\(\int \sqrt {a+b x+c x^2} (A+C x^2) \, dx\) [180]

Optimal result

Integrand size = 22, antiderivative size = 157 \[ \int \sqrt {a+b x+c x^2} \left (A+C x^2\right ) \, dx=\frac {\left (16 A c^2+5 b^2 C-4 a c C\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}-\frac {5 b C \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {C x \left (a+b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (b^2-4 a c\right ) \left (16 A c^2+5 b^2 C-4 a c C\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}} \]

[Out]

-5/24*b*C*(c*x^2+b*x+a)^(3/2)/c^2+1/4*C*x*(c*x^2+b*x+a)^(3/2)/c-1/128*(-4*a*c+b^2)*(16*A*c^2-4*C*a*c+5*C*b^2)*
arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(7/2)+1/64*(16*A*c^2-4*C*a*c+5*C*b^2)*(2*c*x+b)*(c*x^2+b*
x+a)^(1/2)/c^3

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1675, 654, 626, 635, 212} \[ \int \sqrt {a+b x+c x^2} \left (A+C x^2\right ) \, dx=-\frac {\left (b^2-4 a c\right ) \left (-4 a c C+16 A c^2+5 b^2 C\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}}+\frac {(b+2 c x) \sqrt {a+b x+c x^2} \left (-4 a c C+16 A c^2+5 b^2 C\right )}{64 c^3}-\frac {5 b C \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {C x \left (a+b x+c x^2\right )^{3/2}}{4 c} \]

[In]

Int[Sqrt[a + b*x + c*x^2]*(A + C*x^2),x]

[Out]

((16*A*c^2 + 5*b^2*C - 4*a*c*C)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(64*c^3) - (5*b*C*(a + b*x + c*x^2)^(3/2))/
(24*c^2) + (C*x*(a + b*x + c*x^2)^(3/2))/(4*c) - ((b^2 - 4*a*c)*(16*A*c^2 + 5*b^2*C - 4*a*c*C)*ArcTanh[(b + 2*
c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(7/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1675

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {C x \left (a+b x+c x^2\right )^{3/2}}{4 c}+\frac {\int \left (4 A c-a C-\frac {5 b C x}{2}\right ) \sqrt {a+b x+c x^2} \, dx}{4 c} \\ & = -\frac {5 b C \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {C x \left (a+b x+c x^2\right )^{3/2}}{4 c}+\frac {\left (\frac {5 b^2 C}{2}+2 c (4 A c-a C)\right ) \int \sqrt {a+b x+c x^2} \, dx}{8 c^2} \\ & = \frac {\left (16 A c^2+5 b^2 C-4 a c C\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}-\frac {5 b C \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {C x \left (a+b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (\left (b^2-4 a c\right ) \left (16 A c^2+5 b^2 C-4 a c C\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{128 c^3} \\ & = \frac {\left (16 A c^2+5 b^2 C-4 a c C\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}-\frac {5 b C \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {C x \left (a+b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (\left (b^2-4 a c\right ) \left (16 A c^2+5 b^2 C-4 a c C\right )\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{64 c^3} \\ & = \frac {\left (16 A c^2+5 b^2 C-4 a c C\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}-\frac {5 b C \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {C x \left (a+b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (b^2-4 a c\right ) \left (16 A c^2+5 b^2 C-4 a c C\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.92 \[ \int \sqrt {a+b x+c x^2} \left (A+C x^2\right ) \, dx=\frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (48 A c^2 (b+2 c x)+C \left (15 b^3-10 b^2 c x+24 c^2 x \left (a+2 c x^2\right )+b \left (-52 a c+8 c^2 x^2\right )\right )\right )-3 \left (b^2-4 a c\right ) \left (16 A c^2+5 b^2 C-4 a c C\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{192 c^{7/2}} \]

[In]

Integrate[Sqrt[a + b*x + c*x^2]*(A + C*x^2),x]

[Out]

