3.184 \(\int \frac{A+B \tan (c+d x)}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=237 \[ \frac{(-25 B+23 i A) \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{(7 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{A+i B}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{(61 A+35 i B) \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

((-1/2 - I/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d)
+ (A + I*B)/(d*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) - ((7*A + (5*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(
5*a*d*Tan[c + d*x]^(5/2)) + (((23*I)*A - 25*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*a*d*Tan[c + d*x]^(3/2)) + ((61*
A + (35*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*a*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.739402, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {3596, 3598, 12, 3544, 205} \[ \frac{(-25 B+23 i A) \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{(7 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{A+i B}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{(61 A+35 i B) \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((-1/2 - I/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d)
+ (A + I*B)/(d*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) - ((7*A + (5*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(
5*a*d*Tan[c + d*x]^(5/2)) + (((23*I)*A - 25*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*a*d*Tan[c + d*x]^(3/2)) + ((61*
A + (35*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*a*d*Sqrt[Tan[c + d*x]])

Rule 3596

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

Rule 3598

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{A+B \tan (c+d x)}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{A+i B}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (7 A+5 i B)-3 a (i A-B) \tan (c+d x)\right )}{\tan ^{\frac{7}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac{A+i B}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(7 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 (23 i A-25 B)-a^2 (7 A+5 i B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{5 a^3}\\ &=\frac{A+i B}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(7 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{(23 i A-25 B) \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{8} a^3 (61 A+35 i B)+\frac{1}{4} a^3 (23 i A-25 B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^4}\\ &=\frac{A+i B}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(7 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{(23 i A-25 B) \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(61 A+35 i B) \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}+\frac{8 \int -\frac{15 a^4 (i A+B) \sqrt{a+i a \tan (c+d x)}}{16 \sqrt{\tan (c+d x)}} \, dx}{15 a^5}\\ &=\frac{A+i B}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(7 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{(23 i A-25 B) \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(61 A+35 i B) \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}-\frac{(i A+B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{2 a}\\ &=\frac{A+i B}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(7 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{(23 i A-25 B) \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(61 A+35 i B) \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}-\frac{(a (A-i B)) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}+\frac{A+i B}{d \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(7 A+5 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{(23 i A-25 B) \sqrt{a+i a \tan (c+d x)}}{15 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(61 A+35 i B) \sqrt{a+i a \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 5.09103, size = 241, normalized size = 1.02 \[ \frac{(A+B \tan (c+d x)) \left (-\frac{(A-i B) \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )}{\sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-\frac{\csc ^2(c+d x) (-5 (2 A+i B) \cos (c+d x)+(22 A+5 i B) \cos (3 (c+d x))+\sin (c+d x) ((-25 B+59 i A) \cos (2 (c+d x))+9 (5 B-7 i A)))}{15 \sqrt{\tan (c+d x)}}\right )}{2 d \sqrt{a+i a \tan (c+d x)} (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((-(((A - I*B)*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])/Sqrt[((
-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]) - (Csc[c + d*x]^2*(-5*(2*A + I*B)*Cos[c + d*x] + (
22*A + (5*I)*B)*Cos[3*(c + d*x)] + (9*((-7*I)*A + 5*B) + ((59*I)*A - 25*B)*Cos[2*(c + d*x)])*Sin[c + d*x]))/(1
5*Sqrt[Tan[c + d*x]]))*(A + B*Tan[c + d*x]))/(2*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]]
)

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Maple [B]  time = 0.098, size = 821, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x)

[Out]

1/60/d*(a*(1+I*tan(d*x+c)))^(1/2)*(15*I*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))
^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^5*a+30*I*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d
*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^4*a-15*A*2^(1/2)*ln(-(-2*2^(1/2)*
(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^5*a-15*I*B*2
^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*
tan(d*x+c)^3*a+140*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4+30*B*2^(1/2)*ln(-(-2*2^
(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^4*a-39
6*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+15*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2
)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a+244*A*(a*tan(d*x+c)
*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4+180*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*ta
n(d*x+c)^3+16*I*A*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)-144*A*(a*tan(d*x+c)*(1+I*tan(d
*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+40*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+24*
A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/a/tan(d*x+c)^(5/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)
/(-tan(d*x+c)+I)^2/(-I*a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.15469, size = 1694, normalized size = 7.15 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/60*(sqrt(2)*((206*I*A - 70*B)*e^(8*I*d*x + 8*I*c) + (-204*I*A + 180*B)*e^(6*I*d*x + 6*I*c) + (-80*I*A + 40*B
)*e^(4*I*d*x + 4*I*c) + (300*I*A - 180*B)*e^(2*I*d*x + 2*I*c) - 30*I*A + 30*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1
))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 15*(a*d*e^(8*I*d*x + 8*I*c)
- 3*a*d*e^(6*I*d*x + 6*I*c) + 3*a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((2*I*A^2 + 4*A*B - 2*I
*B^2)/(a*d^2))*log((a*d*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(
2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x +
2*I*c) + 1))*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) + 15*(a*d*e^(8*I*d*x + 8*I*c) - 3*a*d*e^(6*I*d*x
+ 6*I*c) + 3*a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2))*log
(-(a*d*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c)
+ I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I
*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)))/(a*d*e^(8*I*d*x + 8*I*c) - 3*a*d*e^(6*I*d*x + 6*I*c) + 3*a*d*e^(
4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(1/2)/tan(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.45037, size = 282, normalized size = 1.19 \begin{align*} \frac{\left (i + 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} +{\left (-\left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + \left (2 i - 2\right ) \, a^{5}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B}{{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a + 12 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} - 28 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} + 32 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} - 18 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} + 4 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

((I + 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)*a^5 + (-(2*I - 2)*(I*a*tan(d*x + c)
+ a)*a^4 + (2*I - 2)*a^5)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d*x + c) + a)*B)/((-2*I*(I*a*
tan(d*x + c) + a)^6*a + 12*I*(I*a*tan(d*x + c) + a)^5*a^2 - 28*I*(I*a*tan(d*x + c) + a)^4*a^3 + 32*I*(I*a*tan(
d*x + c) + a)^3*a^4 - 18*I*(I*a*tan(d*x + c) + a)^2*a^5 + 4*I*(I*a*tan(d*x + c) + a)*a^6)*d)