3.728 \(\int (d+e x)^m \sqrt {a+c x^2} \, dx\)

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

[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)))*(c
*x^2+a)^(1/2)/e/(1+m)/(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)

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Rubi [A]  time = 0.07, 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 = {760, 133} \[ \frac {\sqrt {a+c x^2} (d+e x)^{m+1} F_1\left (m+1;-\frac {1}{2},-\frac {1}{2};m+2;\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e (m+1) \sqrt {1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}} \sqrt {1-\frac {d+e x}{\frac {\sqrt {-a} e}{\sqrt {c}}+d}}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m*Sqrt[a + c*x^2],x]

[Out]

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

Rule 133

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

Rule 760

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 (d+e x)^m \sqrt {a+c x^2} \, dx &=\frac {\sqrt {a+c x^2} \operatorname {Subst}\left (\int 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 {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}}}}}\\ &=\frac {(d+e x)^{1+m} \sqrt {a+c x^2} 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 {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}}}}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 159, normalized size = 1.03 \[ \frac {\sqrt {a+c x^2} (d+e x)^{m+1} F_1\left (m+1;-\frac {1}{2},-\frac {1}{2};m+2;\frac {d+e x}{d-\sqrt {-\frac {a}{c}} e},\frac {d+e x}{d+\sqrt {-\frac {a}{c}} e}\right )}{e (m+1) \sqrt {\frac {e \left (\sqrt {-\frac {a}{c}}-x\right )}{e \sqrt {-\frac {a}{c}}+d}} \sqrt {\frac {e \left (\sqrt {-\frac {a}{c}}+x\right )}{e \sqrt {-\frac {a}{c}}-d}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {c x^{2} + a} {\left (e x + d\right )}^{m}, x\right ) \]

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(sqrt(c*x^2 + a)*(e*x + d)^m, x)

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

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(sqrt(c*x^2 + a)*(e*x + d)^m, x)

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maple [F]  time = 0.71, size = 0, normalized size = 0.00 \[ \int \sqrt {c \,x^{2}+a}\, \left (e x +d \right )^{m}\, dx \]

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

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(sqrt(c*x^2 + a)*(e*x + d)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)^(1/2)*(d + e*x)^m,x)

[Out]

int((a + c*x^2)^(1/2)*(d + e*x)^m, x)

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

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(sqrt(a + c*x**2)*(d + e*x)**m, x)

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