sqrt {a+c x^2}} \, dx\) [729]

Optimal. Leaf size=154 \[ \frac {(d+e x)^{1+m} \sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}} F_1\left (1+m;\frac {1}{2},\frac {1}{2};2+m;\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e (1+m) \sqrt {a+c x^2}} \]

[Out]

(e*x+d)^(1+m)*AppellF1(1+m,1/2,1/2,2+m,(e*x+d)/(d-e*(-a)^(1/2)/c^(1/2)),(e*x+d)/(d+e*(-a)^(1/2)/c^(1/2)))*(1+(
-e*x-d)/(d-e*(-a)^(1/2)/c^(1/2)))^(1/2)*(1+(-e*x-d)/(d+e*(-a)^(1/2)/c^(1/2)))^(1/2)/e/(1+m)/(c*x^2+a)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {774, 138} \begin {gather*} \frac {(d+e x)^{m+1} \sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{\frac {\sqrt {-a} e}{\sqrt {c}}+d}} F_1\left (m+1;\frac {1}{2},\frac {1}{2};m+2;\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e (m+1) \sqrt {a+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m/Sqrt[a + c*x^2],x]

[Out]

((d + e*x)^(1 + m)*Sqrt[1 - (d + e*x)/(d - (Sqrt[-a]*e)/Sqrt[c])]*Sqrt[1 - (d + e*x)/(d + (Sqrt[-a]*e)/Sqrt[c]
)]*AppellF1[1 + m, 1/2, 1/2, 2 + m, (d + e*x)/(d - (Sqrt[-a]*e)/Sqrt[c]), (d + e*x)/(d + (Sqrt[-a]*e)/Sqrt[c])
])/(e*(1 + m)*Sqrt[a + c*x^2])

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 774

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[(a + c*x^
2)^p/(e*(1 - (d + e*x)/(d + e*(q/c)))^p*(1 - (d + e*x)/(d - e*(q/c)))^p), Subst[Int[x^m*Simp[1 - x/(d + e*(q/c
)), x]^p*Simp[1 - x/(d - e*(q/c)), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a
*e^2, 0] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {(d+e x)^m}{\sqrt {a+c x^2}} \, dx &=\frac {\left (\sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}}\right ) \text {Subst}\left (\int \frac {x^m}{\sqrt {1-\frac {x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}}} \, dx,x,d+e x\right )}{e \sqrt {a+c x^2}}\\ &=\frac {(d+e x)^{1+m} \sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}} F_1\left (1+m;\frac {1}{2},\frac {1}{2};2+m;\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e (1+m) \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 159, normalized size = 1.03 \begin {gather*} \frac {\sqrt {\frac {e \left (\sqrt {-\frac {a}{c}}-x\right )}{d+\sqrt {-\frac {a}{c}} e}} \sqrt {\frac {e \left (\sqrt {-\frac {a}{c}}+x\right )}{-d+\sqrt {-\frac {a}{c}} e}} (d+e x)^{1+m} F_1\left (1+m;\frac {1}{2},\frac {1}{2};2+m;\frac {d+e x}{d-\sqrt {-\frac {a}{c}} e},\frac {d+e x}{d+\sqrt {-\frac {a}{c}} e}\right )}{e (1+m) \sqrt {a+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(e*(Sqrt[-(a/c)] - x))/(d + Sqrt[-(a/c)]*e)]*Sqrt[(e*(Sqrt[-(a/c)] + x))/(-d + Sqrt[-(a/c)]*e)]*(d + e*x
)^(1 + m)*AppellF1[1 + m, 1/2, 1/2, 2 + m, (d + e*x)/(d - Sqrt[-(a/c)]*e), (d + e*x)/(d + Sqrt[-(a/c)]*e)])/(e
*(1 + m)*Sqrt[a + c*x^2])

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m}}{\sqrt {c \,x^{2}+a}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m/(c*x^2+a)^(1/2),x)

[Out]

int((e*x+d)^m/(c*x^2+a)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{m}}{\sqrt {a + c x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m/(c*x**2+a)**(1/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^m}{\sqrt {c\,x^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^m/(a + c*x^2)^(1/2),x)

[Out]

int((d + e*x)^m/(a + c*x^2)^(1/2), x)

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