3.19.69 \(\int \frac {d+c x}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx\) [1869]

Optimal. Leaf size=128 \[ \frac {b^4 c}{10 a^2 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{5/2}}-\frac {b^2 d}{3 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {d \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{a}+\frac {c \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}{6 a^2} \]

[Out]

1/10*b^4*c/a^2/(a*x+(a^2*x^2+b^2)^(1/2))^(5/2)-1/3*b^2*d/a/(a*x+(a^2*x^2+b^2)^(1/2))^(3/2)+d*(a*x+(a^2*x^2+b^2
)^(1/2))^(1/2)/a+1/6*c*(a*x+(a^2*x^2+b^2)^(1/2))^(3/2)/a^2

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Rubi [A]
time = 0.09, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2144, 1642} \begin {gather*} \frac {c \left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}}{6 a^2}-\frac {b^2 d}{3 a \left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}}+\frac {d \sqrt {\sqrt {a^2 x^2+b^2}+a x}}{a}+\frac {b^4 c}{10 a^2 \left (\sqrt {a^2 x^2+b^2}+a x\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + c*x)/Sqrt[a*x + Sqrt[b^2 + a^2*x^2]],x]

[Out]

(b^4*c)/(10*a^2*(a*x + Sqrt[b^2 + a^2*x^2])^(5/2)) - (b^2*d)/(3*a*(a*x + Sqrt[b^2 + a^2*x^2])^(3/2)) + (d*Sqrt
[a*x + Sqrt[b^2 + a^2*x^2]])/a + (c*(a*x + Sqrt[b^2 + a^2*x^2])^(3/2))/(6*a^2)

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2144

Int[((g_.) + (h_.)*(x_))^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dist[1/(2^(
m + 1)*e^(m + 1)), Subst[Int[x^(n - m - 2)*(a*f^2 + x^2)*((-a)*f^2*h + 2*e*g*x + h*x^2)^m, x], x, e*x + f*Sqrt
[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {d+c x}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (b^2+x^2\right ) \left (-b^2 c+2 a d x+c x^2\right )}{x^{7/2}} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{4 a^2}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {b^4 c}{x^{7/2}}+\frac {2 a b^2 d}{x^{5/2}}+\frac {2 a d}{\sqrt {x}}+c \sqrt {x}\right ) \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{4 a^2}\\ &=\frac {b^4 c}{10 a^2 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{5/2}}-\frac {b^2 d}{3 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {d \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{a}+\frac {c \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}{6 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 110, normalized size = 0.86 \begin {gather*} \frac {4 b^4 c+20 a^3 x^2 (3 d+c x) \left (a x+\sqrt {b^2+a^2 x^2}\right )+10 a b^2 \left (2 a x (2 d+c x)+(d+c x) \sqrt {b^2+a^2 x^2}\right )}{15 a^2 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^4*c + 20*a^3*x^2*(3*d + c*x)*(a*x + Sqrt[b^2 + a^2*x^2]) + 10*a*b^2*(2*a*x*(2*d + c*x) + (d + c*x)*Sqrt[b
^2 + a^2*x^2]))/(15*a^2*(a*x + Sqrt[b^2 + a^2*x^2])^(5/2))

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {c x +d}{\sqrt {a x +\sqrt {a^{2} x^{2}+b^{2}}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((c*x+d)/(a*x+(a^2*x^2+b^2)^(1/2))^(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((c*x+d)/(a*x+(a^2*x^2+b^2)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.40, size = 98, normalized size = 0.77 \begin {gather*} -\frac {2 \, {\left (3 \, a^{3} c x^{3} + 5 \, a^{3} d x^{2} + a b^{2} c x - 5 \, a b^{2} d - {\left (3 \, a^{2} c x^{2} + 5 \, a^{2} d x + 2 \, b^{2} c\right )} \sqrt {a^{2} x^{2} + b^{2}}\right )} \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}{15 \, a^{2} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*a^3*c*x^3 + 5*a^3*d*x^2 + a*b^2*c*x - 5*a*b^2*d - (3*a^2*c*x^2 + 5*a^2*d*x + 2*b^2*c)*sqrt(a^2*x^2 +
b^2))*sqrt(a*x + sqrt(a^2*x^2 + b^2))/(a^2*b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*x + d)/sqrt(a*x + sqrt(a**2*x**2 + b**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((c*x+d)/(a*x+(a^2*x^2+b^2)^(1/2))^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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