Integrand size = 30, antiderivative size = 155 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a \sqrt {c-i d} f}+\frac {(i c-2 d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a (c+i d)^{3/2} f}-\frac {\sqrt {c+d \tan (e+f x)}}{2 (i c-d) f (a+i a \tan (e+f x))} \]
1/2*(I*c-2*d)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/a/(c+I*d)^(3/2 )/f-1/2*I*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a/f/(c-I*d)^(1/2)- 1/2*(c+d*tan(f*x+e))^(1/2)/(I*c-d)/f/(a+I*a*tan(f*x+e))
Time = 1.06 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \left (\frac {(c+i d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}-\frac {(c+2 i d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}+\frac {i \sqrt {c+d \tan (e+f x)}}{-i+\tan (e+f x)}\right )}{2 a (c+i d) f} \]
((-1/2*I)*(((c + I*d)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sqr t[c - I*d] - ((c + (2*I)*d)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d] ])/Sqrt[c + I*d] + (I*Sqrt[c + d*Tan[e + f*x]])/(-I + Tan[e + f*x])))/(a*( c + I*d)*f)
Time = 0.71 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4035, 27, 3042, 4022, 3042, 4020, 25, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 4035 |
| \(\displaystyle \frac {\int \frac {a (2 i c-3 d)+i a d \tan (e+f x)}{2 \sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{2 f (-d+i c) (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (2 i c-3 d)+i a d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{4 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{2 f (-d+i c) (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (2 i c-3 d)+i a d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{4 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{2 f (-d+i c) (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {a (-2 d+i c) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+a (-d+i c) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{4 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{2 f (-d+i c) (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (-2 d+i c) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+a (-d+i c) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{4 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{2 f (-d+i c) (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {\frac {i a (-d+i c) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}-\frac {i a (-2 d+i c) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}}{4 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{2 f (-d+i c) (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {i a (-2 d+i c) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}-\frac {i a (-d+i c) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}}{4 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{2 f (-d+i c) (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {2 a (-2 d+i c) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 a (-d+i c) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{4 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{2 f (-d+i c) (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 a (-d+i c) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {2 a (-2 d+i c) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{4 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{2 f (-d+i c) (a+i a \tan (e+f x))}\) |
((2*a*(I*c - d)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (2 *a*(I*c - 2*d)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f))/(4*a ^2*(I*c - d)) - Sqrt[c + d*Tan[e + f*x]]/(2*(I*c - d)*f*(a + I*a*Tan[e + f *x]))
3.12.22.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-a)*((c + d*Tan[e + f*x])^(n + 1)/(2*f*(b* c - a*d)*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a*(b*c - a*d)) Int[(c + d *Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !GtQ[n, 0]
Time = 0.80 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97
| method | result | size |
| derivativedivides | \(\frac {2 d^{2} \left (\frac {-\frac {d \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c -2 d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2}}+\frac {i \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2} \sqrt {i d -c}}\right )}{f a}\) | \(150\) |
| default | \(\frac {2 d^{2} \left (\frac {-\frac {d \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c -2 d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2}}+\frac {i \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2} \sqrt {i d -c}}\right )}{f a}\) | \(150\) |
2/f/a*d^2*(1/4/d^2*(-1/(c+I*d)*d*(c+d*tan(f*x+e))^(1/2)/(-d*tan(f*x+e)+I*d )-(I*c-2*d)/(c+I*d)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^ (1/2)))+1/4*I/d^2/(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2 )))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 990 vs. \(2 (119) = 238\).
