3.595 \(\int \frac {a+c x^2}{(d+e x)^3 \sqrt {f+g x}} \, dx\)

Optimal. Leaf size=178 \[ -\frac {\sqrt {f+g x} \left (a+\frac {c d^2}{e^2}\right )}{2 (d+e x)^2 (e f-d g)}-\frac {\left (3 a e^2 g^2+c \left (3 d^2 g^2-8 d e f g+8 e^2 f^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{5/2} (e f-d g)^{5/2}}+\frac {\sqrt {f+g x} \left (3 a e^2 g+c d (8 e f-5 d g)\right )}{4 e^2 (d+e x) (e f-d g)^2} \]

[Out]

-1/4*(3*a*e^2*g^2+c*(3*d^2*g^2-8*d*e*f*g+8*e^2*f^2))*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))/e^(5/2)/(
-d*g+e*f)^(5/2)-1/2*(a+c*d^2/e^2)*(g*x+f)^(1/2)/(-d*g+e*f)/(e*x+d)^2+1/4*(3*a*e^2*g+c*d*(-5*d*g+8*e*f))*(g*x+f
)^(1/2)/e^2/(-d*g+e*f)^2/(e*x+d)

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Rubi [A]  time = 0.30, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {898, 1157, 385, 208} \[ -\frac {\left (3 a e^2 g^2+c \left (3 d^2 g^2-8 d e f g+8 e^2 f^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{5/2} (e f-d g)^{5/2}}-\frac {\sqrt {f+g x} \left (a+\frac {c d^2}{e^2}\right )}{2 (d+e x)^2 (e f-d g)}+\frac {\sqrt {f+g x} \left (3 a e^2 g+c d (8 e f-5 d g)\right )}{4 e^2 (d+e x) (e f-d g)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)/((d + e*x)^3*Sqrt[f + g*x]),x]

[Out]

-((a + (c*d^2)/e^2)*Sqrt[f + g*x])/(2*(e*f - d*g)*(d + e*x)^2) + ((3*a*e^2*g + c*d*(8*e*f - 5*d*g))*Sqrt[f + g
*x])/(4*e^2*(e*f - d*g)^2*(d + e*x)) - ((3*a*e^2*g^2 + c*(8*e^2*f^2 - 8*d*e*f*g + 3*d^2*g^2))*ArcTanh[(Sqrt[e]
*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(4*e^(5/2)*(e*f - d*g)^(5/2))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 898

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 + a*e^2)/e^2 - (2*c
*d*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rubi steps

\begin {align*} \int \frac {a+c x^2}{(d+e x)^3 \sqrt {f+g x}} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {\frac {c f^2+a g^2}{g^2}-\frac {2 c f x^2}{g^2}+\frac {c x^4}{g^2}}{\left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^3} \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=-\frac {\left (a+\frac {c d^2}{e^2}\right ) \sqrt {f+g x}}{2 (e f-d g) (d+e x)^2}+\frac {\operatorname {Subst}\left (\int \frac {-3 a+\frac {c d^2}{e^2}-\frac {4 c f^2}{g^2}+\frac {4 c (e f-d g) x^2}{e g^2}}{\left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^2} \, dx,x,\sqrt {f+g x}\right )}{2 (e f-d g)}\\ &=-\frac {\left (a+\frac {c d^2}{e^2}\right ) \sqrt {f+g x}}{2 (e f-d g) (d+e x)^2}+\frac {\left (3 a e^2 g+c d (8 e f-5 d g)\right ) \sqrt {f+g x}}{4 e^2 (e f-d g)^2 (d+e x)}+\frac {\left (3 a e^2 g^2+c \left (8 e^2 f^2-8 d e f g+3 d^2 g^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {-e f+d g}{g}+\frac {e x^2}{g}} \, dx,x,\sqrt {f+g x}\right )}{4 e^2 g (e f-d g)^2}\\ &=-\frac {\left (a+\frac {c d^2}{e^2}\right ) \sqrt {f+g x}}{2 (e f-d g) (d+e x)^2}+\frac {\left (3 a e^2 g+c d (8 e f-5 d g)\right ) \sqrt {f+g x}}{4 e^2 (e f-d g)^2 (d+e x)}-\frac {\left (3 a e^2 g^2+c \left (8 e^2 f^2-8 d e f g+3 d^2 g^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{5/2} (e f-d g)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.82, size = 207, normalized size = 1.16 \[ \frac {2 \left (\frac {\sqrt {e} g^2 \sqrt {f+g x} \left (a e^2+c d^2\right ) \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {e (f+g x)}{e f-d g}\right )}{(d g-e f)^3}-\frac {c d \left (\sqrt {e} \sqrt {f+g x} (d g-e f)+g (d+e x) \sqrt {d g-e f} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {d g-e f}}\right )\right )}{(d+e x) (e f-d g)^2}-\frac {c \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e f-d g}}\right )}{e^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^2)/((d + e*x)^3*Sqrt[f + g*x]),x]

