3.25.16 \(\int \frac {d+c x^2}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx\)

Optimal. Leaf size=194 \[ \frac {2 \sqrt {a^2 x^2+b^2} \left (12 a^4 c x^5+60 a^4 d x^3-3 a^2 b^2 c x^3+25 a^2 b^2 d x-4 b^4 c x\right )}{15 a^2 \left (\sqrt {a^2 x^2+b^2}+a x\right )^{7/2}}+\frac {2 \left (84 a^6 c x^6+420 a^6 d x^4+21 a^4 b^2 c x^4+385 a^4 b^2 d x^2-49 a^2 b^4 c x^2+35 a^2 b^4 d-8 b^6 c\right )}{105 a^3 \left (\sqrt {a^2 x^2+b^2}+a x\right )^{7/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.32, antiderivative size = 196, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {6742, 2117, 14, 2119, 448} \begin {gather*} -\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^2 c \sqrt {\sqrt {a^2 x^2+b^2}+a x}}{4 a^3}+\frac {c \left (\sqrt {a^2 x^2+b^2}+a x\right )^{5/2}}{20 a^3}-\frac {b^6 c}{28 a^3 \left (\sqrt {a^2 x^2+b^2}+a x\right )^{7/2}}+\frac {b^4 c}{12 a^3 \left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/28*(b^6*c)/(a^3*(a*x + Sqrt[b^2 + a^2*x^2])^(7/2)) + (b^4*c)/(12*a^3*(a*x + Sqrt[b^2 + a^2*x^2])^(3/2)) - (
b^2*d)/(3*a*(a*x + Sqrt[b^2 + a^2*x^2])^(3/2)) - (b^2*c*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]])/(4*a^3) + (d*Sqrt[a*x
+ Sqrt[b^2 + a^2*x^2]])/a + (c*(a*x + Sqrt[b^2 + a^2*x^2])^(5/2))/(20*a^3)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rule 2117