Time = 0.29 (sec) , antiderivative size = 990, normalized size of antiderivative = 6.39 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]
-1/8*(2*(I*a*c - a*d)*f*sqrt(1/4*I/((-I*a^2*c - a^2*d)*f^2))*e^(2*I*f*x + 2*I*e)*log(-2*(2*((I*a*c + a*d)*f*e^(2*I*f*x + 2*I*e) + (I*a*c + a*d)*f)*s qrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*s qrt(1/4*I/((-I*a^2*c - a^2*d)*f^2)) - (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e ^(-2*I*f*x - 2*I*e)) + 2*(-I*a*c + a*d)*f*sqrt(1/4*I/((-I*a^2*c - a^2*d)*f ^2))*e^(2*I*f*x + 2*I*e)*log(-2*(2*((-I*a*c - a*d)*f*e^(2*I*f*x + 2*I*e) + (-I*a*c - a*d)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I* f*x + 2*I*e) + 1))*sqrt(1/4*I/((-I*a^2*c - a^2*d)*f^2)) - (c - I*d)*e^(2*I *f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + (I*a*c - a*d)*f*sqrt(-(-I*c^2 + 4*c*d + 4*I*d^2)/((-I*a^2*c^3 + 3*a^2*c^2*d + 3*I*a^2*c*d^2 - a^2*d^3)*f^ 2))*e^(2*I*f*x + 2*I*e)*log(-1/2*(-I*c^2 + 3*c*d + 2*I*d^2 + ((a*c^2 + 2*I *a*c*d - a*d^2)*f*e^(2*I*f*x + 2*I*e) + (a*c^2 + 2*I*a*c*d - a*d^2)*f)*sqr t(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqr t(-(-I*c^2 + 4*c*d + 4*I*d^2)/((-I*a^2*c^3 + 3*a^2*c^2*d + 3*I*a^2*c*d^2 - a^2*d^3)*f^2)) + (-I*c^2 + 2*c*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I* e)/((a*c^2 + 2*I*a*c*d - a*d^2)*f)) + (-I*a*c + a*d)*f*sqrt(-(-I*c^2 + 4*c *d + 4*I*d^2)/((-I*a^2*c^3 + 3*a^2*c^2*d + 3*I*a^2*c*d^2 - a^2*d^3)*f^2))* e^(2*I*f*x + 2*I*e)*log(-1/2*(-I*c^2 + 3*c*d + 2*I*d^2 - ((a*c^2 + 2*I*a*c *d - a*d^2)*f*e^(2*I*f*x + 2*I*e) + (a*c^2 + 2*I*a*c*d - a*d^2)*f)*sqrt((( c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt...
\[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=- \frac {i \int \frac {1}{\sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} - i \sqrt {c + d \tan {\left (e + f x \right )}}}\, dx}{a} \]
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=\text {Exception raised: RuntimeError} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (119) = 238\).
Time = 0.52 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.48 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {1}{2} \, d^{2} {\left (\frac {4 \, {\left (c + 2 i \, d\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} + i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{-2 \, {\left (i \, a c d^{2} f - a d^{3} f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {\sqrt {d \tan \left (f x + e\right ) + c}}{{\left (a c d f + i \, a d^{2} f\right )} {\left (d \tan \left (f x + e\right ) - i \, d\right )}} - \frac {2 i \, \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{a \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d^{2} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}}\right )} \]
-1/2*d^2*(4*(c + 2*I*d)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) + I*sqrt( -2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2 ))))/((-2*I*a*c*d^2*f + 2*a*d^3*f)*sqrt(-2*c + 2*sqrt(c^2 + d^2))*(I*d/(c - sqrt(c^2 + d^2)) + 1)) - sqrt(d*tan(f*x + e) + c)/((a*c*d*f + I*a*d^2*f) *(d*tan(f*x + e) - I*d)) - 2*I*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt (c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) - I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^ 2 + d^2))))/(a*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d^2*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)))
Time = 10.16 (sec) , antiderivative size = 12379, normalized size of antiderivative = 79.86 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]
log(a*d^6*f*1i - ((-(c*d^6*48i + 48*d^7 + 96*c^2*d^5 - c^3*d^4*32i - a^2*c ^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^ 2 + 2*a^2*c^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f ^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4 )/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5) *1i)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c^2*d ^4))^(1/2)*1i + a^2*d^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*d^5)*1i)/(a^2 *c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 3 2*c^5*d^4)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4) + ( (4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4)) *(256*d^6 + 256*c^2*d^4))^(1/2)*1i + 2*a^2*c*d*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144* c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d ^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^ 4 + 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c^2*d^4))^(1/2))/(512*(d^6 + c^2*d^ 4)*(a^2*d^2*f^2*1i - a^2*c^2*f^2*1i + 2*a^2*c*d*f^2)))^(1/2)*(24*a^3*d^7*f ^3 + a^3*c*d^6*f^3*16i + 32*a^3*c^2*d^5*f^3 + a^3*c^3*d^4*f^3*16i + 8*a^3* c^4*d^3*f^3 - 2*(c + d*tan(e + f*x))^(1/2)*(a^2*c^2*d^3*f^2*64i - 32*a^...