3.28 \(\int (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=140 \[ -\frac {4 a^4 (A-i B) \tan (c+d x)}{d}-\frac {8 a^4 (B+i A) \log (\cos (c+d x))}{d}+8 a^4 x (A-i B)+\frac {(B+i A) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}+\frac {a (B+i A) (a+i a \tan (c+d x))^3}{3 d}+\frac {B (a+i a \tan (c+d x))^4}{4 d} \]

[Out]

8*a^4*(A-I*B)*x-8*a^4*(I*A+B)*ln(cos(d*x+c))/d-4*a^4*(A-I*B)*tan(d*x+c)/d+1/3*a*(I*A+B)*(a+I*a*tan(d*x+c))^3/d
+1/4*B*(a+I*a*tan(d*x+c))^4/d+(I*A+B)*(a^2+I*a^2*tan(d*x+c))^2/d

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Rubi [A]  time = 0.11, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3527, 3478, 3477, 3475} \[ -\frac {4 a^4 (A-i B) \tan (c+d x)}{d}+\frac {(B+i A) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\frac {8 a^4 (B+i A) \log (\cos (c+d x))}{d}+8 a^4 x (A-i B)+\frac {a (B+i A) (a+i a \tan (c+d x))^3}{3 d}+\frac {B (a+i a \tan (c+d x))^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

8*a^4*(A - I*B)*x - (8*a^4*(I*A + B)*Log[Cos[c + d*x]])/d - (4*a^4*(A - I*B)*Tan[c + d*x])/d + (a*(I*A + B)*(a
+ I*a*Tan[c + d*x])^3)/(3*d) + (B*(a + I*a*Tan[c + d*x])^4)/(4*d) + ((I*A + B)*(a^2 + I*a^2*Tan[c + d*x])^2)/
d

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3527

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*
(a + b*Tan[e + f*x])^m)/(f*m), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

Rubi steps

\begin {align*} \int (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx &=\frac {B (a+i a \tan (c+d x))^4}{4 d}-(-A+i B) \int (a+i a \tan (c+d x))^4 \, dx\\ &=\frac {a (i A+B) (a+i a \tan (c+d x))^3}{3 d}+\frac {B (a+i a \tan (c+d x))^4}{4 d}+(2 a (A-i B)) \int (a+i a \tan (c+d x))^3 \, dx\\ &=\frac {a (i A+B) (a+i a \tan (c+d x))^3}{3 d}+\frac {B (a+i a \tan (c+d x))^4}{4 d}+\frac {(i A+B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}+\left (4 a^2 (A-i B)\right ) \int (a+i a \tan (c+d x))^2 \, dx\\ &=8 a^4 (A-i B) x-\frac {4 a^4 (A-i B) \tan (c+d x)}{d}+\frac {a (i A+B) (a+i a \tan (c+d x))^3}{3 d}+\frac {B (a+i a \tan (c+d x))^4}{4 d}+\frac {(i A+B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}+\left (8 a^4 (i A+B)\right ) \int \tan (c+d x) \, dx\\ &=8 a^4 (A-i B) x-\frac {8 a^4 (i A+B) \log (\cos (c+d x))}{d}-\frac {4 a^4 (A-i B) \tan (c+d x)}{d}+\frac {a (i A+B) (a+i a \tan (c+d x))^3}{3 d}+\frac {B (a+i a \tan (c+d x))^4}{4 d}+\frac {(i A+B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}\\ \end {align*}

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Mathematica [B]  time = 3.70, size = 448, normalized size = 3.20 \[ \frac {a^4 \sec (c) \sec ^4(c+d x) \left (3 \cos (c) \left ((-6 B-6 i A) \log \left (\cos ^2(c+d x)\right )+12 A d x-4 i A-12 i B d x-7 B\right )+6 \cos (c+2 d x) \left ((-2 B-2 i A) \log \left (\cos ^2(c+d x)\right )+4 A d x-i A-4 i B d x-2 B\right )-32 A \sin (c+2 d x)+12 A \sin (3 c+2 d x)-11 A \sin (3 c+4 d x)-6 i A \cos (3 c+2 d x)+24 A d x \cos (3 c+2 d x)+6 A d x \cos (3 c+4 d x)+6 A d x \cos (5 c+4 d x)-12 i A \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )-3 i A \cos (3 c+4 d x) \log \left (\cos ^2(c+d x)\right )-3 i A \cos (5 c+4 d x) \log \left (\cos ^2(c+d x)\right )+33 A \sin (c)+38 i B \sin (c+2 d x)-18 i B \sin (3 c+2 d x)+14 i B \sin (3 c+4 d x)-12 B \cos (3 c+2 d x)-24 i B d x \cos (3 c+2 d x)-6 i B d x \cos (3 c+4 d x)-6 i B d x \cos (5 c+4 d x)-12 B \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )-3 B \cos (3 c+4 d x) \log \left (\cos ^2(c+d x)\right )-3 B \cos (5 c+4 d x) \log \left (\cos ^2(c+d x)\right )-42 i B \sin (c)\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