[Out]

(2*(-((c*d*(Sqrt[e]*(-(e*f) + d*g)*Sqrt[f + g*x] + g*Sqrt[-(e*f) + d*g]*(d + e*x)*ArcTan[(Sqrt[e]*Sqrt[f + g*x
])/Sqrt[-(e*f) + d*g]]))/((e*f - d*g)^2*(d + e*x))) - (c*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/Sqr
t[e*f - d*g] + (Sqrt[e]*(c*d^2 + a*e^2)*g^2*Sqrt[f + g*x]*Hypergeometric2F1[1/2, 3, 3/2, (e*(f + g*x))/(e*f -
d*g)])/(-(e*f) + d*g)^3))/e^(5/2)

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fricas [B]  time = 1.07, size = 896, normalized size = 5.03 \[ \left [\frac {{\left (8 \, c d^{2} e^{2} f^{2} - 8 \, c d^{3} e f g + 3 \, {\left (c d^{4} + a d^{2} e^{2}\right )} g^{2} + {\left (8 \, c e^{4} f^{2} - 8 \, c d e^{3} f g + 3 \, {\left (c d^{2} e^{2} + a e^{4}\right )} g^{2}\right )} x^{2} + 2 \, {\left (8 \, c d e^{3} f^{2} - 8 \, c d^{2} e^{2} f g + 3 \, {\left (c d^{3} e + a d e^{3}\right )} g^{2}\right )} x\right )} \sqrt {e^{2} f - d e g} \log \left (\frac {e g x + 2 \, e f - d g - 2 \, \sqrt {e^{2} f - d e g} \sqrt {g x + f}}{e x + d}\right ) + 2 \, {\left (2 \, {\left (3 \, c d^{2} e^{3} - a e^{5}\right )} f^{2} - {\left (9 \, c d^{3} e^{2} - 7 \, a d e^{4}\right )} f g + {\left (3 \, c d^{4} e - 5 \, a d^{2} e^{3}\right )} g^{2} + {\left (8 \, c d e^{4} f^{2} - {\left (13 \, c d^{2} e^{3} - 3 \, a e^{5}\right )} f g + {\left (5 \, c d^{3} e^{2} - 3 \, a d e^{4}\right )} g^{2}\right )} x\right )} \sqrt {g x + f}}{8 \, {\left (d^{2} e^{6} f^{3} - 3 \, d^{3} e^{5} f^{2} g + 3 \, d^{4} e^{4} f g^{2} - d^{5} e^{3} g^{3} + {\left (e^{8} f^{3} - 3 \, d e^{7} f^{2} g + 3 \, d^{2} e^{6} f g^{2} - d^{3} e^{5} g^{3}\right )} x^{2} + 2 \, {\left (d e^{7} f^{3} - 3 \, d^{2} e^{6} f^{2} g + 3 \, d^{3} e^{5} f g^{2} - d^{4} e^{4} g^{3}\right )} x\right )}}, \frac {{\left (8 \, c d^{2} e^{2} f^{2} - 8 \, c d^{3} e f g + 3 \, {\left (c d^{4} + a d^{2} e^{2}\right )} g^{2} + {\left (8 \, c e^{4} f^{2} - 8 \, c d e^{3} f g + 3 \, {\left (c d^{2} e^{2} + a e^{4}\right )} g^{2}\right )} x^{2} + 2 \, {\left (8 \, c d e^{3} f^{2} - 8 \, c d^{2} e^{2} f g + 3 \, {\left (c d^{3} e + a d e^{3}\right )} g^{2}\right )} x\right )} \sqrt {-e^{2} f + d e g} \arctan \left (\frac {\sqrt {-e^{2} f + d e g} \sqrt {g x + f}}{e g x + e f}\right ) + {\left (2 \, {\left (3 \, c d^{2} e^{3} - a e^{5}\right )} f^{2} - {\left (9 \, c d^{3} e^{2} - 7 \, a d e^{4}\right )} f g + {\left (3 \, c d^{4} e - 5 \, a d^{2} e^{3}\right )} g^{2} + {\left (8 \, c d e^{4} f^{2} - {\left (13 \, c d^{2} e^{3} - 3 \, a e^{5}\right )} f g + {\left (5 \, c d^{3} e^{2} - 3 \, a d e^{4}\right )} g^{2}\right )} x\right )} \sqrt {g x + f}}{4 \, {\left (d^{2} e^{6} f^{3} - 3 \, d^{3} e^{5} f^{2} g + 3 \, d^{4} e^{4} f g^{2} - d^{5} e^{3} g^{3} + {\left (e^{8} f^{3} - 3 \, d e^{7} f^{2} g + 3 \, d^{2} e^{6} f g^{2} - d^{3} e^{5} g^{3}\right )} x^{2} + 2 \, {\left (d e^{7} f^{3} - 3 \, d^{2} e^{6} f^{2} g + 3 \, d^{3} e^{5} f g^{2} - d^{4} e^{4} g^{3}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)/(e*x+d)^3/(g*x+f)^(1/2),x, algorithm="fricas")