(Sqrt[c]*Sqrt[a + x*(b + c*x)]*(48*A*c^2*(b + 2*c*x) + C*(15*b^3 - 10*b^2*c*x + 24*c^2*x*(a + 2*c*x^2) + b*(-5
2*a*c + 8*c^2*x^2))) - 3*(b^2 - 4*a*c)*(16*A*c^2 + 5*b^2*C - 4*a*c*C)*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a +
x*(b + c*x)])])/(192*c^(7/2))

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {\left (48 c^{3} C \,x^{3}+8 b \,c^{2} C \,x^{2}+96 A \,c^{3} x +24 a \,c^{2} C x -10 C \,b^{2} c x +48 A b \,c^{2}-52 C a b c +15 C \,b^{3}\right ) \sqrt {c \,x^{2}+b x +a}}{192 c^{3}}+\frac {\left (64 A a \,c^{3}-16 A \,b^{2} c^{2}-16 C \,a^{2} c^{2}+24 C a \,b^{2} c -5 C \,b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {7}{2}}}\)	\(151\)
default	\(A \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )+C \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )\)	\(253\)

[In]

int((c*x^2+b*x+a)^(1/2)*(C*x^2+A),x,method=_RETURNVERBOSE)

[Out]

1/192*(48*C*c^3*x^3+8*C*b*c^2*x^2+96*A*c^3*x+24*C*a*c^2*x-10*C*b^2*c*x+48*A*b*c^2-52*C*a*b*c+15*C*b^3)/c^3*(c*
x^2+b*x+a)^(1/2)+1/128*(64*A*a*c^3-16*A*b^2*c^2-16*C*a^2*c^2+24*C*a*b^2*c-5*C*b^4)/c^(7/2)*ln((1/2*b+c*x)/c^(1
/2)+(c*x^2+b*x+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 355, normalized size of antiderivative = 2.26 \[ \int \sqrt {a+b x+c x^2} \left (A+C x^2\right ) \, dx=\left [-\frac {3 \, {\left (5 \, C b^{4} - 24 \, C a b^{2} c - 64 \, A a c^{3} + 16 \, {\left (C a^{2} + A b^{2}\right )} c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (48 \, C c^{4} x^{3} + 8 \, C b c^{3} x^{2} + 15 \, C b^{3} c - 52 \, C a b c^{2} + 48 \, A b c^{3} - 2 \, {\left (5 \, C b^{2} c^{2} - 12 \, C a c^{3} - 48 \, A c^{4}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{4}}, \frac {3 \, {\left (5 \, C b^{4} - 24 \, C a b^{2} c - 64 \, A a c^{3} + 16 \, {\left (C a^{2} + A b^{2}\right )} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (48 \, C c^{4} x^{3} + 8 \, C b c^{3} x^{2} + 15 \, C b^{3} c - 52 \, C a b c^{2} + 48 \, A b c^{3} - 2 \, {\left (5 \, C b^{2} c^{2} - 12 \, C a c^{3} - 48 \, A c^{4}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{4}}\right ] \]

[In]

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

[Out]

[-1/768*(3*(5*C*b^4 - 24*C*a*b^2*c - 64*A*a*c^3 + 16*(C*a^2 + A*b^2)*c^2)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b
^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(48*C*c^4*x^3 + 8*C*b*c^3*x^2 + 15*C*b^3*c - 52*
C*a*b*c^2 + 48*A*b*c^3 - 2*(5*C*b^2*c^2 - 12*C*a*c^3 - 48*A*c^4)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/384*(3*(5*C*
b^4 - 24*C*a*b^2*c - 64*A*a*c^3 + 16*(C*a^2 + A*b^2)*c^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b
)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(48*C*c^4*x^3 + 8*C*b*c^3*x^2 + 15*C*b^3*c - 52*C*a*b*c^2 + 48*A*b*c^3
- 2*(5*C*b^2*c^2 - 12*C*a*c^3 - 48*A*c^4)*x)*sqrt(c*x^2 + b*x + a))/c^4]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (151) = 302\).