(a^4*Sec[c]*Sec[c + d*x]^4*((-6*I)*A*Cos[3*c + 2*d*x] - 12*B*Cos[3*c + 2*d*x] + 24*A*d*x*Cos[3*c + 2*d*x] - (2
4*I)*B*d*x*Cos[3*c + 2*d*x] + 6*A*d*x*Cos[3*c + 4*d*x] - (6*I)*B*d*x*Cos[3*c + 4*d*x] + 6*A*d*x*Cos[5*c + 4*d*
x] - (6*I)*B*d*x*Cos[5*c + 4*d*x] - (12*I)*A*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] - 12*B*Cos[3*c + 2*d*x]*Log[
Cos[c + d*x]^2] - (3*I)*A*Cos[3*c + 4*d*x]*Log[Cos[c + d*x]^2] - 3*B*Cos[3*c + 4*d*x]*Log[Cos[c + d*x]^2] - (3
*I)*A*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]^2] - 3*B*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]^2] + 3*Cos[c]*((-4*I)*A - 7
*B + 12*A*d*x - (12*I)*B*d*x + ((-6*I)*A - 6*B)*Log[Cos[c + d*x]^2]) + 6*Cos[c + 2*d*x]*((-I)*A - 2*B + 4*A*d*
x - (4*I)*B*d*x + ((-2*I)*A - 2*B)*Log[Cos[c + d*x]^2]) + 33*A*Sin[c] - (42*I)*B*Sin[c] - 32*A*Sin[c + 2*d*x]
+ (38*I)*B*Sin[c + 2*d*x] + 12*A*Sin[3*c + 2*d*x] - (18*I)*B*Sin[3*c + 2*d*x] - 11*A*Sin[3*c + 4*d*x] + (14*I)
*B*Sin[3*c + 4*d*x]))/(12*d)

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fricas [A]  time = 0.48, size = 230, normalized size = 1.64 \[ \frac {{\left (-72 i \, A - 120 \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-180 i \, A - 252 \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-152 i \, A - 200 \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-44 i \, A - 56 \, B\right )} a^{4} + {\left ({\left (-24 i \, A - 24 \, B\right )} a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (-96 i \, A - 96 \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-144 i \, A - 144 \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-96 i \, A - 96 \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-24 i \, A - 24 \, B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*((-72*I*A - 120*B)*a^4*e^(6*I*d*x + 6*I*c) + (-180*I*A - 252*B)*a^4*e^(4*I*d*x + 4*I*c) + (-152*I*A - 200*
B)*a^4*e^(2*I*d*x + 2*I*c) + (-44*I*A - 56*B)*a^4 + ((-24*I*A - 24*B)*a^4*e^(8*I*d*x + 8*I*c) + (-96*I*A - 96*
B)*a^4*e^(6*I*d*x + 6*I*c) + (-144*I*A - 144*B)*a^4*e^(4*I*d*x + 4*I*c) + (-96*I*A - 96*B)*a^4*e^(2*I*d*x + 2*
I*c) + (-24*I*A - 24*B)*a^4)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) +
6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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giac [B]  time = 0.75, size = 408, normalized size = 2.91 \[ \frac {-24 i \, A a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 24 \, B a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 96 i \, A a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 96 \, B a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 144 i \, A a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 144 \, B a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 96 i \, A a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 96 \, B a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 72 i \, A a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 120 \, B a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 180 i \, A a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 252 \, B a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 152 i \, A a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 200 \, B a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 24 i \, A a^{4} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 24 \, B a^{4} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 44 i \, A a^{4} - 56 \, B a^{4}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/3*(-24*I*A*a^4*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 24*B*a^4*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*
x + 2*I*c) + 1) - 96*I*A*a^4*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 96*B*a^4*e^(6*I*d*x + 6*I*c)*l
og(e^(2*I*d*x + 2*I*c) + 1) - 144*I*A*a^4*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 144*B*a^4*e^(4*I*
d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 96*I*A*a^4*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 96*B
*a^4*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 72*I*A*a^4*e^(6*I*d*x + 6*I*c) - 120*B*a^4*e^(6*I*d*x
+ 6*I*c) - 180*I*A*a^4*e^(4*I*d*x + 4*I*c) - 252*B*a^4*e^(4*I*d*x + 4*I*c) - 152*I*A*a^4*e^(2*I*d*x + 2*I*c) -
200*B*a^4*e^(2*I*d*x + 2*I*c) - 24*I*A*a^4*log(e^(2*I*d*x + 2*I*c) + 1) - 24*B*a^4*log(e^(2*I*d*x + 2*I*c) +
1) - 44*I*A*a^4 - 56*B*a^4)/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e
^(2*I*d*x + 2*I*c) + d)

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maple [A]  time = 0.02, size = 194, normalized size = 1.39 \[ -\frac {4 i a^{4} B \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{4} B \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {2 i a^{4} A \left (\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {a^{4} A \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {8 i a^{4} B \tan \left (d x +c \right )}{d}-\frac {7 a^{4} B \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {7 A \,a^{4} \tan \left (d x +c \right )}{d}+\frac {4 i a^{4} A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {4 a^{4} B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {8 i a^{4} B \arctan \left (\tan \left (d x +c \right )\right )}{d}+\frac {8 a^{4} A \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)