[Out]

[1/8*((8*c*d^2*e^2*f^2 - 8*c*d^3*e*f*g + 3*(c*d^4 + a*d^2*e^2)*g^2 + (8*c*e^4*f^2 - 8*c*d*e^3*f*g + 3*(c*d^2*e
^2 + a*e^4)*g^2)*x^2 + 2*(8*c*d*e^3*f^2 - 8*c*d^2*e^2*f*g + 3*(c*d^3*e + a*d*e^3)*g^2)*x)*sqrt(e^2*f - d*e*g)*
log((e*g*x + 2*e*f - d*g - 2*sqrt(e^2*f - d*e*g)*sqrt(g*x + f))/(e*x + d)) + 2*(2*(3*c*d^2*e^3 - a*e^5)*f^2 -
(9*c*d^3*e^2 - 7*a*d*e^4)*f*g + (3*c*d^4*e - 5*a*d^2*e^3)*g^2 + (8*c*d*e^4*f^2 - (13*c*d^2*e^3 - 3*a*e^5)*f*g
+ (5*c*d^3*e^2 - 3*a*d*e^4)*g^2)*x)*sqrt(g*x + f))/(d^2*e^6*f^3 - 3*d^3*e^5*f^2*g + 3*d^4*e^4*f*g^2 - d^5*e^3*
g^3 + (e^8*f^3 - 3*d*e^7*f^2*g + 3*d^2*e^6*f*g^2 - d^3*e^5*g^3)*x^2 + 2*(d*e^7*f^3 - 3*d^2*e^6*f^2*g + 3*d^3*e
^5*f*g^2 - d^4*e^4*g^3)*x), 1/4*((8*c*d^2*e^2*f^2 - 8*c*d^3*e*f*g + 3*(c*d^4 + a*d^2*e^2)*g^2 + (8*c*e^4*f^2 -
8*c*d*e^3*f*g + 3*(c*d^2*e^2 + a*e^4)*g^2)*x^2 + 2*(8*c*d*e^3*f^2 - 8*c*d^2*e^2*f*g + 3*(c*d^3*e + a*d*e^3)*g
^2)*x)*sqrt(-e^2*f + d*e*g)*arctan(sqrt(-e^2*f + d*e*g)*sqrt(g*x + f)/(e*g*x + e*f)) + (2*(3*c*d^2*e^3 - a*e^5
)*f^2 - (9*c*d^3*e^2 - 7*a*d*e^4)*f*g + (3*c*d^4*e - 5*a*d^2*e^3)*g^2 + (8*c*d*e^4*f^2 - (13*c*d^2*e^3 - 3*a*e
^5)*f*g + (5*c*d^3*e^2 - 3*a*d*e^4)*g^2)*x)*sqrt(g*x + f))/(d^2*e^6*f^3 - 3*d^3*e^5*f^2*g + 3*d^4*e^4*f*g^2 -
d^5*e^3*g^3 + (e^8*f^3 - 3*d*e^7*f^2*g + 3*d^2*e^6*f*g^2 - d^3*e^5*g^3)*x^2 + 2*(d*e^7*f^3 - 3*d^2*e^6*f^2*g +
3*d^3*e^5*f*g^2 - d^4*e^4*g^3)*x)]