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

Rule 2119

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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {d+c x^2}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx &=\int \left (\frac {d}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {c x^2}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}}\right ) \, dx\\ &=c \int \frac {x^2}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx+d \int \frac {1}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx\\ &=\frac {c \operatorname {Subst}\left (\int \frac {\left (-b^2+x^2\right )^2 \left (b^2+x^2\right )}{x^{9/2}} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{8 a^3}+\frac {d \operatorname {Subst}\left (\int \frac {b^2+x^2}{x^{5/2}} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{2 a}\\ &=\frac {c \operatorname {Subst}\left (\int \left (\frac {b^6}{x^{9/2}}-\frac {b^4}{x^{5/2}}-\frac {b^2}{\sqrt {x}}+x^{3/2}\right ) \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{8 a^3}+\frac {d \operatorname {Subst}\left (\int \left (\frac {b^2}{x^{5/2}}+\frac {1}{\sqrt {x}}\right ) \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{2 a}\\ &=-\frac {b^6 c}{28 a^3 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{7/2}}+\frac {b^4 c}{12 a^3 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}-\frac {b^2 d}{3 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}-\frac {b^2 c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{4 a^3}+\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 )^{5/2}}{20 a^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 7.39, size = 952, normalized size = 4.91 \begin {gather*} \frac {2 \sqrt {b^2+a^2 x^2} \left (\frac {21 a^2 d \left (7 b^4+3 a x \left (7 a x+6 \sqrt {b^2+a^2 x^2}\right ) b^2+6 a^3 x^3 \left (a x+\sqrt {b^2+a^2 x^2}\right )\right ) \left (b^2+2 a x \left (a x+\sqrt {b^2+a^2 x^2}\right )\right )^5}{b^{10}+a x \left (41 a x+9 \sqrt {b^2+a^2 x^2}\right ) b^8+40 a^3 x^3 \left (7 a x+3 \sqrt {b^2+a^2 x^2}\right ) b^6+16 a^5 x^5 \left (43 a x+27 \sqrt {b^2+a^2 x^2}\right ) b^4+64 a^7 x^7 \left (11 a x+9 \sqrt {b^2+a^2 x^2}\right ) b^2+256 a^9 x^9 \left (a x+\sqrt {b^2+a^2 x^2}\right )}+\frac {5 c \left (a x+\sqrt {b^2+a^2 x^2}\right )^2 \left (-8 b^8-7 a x \left (7 a x+4 \sqrt {b^2+a^2 x^2}\right ) b^6+7 a^3 x^3 \left (a x-4 \sqrt {b^2+a^2 x^2}\right ) b^4+28 a^5 x^5 \left (4 a x+3 \sqrt {b^2+a^2 x^2}\right ) b^2+56 a^7 x^7 \left (a x+\sqrt {b^2+a^2 x^2}\right )\right ) \left (b^2+2 a x \left (a x+\sqrt {b^2+a^2 x^2}\right )\right )^4}{b^{12}+a x \left (61 a x+11 \sqrt {b^2+a^2 x^2}\right ) b^{10}+20 a^3 x^3 \left (31 a x+11 \sqrt {b^2+a^2 x^2}\right ) b^8+112 a^5 x^5 \left (21 a x+11 \sqrt {b^2+a^2 x^2}\right ) b^6+256 a^7 x^7 \left (16 a x+11 \sqrt {b^2+a^2 x^2}\right ) b^4+256 a^9 x^9 \left (13 a x+11 \sqrt {b^2+a^2 x^2}\right ) b^2+1024 a^{11} x^{11} \left (a x+\sqrt {b^2+a^2 x^2}\right )}-\frac {21 a^2 d \left (a x+\sqrt {b^2+a^2 x^2}\right )^2 \left (2 b^4+3 a x \left (2 a x+\sqrt {b^2+a^2 x^2}\right ) b^2+6 a^3 x^3 \left (a x+\sqrt {b^2+a^2 x^2}\right )\right )}{b^2+a x \left (a x+\sqrt {b^2+a^2 x^2}\right )}+\frac {c \left (16 b^8+14 a x \left (7 a x+4 \sqrt {b^2+a^2 x^2}\right ) b^6+7 a^3 x^3 \left (4 a x+11 \sqrt {b^2+a^2 x^2}\right ) b^4-28 a^5 x^5 \left (11 a x+6 \sqrt {b^2+a^2 x^2}\right ) b^2-280 a^7 x^7 \left (a x+\sqrt {b^2+a^2 x^2}\right )\right )}{b^2+a x \left (a x+\sqrt {b^2+a^2 x^2}\right )}\right )}{315 a^3 b^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^2)/Sqrt[a*x + Sqrt[b^2 + a^2*x^2]],x]

[Out]

