3.829 \(\int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{17/2}} \, dx\)

Optimal. Leaf size=314 \[ -\frac {8 (-12 B+5 i A) (a+i a \tan (e+f x))^{7/2}}{765765 c^5 f (c-i c \tan (e+f x))^{7/2}}-\frac {8 (-12 B+5 i A) (a+i a \tan (e+f x))^{7/2}}{109395 c^4 f (c-i c \tan (e+f x))^{9/2}}-\frac {4 (-12 B+5 i A) (a+i a \tan (e+f x))^{7/2}}{12155 c^3 f (c-i c \tan (e+f x))^{11/2}}-\frac {4 (-12 B+5 i A) (a+i a \tan (e+f x))^{7/2}}{3315 c^2 f (c-i c \tan (e+f x))^{13/2}}-\frac {(-12 B+5 i A) (a+i a \tan (e+f x))^{7/2}}{255 c f (c-i c \tan (e+f x))^{15/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{17 f (c-i c \tan (e+f x))^{17/2}} \]

[Out]

-1/17*(I*A+B)*(a+I*a*tan(f*x+e))^(7/2)/f/(c-I*c*tan(f*x+e))^(17/2)-1/255*(5*I*A-12*B)*(a+I*a*tan(f*x+e))^(7/2)
/c/f/(c-I*c*tan(f*x+e))^(15/2)-4/3315*(5*I*A-12*B)*(a+I*a*tan(f*x+e))^(7/2)/c^2/f/(c-I*c*tan(f*x+e))^(13/2)-4/
12155*(5*I*A-12*B)*(a+I*a*tan(f*x+e))^(7/2)/c^3/f/(c-I*c*tan(f*x+e))^(11/2)-8/109395*(5*I*A-12*B)*(a+I*a*tan(f
*x+e))^(7/2)/c^4/f/(c-I*c*tan(f*x+e))^(9/2)-8/765765*(5*I*A-12*B)*(a+I*a*tan(f*x+e))^(7/2)/c^5/f/(c-I*c*tan(f*
x+e))^(7/2)

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Rubi [A]  time = 0.35, antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ -\frac {8 (-12 B+5 i A) (a+i a \tan (e+f x))^{7/2}}{765765 c^5 f (c-i c \tan (e+f x))^{7/2}}-\frac {8 (-12 B+5 i A) (a+i a \tan (e+f x))^{7/2}}{109395 c^4 f (c-i c \tan (e+f x))^{9/2}}-\frac {4 (-12 B+5 i A) (a+i a \tan (e+f x))^{7/2}}{12155 c^3 f (c-i c \tan (e+f x))^{11/2}}-\frac {4 (-12 B+5 i A) (a+i a \tan (e+f x))^{7/2}}{3315 c^2 f (c-i c \tan (e+f x))^{13/2}}-\frac {(-12 B+5 i A) (a+i a \tan (e+f x))^{7/2}}{255 c f (c-i c \tan (e+f x))^{15/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{17 f (c-i c \tan (e+f x))^{17/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(17/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(17*f*(c - I*c*Tan[e + f*x])^(17/2)) - (((5*I)*A - 12*B)*(a + I*a*Ta
n[e + f*x])^(7/2))/(255*c*f*(c - I*c*Tan[e + f*x])^(15/2)) - (4*((5*I)*A - 12*B)*(a + I*a*Tan[e + f*x])^(7/2))
/(3315*c^2*f*(c - I*c*Tan[e + f*x])^(13/2)) - (4*((5*I)*A - 12*B)*(a + I*a*Tan[e + f*x])^(7/2))/(12155*c^3*f*(
c - I*c*Tan[e + f*x])^(11/2)) - (8*((5*I)*A - 12*B)*(a + I*a*Tan[e + f*x])^(7/2))/(109395*c^4*f*(c - I*c*Tan[e
+ f*x])^(9/2)) - (8*((5*I)*A - 12*B)*(a + I*a*Tan[e + f*x])^(7/2))/(765765*c^5*f*(c - I*c*Tan[e + f*x])^(7/2)
)