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giac [A]  time = 0.21, size = 278, normalized size = 1.56 \[ \frac {{\left (3 \, c d^{2} g^{2} - 8 \, c d f g e + 8 \, c f^{2} e^{2} + 3 \, a g^{2} e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right )}{4 \, {\left (d^{2} g^{2} e^{2} - 2 \, d f g e^{3} + f^{2} e^{4}\right )} \sqrt {d g e - f e^{2}}} - \frac {3 \, \sqrt {g x + f} c d^{3} g^{3} + 5 \, {\left (g x + f\right )}^{\frac {3}{2}} c d^{2} g^{2} e - 11 \, \sqrt {g x + f} c d^{2} f g^{2} e - 8 \, {\left (g x + f\right )}^{\frac {3}{2}} c d f g e^{2} + 8 \, \sqrt {g x + f} c d f^{2} g e^{2} - 5 \, \sqrt {g x + f} a d g^{3} e^{2} - 3 \, {\left (g x + f\right )}^{\frac {3}{2}} a g^{2} e^{3} + 5 \, \sqrt {g x + f} a f g^{2} e^{3}}{4 \, {\left (d^{2} g^{2} e^{2} - 2 \, d f g e^{3} + f^{2} e^{4}\right )} {\left (d g + {\left (g x + f\right )} e - f e\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)/(e*x+d)^3/(g*x+f)^(1/2),x, algorithm="giac")

[Out]

1/4*(3*c*d^2*g^2 - 8*c*d*f*g*e + 8*c*f^2*e^2 + 3*a*g^2*e^2)*arctan(sqrt(g*x + f)*e/sqrt(d*g*e - f*e^2))/((d^2*
g^2*e^2 - 2*d*f*g*e^3 + f^2*e^4)*sqrt(d*g*e - f*e^2)) - 1/4*(3*sqrt(g*x + f)*c*d^3*g^3 + 5*(g*x + f)^(3/2)*c*d
^2*g^2*e - 11*sqrt(g*x + f)*c*d^2*f*g^2*e - 8*(g*x + f)^(3/2)*c*d*f*g*e^2 + 8*sqrt(g*x + f)*c*d*f^2*g*e^2 - 5*
sqrt(g*x + f)*a*d*g^3*e^2 - 3*(g*x + f)^(3/2)*a*g^2*e^3 + 5*sqrt(g*x + f)*a*f*g^2*e^3)/((d^2*g^2*e^2 - 2*d*f*g
*e^3 + f^2*e^4)*(d*g + (g*x + f)*e - f*e)^2)

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maple [B]  time = 0.02, size = 384, normalized size = 2.16 \[ \frac {3 a \,g^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \sqrt {\left (d g -e f \right ) e}}+\frac {3 c \,d^{2} g^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \sqrt {\left (d g -e f \right ) e}\, e^{2}}-\frac {2 c d f g \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \sqrt {\left (d g -e f \right ) e}\, e}+\frac {2 c \,f^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \sqrt {\left (d g -e f \right ) e}}+\frac {\frac {\left (3 a \,e^{2} g -5 c \,d^{2} g +8 c d e f \right ) \left (g x +f \right )^{\frac {3}{2}} g}{4 \left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) e}+\frac {\left (5 a \,e^{2} g -3 c \,d^{2} g +8 c d e f \right ) \sqrt {g x +f}\, g}{4 \left (d g -e f \right ) e^{2}}}{\left (d g -e f +\left (g x +f \right ) e \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)/(e*x+d)^3/(g*x+f)^(1/2),x)