(2*Sqrt[b^2 + a^2*x^2]*((-21*a^2*d*(a*x + Sqrt[b^2 + a^2*x^2])^2*(2*b^4 + 6*a^3*x^3*(a*x + Sqrt[b^2 + a^2*x^2]
) + 3*a*b^2*x*(2*a*x + Sqrt[b^2 + a^2*x^2])))/(b^2 + a*x*(a*x + Sqrt[b^2 + a^2*x^2])) + (c*(16*b^8 - 280*a^7*x
^7*(a*x + Sqrt[b^2 + a^2*x^2]) + 14*a*b^6*x*(7*a*x + 4*Sqrt[b^2 + a^2*x^2]) - 28*a^5*b^2*x^5*(11*a*x + 6*Sqrt[
b^2 + a^2*x^2]) + 7*a^3*b^4*x^3*(4*a*x + 11*Sqrt[b^2 + a^2*x^2])))/(b^2 + a*x*(a*x + Sqrt[b^2 + a^2*x^2])) + (
5*c*(a*x + Sqrt[b^2 + a^2*x^2])^2*(b^2 + 2*a*x*(a*x + Sqrt[b^2 + a^2*x^2]))^4*(-8*b^8 + 7*a^3*b^4*x^3*(a*x - 4
*Sqrt[b^2 + a^2*x^2]) + 56*a^7*x^7*(a*x + Sqrt[b^2 + a^2*x^2]) + 28*a^5*b^2*x^5*(4*a*x + 3*Sqrt[b^2 + a^2*x^2]
) - 7*a*b^6*x*(7*a*x + 4*Sqrt[b^2 + a^2*x^2])))/(b^12 + 1024*a^11*x^11*(a*x + Sqrt[b^2 + a^2*x^2]) + 256*a^9*b
^2*x^9*(13*a*x + 11*Sqrt[b^2 + a^2*x^2]) + 256*a^7*b^4*x^7*(16*a*x + 11*Sqrt[b^2 + a^2*x^2]) + 112*a^5*b^6*x^5
*(21*a*x + 11*Sqrt[b^2 + a^2*x^2]) + 20*a^3*b^8*x^3*(31*a*x + 11*Sqrt[b^2 + a^2*x^2]) + a*b^10*x*(61*a*x + 11*
Sqrt[b^2 + a^2*x^2])) + (21*a^2*d*(b^2 + 2*a*x*(a*x + Sqrt[b^2 + a^2*x^2]))^5*(7*b^4 + 6*a^3*x^3*(a*x + Sqrt[b
^2 + a^2*x^2]) + 3*a*b^2*x*(7*a*x + 6*Sqrt[b^2 + a^2*x^2])))/(b^10 + 256*a^9*x^9*(a*x + Sqrt[b^2 + a^2*x^2]) +
40*a^3*b^6*x^3*(7*a*x + 3*Sqrt[b^2 + a^2*x^2]) + 64*a^7*b^2*x^7*(11*a*x + 9*Sqrt[b^2 + a^2*x^2]) + a*b^8*x*(4
1*a*x + 9*Sqrt[b^2 + a^2*x^2]) + 16*a^5*b^4*x^5*(43*a*x + 27*Sqrt[b^2 + a^2*x^2]))))/(315*a^3*b^2*(a*x + Sqrt[
b^2 + a^2*x^2])^(5/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.26, size = 194, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {b^2+a^2 x^2} \left (-4 b^4 c x+25 a^2 b^2 d x-3 a^2 b^2 c x^3+60 a^4 d x^3+12 a^4 c x^5\right )}{15 a^2 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{7/2}}+\frac {2 \left (-8 b^6 c+35 a^2 b^4 d-49 a^2 b^4 c x^2+385 a^4 b^2 d x^2+21 a^4 b^2 c x^4+420 a^6 d x^4+84 a^6 c x^6\right )}{105 a^3 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b^2 + a^2*x^2]*(-4*b^4*c*x + 25*a^2*b^2*d*x - 3*a^2*b^2*c*x^3 + 60*a^4*d*x^3 + 12*a^4*c*x^5))/(15*a^2*
(a*x + Sqrt[b^2 + a^2*x^2])^(7/2)) + (2*(-8*b^6*c + 35*a^2*b^4*d - 49*a^2*b^4*c*x^2 + 385*a^4*b^2*d*x^2 + 21*a
^4*b^2*c*x^4 + 420*a^6*d*x^4 + 84*a^6*c*x^6))/(105*a^3*(a*x + Sqrt[b^2 + a^2*x^2])^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 112, normalized size = 0.58 \begin {gather*} -\frac {2 \, {\left (15 \, a^{4} c x^{4} + 8 \, b^{4} c - 35 \, a^{2} b^{2} d + {\left (a^{2} b^{2} c + 35 \, a^{4} d\right )} x^{2} - {\left (15 \, a^{3} c x^{3} + {\left (4 \, a b^{2} c + 35 \, a^{3} d\right )} x\right )} \sqrt {a^{2} x^{2} + b^{2}}\right )} \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}{105 \, a^{3} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*a^4*c*x^4 + 8*b^4*c - 35*a^2*b^2*d + (a^2*b^2*c + 35*a^4*d)*x^2 - (15*a^3*c*x^3 + (4*a*b^2*c + 35*a
^3*d)*x)*sqrt(a^2*x^2 + b^2))*sqrt(a*x + sqrt(a^2*x^2 + b^2))/(a^3*b^2)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {c x^{2} + d}{\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]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {c x^{2} + d}{\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]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {c x^{2} + 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**2+d)/(a*x+(a**2*x**2+b**2)**(1/2))**(1/2),x)

[Out]

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

________________________________________________________________________________________