Time = 0.43 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.96 \[ \int \sqrt {a+b x+c x^2} \left (A+C x^2\right ) \, dx=\begin {cases} \sqrt {a + b x + c x^{2}} \left (\frac {C b x^{2}}{24 c} + \frac {C x^{3}}{4} + \frac {x \left (A c + \frac {C a}{4} - \frac {5 C b^{2}}{48 c}\right )}{2 c} + \frac {A b - \frac {C a b}{12 c} - \frac {3 b \left (A c + \frac {C a}{4} - \frac {5 C b^{2}}{48 c}\right )}{4 c}}{c}\right ) + \left (A a - \frac {a \left (A c + \frac {C a}{4} - \frac {5 C b^{2}}{48 c}\right )}{2 c} - \frac {b \left (A b - \frac {C a b}{12 c} - \frac {3 b \left (A c + \frac {C a}{4} - \frac {5 C b^{2}}{48 c}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (- \frac {2 C a \left (a + b x\right )^{\frac {5}{2}}}{5 b^{2}} + \frac {C \left (a + b x\right )^{\frac {7}{2}}}{7 b^{2}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (A b^{2} + C a^{2}\right )}{3 b^{2}}\right )}{b} & \text {for}\: b \neq 0 \\\sqrt {a} \left (A x + \frac {C x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a + b*x + c*x**2)*(C*b*x**2/(24*c) + C*x**3/4 + x*(A*c + C*a/4 - 5*C*b**2/(48*c))/(2*c) + (A*b
- C*a*b/(12*c) - 3*b*(A*c + C*a/4 - 5*C*b**2/(48*c))/(4*c))/c) + (A*a - a*(A*c + C*a/4 - 5*C*b**2/(48*c))/(2*
c) - b*(A*b - C*a*b/(12*c) - 3*b*(A*c + C*a/4 - 5*C*b**2/(48*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sq
rt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c)
+ x)**2), True)), Ne(c, 0)), (2*(-2*C*a*(a + b*x)**(5/2)/(5*b**2) + C*(a + b*x)**(7/2)/(7*b**2) + (a + b*x)**
(3/2)*(A*b**2 + C*a**2)/(3*b**2))/b, Ne(b, 0)), (sqrt(a)*(A*x + C*x**3/3), True))

Maxima [F(-2)]

Exception generated. \[ \int \sqrt {a+b x+c x^2} \left (A+C x^2\right ) \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c*x^2+b*x+a)^(1/2)*(C*x^2+A),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(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.01 \[ \int \sqrt {a+b x+c x^2} \left (A+C x^2\right ) \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, C x + \frac {C b}{c}\right )} x - \frac {5 \, C b^{2} c - 12 \, C a c^{2} - 48 \, A c^{3}}{c^{3}}\right )} x + \frac {15 \, C b^{3} - 52 \, C a b c + 48 \, A b c^{2}}{c^{3}}\right )} + \frac {{\left (5 \, C b^{4} - 24 \, C a b^{2} c + 16 \, C a^{2} c^{2} + 16 \, A b^{2} c^{2} - 64 \, A a c^{3}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {7}{2}}} \]

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*C*x + C*b/c)*x - (5*C*b^2*c - 12*C*a*c^2 - 48*A*c^3)/c^3)*x + (15*C*b^3 -
52*C*a*b*c + 48*A*b*c^2)/c^3) + 1/128*(5*C*b^4 - 24*C*a*b^2*c + 16*C*a^2*c^2 + 16*A*b^2*c^2 - 64*A*a*c^3)*log
(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(7/2)

Mupad [B] (verification not implemented)

Time = 13.75 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.53 \[ \int \sqrt {a+b x+c x^2} \left (A+C x^2\right ) \, dx=A\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}-\frac {C\,a\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}+\frac {A\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}-\frac {5\,C\,b\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}+\frac {C\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \]

[In]

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

[Out]

A*(x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) - (C*a*((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c
^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c) + (A*log((b/2 + c*x)/c^(1/2) + (a + b*x +
c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2)) - (5*C*b*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3
- 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/(8*c) + (
C*x*(a + b*x + c*x^2)^(3/2))/(4*c)