[Out]

-4/3*I/d*a^4*B*tan(d*x+c)^3+1/4/d*a^4*B*tan(d*x+c)^4-2*I/d*a^4*A*tan(d*x+c)^2+1/3/d*a^4*A*tan(d*x+c)^3+8*I/d*a
^4*B*tan(d*x+c)-7/2/d*a^4*B*tan(d*x+c)^2-7/d*A*a^4*tan(d*x+c)+4*I/d*a^4*A*ln(1+tan(d*x+c)^2)+4/d*a^4*B*ln(1+ta
n(d*x+c)^2)-8*I/d*a^4*B*arctan(tan(d*x+c))+8/d*a^4*A*arctan(tan(d*x+c))

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maxima [A]  time = 1.63, size = 113, normalized size = 0.81 \[ \frac {3 \, B a^{4} \tan \left (d x + c\right )^{4} + 4 \, {\left (A - 4 i \, B\right )} a^{4} \tan \left (d x + c\right )^{3} + {\left (-24 i \, A - 42 \, B\right )} a^{4} \tan \left (d x + c\right )^{2} + 96 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{4} + 12 \, {\left (4 i \, A + 4 \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 12 \, {\left (7 \, A - 8 i \, B\right )} a^{4} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(3*B*a^4*tan(d*x + c)^4 + 4*(A - 4*I*B)*a^4*tan(d*x + c)^3 + (-24*I*A - 42*B)*a^4*tan(d*x + c)^2 + 96*(d*
x + c)*(A - I*B)*a^4 + 12*(4*I*A + 4*B)*a^4*log(tan(d*x + c)^2 + 1) - 12*(7*A - 8*I*B)*a^4*tan(d*x + c))/d

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mupad [B]  time = 6.12, size = 181, normalized size = 1.29 \[ \frac {B\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (8\,B\,a^4+A\,a^4\,8{}\mathrm {i}\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a^4\,\left (A-B\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2}+\frac {B\,a^4}{2}+\frac {a^4\,\left (3\,B+A\,1{}\mathrm {i}\right )}{2}\right )}{d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-3\,a^4\,\left (A-B\,1{}\mathrm {i}\right )+a^4\,\left (B+A\,3{}\mathrm {i}\right )\,1{}\mathrm {i}+B\,a^4\,1{}\mathrm {i}+a^4\,\left (3\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {B\,a^4\,1{}\mathrm {i}}{3}+\frac {a^4\,\left (3\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(log(tan(c + d*x) + 1i)*(A*a^4*8i + 8*B*a^4))/d - (tan(c + d*x)^3*((B*a^4*1i)/3 + (a^4*(A*1i + 3*B)*1i)/3))/d
- (tan(c + d*x)^2*((a^4*(A - B*1i)*3i)/2 + (B*a^4)/2 + (a^4*(A*1i + 3*B))/2))/d + (tan(c + d*x)*(a^4*(A*3i + B
)*1i - 3*a^4*(A - B*1i) + B*a^4*1i + a^4*(A*1i + 3*B)*1i))/d + (B*a^4*tan(c + d*x)^4)/(4*d)

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sympy [B]  time = 0.98, size = 238, normalized size = 1.70 \[ - \frac {8 i a^{4} \left (A - i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {44 i A a^{4} + 56 B a^{4} + \left (152 i A a^{4} e^{2 i c} + 200 B a^{4} e^{2 i c}\right ) e^{2 i d x} + \left (180 i A a^{4} e^{4 i c} + 252 B a^{4} e^{4 i c}\right ) e^{4 i d x} + \left (72 i A a^{4} e^{6 i c} + 120 B a^{4} e^{6 i c}\right ) e^{6 i d x}}{- 3 d e^{8 i c} e^{8 i d x} - 12 d e^{6 i c} e^{6 i d x} - 18 d e^{4 i c} e^{4 i d x} - 12 d e^{2 i c} e^{2 i d x} - 3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

-8*I*a**4*(A - I*B)*log(exp(2*I*d*x) + exp(-2*I*c))/d + (44*I*A*a**4 + 56*B*a**4 + (152*I*A*a**4*exp(2*I*c) +
200*B*a**4*exp(2*I*c))*exp(2*I*d*x) + (180*I*A*a**4*exp(4*I*c) + 252*B*a**4*exp(4*I*c))*exp(4*I*d*x) + (72*I*A
*a**4*exp(6*I*c) + 120*B*a**4*exp(6*I*c))*exp(6*I*d*x))/(-3*d*exp(8*I*c)*exp(8*I*d*x) - 12*d*exp(6*I*c)*exp(6*
I*d*x) - 18*d*exp(4*I*c)*exp(4*I*d*x) - 12*d*exp(2*I*c)*exp(2*I*d*x) - 3*d)

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