[Out]

2*(1/8*g*(3*a*e^2*g-5*c*d^2*g+8*c*d*e*f)/e/(d^2*g^2-2*d*e*f*g+e^2*f^2)*(g*x+f)^(3/2)+1/8*(5*a*e^2*g-3*c*d^2*g+
8*c*d*e*f)/e^2*g/(d*g-e*f)*(g*x+f)^(1/2))/(e*(g*x+f)+d*g-e*f)^2+3/4/(d^2*g^2-2*d*e*f*g+e^2*f^2)/((d*g-e*f)*e)^
(1/2)*arctan((g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)*e)*a*g^2+3/4/(d^2*g^2-2*d*e*f*g+e^2*f^2)/e^2/((d*g-e*f)*e)^(1/2
)*arctan((g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)*e)*c*d^2*g^2-2/(d^2*g^2-2*d*e*f*g+e^2*f^2)/e/((d*g-e*f)*e)^(1/2)*ar
ctan((g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)*e)*c*d*f*g+2/(d^2*g^2-2*d*e*f*g+e^2*f^2)/((d*g-e*f)*e)^(1/2)*arctan((g*
x+f)^(1/2)/((d*g-e*f)*e)^(1/2)*e)*c*f^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)/(e*x+d)^3/(g*x+f)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d*g-e*f>0)', see `assume?` for
more details)Is d*g-e*f positive or negative?

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mupad [B]  time = 2.91, size = 224, normalized size = 1.26 \[ \frac {\frac {\sqrt {f+g\,x}\,\left (-3\,c\,d^2\,g^2+8\,c\,f\,d\,e\,g+5\,a\,e^2\,g^2\right )}{4\,e^2\,\left (d\,g-e\,f\right )}+\frac {{\left (f+g\,x\right )}^{3/2}\,\left (-5\,c\,d^2\,g^2+8\,c\,f\,d\,e\,g+3\,a\,e^2\,g^2\right )}{4\,e\,{\left (d\,g-e\,f\right )}^2}}{e^2\,{\left (f+g\,x\right )}^2-\left (f+g\,x\right )\,\left (2\,e^2\,f-2\,d\,e\,g\right )+d^2\,g^2+e^2\,f^2-2\,d\,e\,f\,g}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}}{\sqrt {d\,g-e\,f}}\right )\,\left (3\,c\,d^2\,g^2-8\,c\,d\,e\,f\,g+8\,c\,e^2\,f^2+3\,a\,e^2\,g^2\right )}{4\,e^{5/2}\,{\left (d\,g-e\,f\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)/((f + g*x)^(1/2)*(d + e*x)^3),x)

[Out]

(((f + g*x)^(1/2)*(5*a*e^2*g^2 - 3*c*d^2*g^2 + 8*c*d*e*f*g))/(4*e^2*(d*g - e*f)) + ((f + g*x)^(3/2)*(3*a*e^2*g
^2 - 5*c*d^2*g^2 + 8*c*d*e*f*g))/(4*e*(d*g - e*f)^2))/(e^2*(f + g*x)^2 - (f + g*x)*(2*e^2*f - 2*d*e*g) + d^2*g
^2 + e^2*f^2 - 2*d*e*f*g) + (atan((e^(1/2)*(f + g*x)^(1/2))/(d*g - e*f)^(1/2))*(3*a*e^2*g^2 + 3*c*d^2*g^2 + 8*
c*e^2*f^2 - 8*c*d*e*f*g))/(4*e^(5/2)*(d*g - e*f)^(5/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+a)/(e*x+d)**3/(g*x+f)**(1/2),x)

[Out]

Timed out

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