Rule 37

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 3588

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

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{17/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2} (A+B x)}{(c-i c x)^{19/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{17 f (c-i c \tan (e+f x))^{17/2}}+\frac {(a (5 A+12 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{17/2}} \, dx,x,\tan (e+f x)\right )}{17 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{17 f (c-i c \tan (e+f x))^{17/2}}-\frac {(5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{255 c f (c-i c \tan (e+f x))^{15/2}}+\frac {(4 a (5 A+12 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{15/2}} \, dx,x,\tan (e+f x)\right )}{255 c f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{17 f (c-i c \tan (e+f x))^{17/2}}-\frac {(5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{255 c f (c-i c \tan (e+f x))^{15/2}}-\frac {4 (5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{3315 c^2 f (c-i c \tan (e+f x))^{13/2}}+\frac {(4 a (5 A+12 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{13/2}} \, dx,x,\tan (e+f x)\right )}{1105 c^2 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{17 f (c-i c \tan (e+f x))^{17/2}}-\frac {(5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{255 c f (c-i c \tan (e+f x))^{15/2}}-\frac {4 (5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{3315 c^2 f (c-i c \tan (e+f x))^{13/2}}-\frac {4 (5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{12155 c^3 f (c-i c \tan (e+f x))^{11/2}}+\frac {(8 a (5 A+12 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{12155 c^3 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{17 f (c-i c \tan (e+f x))^{17/2}}-\frac {(5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{255 c f (c-i c \tan (e+f x))^{15/2}}-\frac {4 (5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{3315 c^2 f (c-i c \tan (e+f x))^{13/2}}-\frac {4 (5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{12155 c^3 f (c-i c \tan (e+f x))^{11/2}}-\frac {8 (5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{109395 c^4 f (c-i c \tan (e+f x))^{9/2}}+\frac {(8 a (5 A+12 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{109395 c^4 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{17 f (c-i c \tan (e+f x))^{17/2}}-\frac {(5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{255 c f (c-i c \tan (e+f x))^{15/2}}-\frac {4 (5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{3315 c^2 f (c-i c \tan (e+f x))^{13/2}}-\frac {4 (5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{12155 c^3 f (c-i c \tan (e+f x))^{11/2}}-\frac {8 (5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{109395 c^4 f (c-i c \tan (e+f x))^{9/2}}-\frac {8 (5 i A-12 B) (a+i a \tan (e+f x))^{7/2}}{765765 c^5 f (c-i c \tan (e+f x))^{7/2}}\\ \end {align*}

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Mathematica [B]  time = 17.71, size = 655, normalized size = 2.09 \[ \frac {\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt {\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((B-i A) \cos (6 f x) \left (\frac {\cos (3 e)}{448 c^9}+\frac {i \sin (3 e)}{448 c^9}\right )+(A+i B) \sin (6 f x) \left (\frac {\cos (3 e)}{448 c^9}+\frac {i \sin (3 e)}{448 c^9}\right )+(15 B-22 i A) \cos (8 f x) \left (\frac {\cos (5 e)}{2016 c^9}+\frac {i \sin (5 e)}{2016 c^9}\right )+(51 B-145 i A) \cos (10 f x) \left (\frac {\cos (7 e)}{6336 c^9}+\frac {i \sin (7 e)}{6336 c^9}\right )+(B-60 i A) \cos (12 f x) \left (\frac {\cos (9 e)}{2288 c^9}+\frac {i \sin (9 e)}{2288 c^9}\right )+(215 A-69 i B) \cos (14 f x) \left (\frac {\sin (11 e)}{12480 c^9}-\frac {i \cos (11 e)}{12480 c^9}\right )+(50 A-33 i B) \cos (16 f x) \left (\frac {\sin (13 e)}{8160 c^9}-\frac {i \cos (13 e)}{8160 c^9}\right )+(A-i B) \cos (18 f x) \left (\frac {\sin (15 e)}{1088 c^9}-\frac {i \cos (15 e)}{1088 c^9}\right )+(22 A+15 i B) \sin (8 f x) \left (\frac {\cos (5 e)}{2016 c^9}+\frac {i \sin (5 e)}{2016 c^9}\right )+(145 A+51 i B) \sin (10 f x) \left (\frac {\cos (7 e)}{6336 c^9}+\frac {i \sin (7 e)}{6336 c^9}\right )+(60 A+i B) \sin (12 f x) \left (\frac {\cos (9 e)}{2288 c^9}+\frac {i \sin (9 e)}{2288 c^9}\right )+(215 A-69 i B) \sin (14 f x) \left (\frac {\cos (11 e)}{12480 c^9}+\frac {i \sin (11 e)}{12480 c^9}\right )+(50 A-33 i B) \sin (16 f x) \left (\frac {\cos (13 e)}{8160 c^9}+\frac {i \sin (13 e)}{8160 c^9}\right )+(A-i B) \sin (18 f x) \left (\frac {\cos (15 e)}{1088 c^9}+\frac {i \sin (15 e)}{1088 c^9}\right )\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(17/2),x]

[Out]

(Cos[e + f*x]^4*(((-I)*A + B)*Cos[6*f*x]*(Cos[3*e]/(448*c^9) + ((I/448)*Sin[3*e])/c^9) + ((-22*I)*A + 15*B)*Co
s[8*f*x]*(Cos[5*e]/(2016*c^9) + ((I/2016)*Sin[5*e])/c^9) + ((-145*I)*A + 51*B)*Cos[10*f*x]*(Cos[7*e]/(6336*c^9
) + ((I/6336)*Sin[7*e])/c^9) + ((-60*I)*A + B)*Cos[12*f*x]*(Cos[9*e]/(2288*c^9) + ((I/2288)*Sin[9*e])/c^9) + (
215*A - (69*I)*B)*Cos[14*f*x]*(((-1/12480*I)*Cos[11*e])/c^9 + Sin[11*e]/(12480*c^9)) + (50*A - (33*I)*B)*Cos[1
6*f*x]*(((-1/8160*I)*Cos[13*e])/c^9 + Sin[13*e]/(8160*c^9)) + (A - I*B)*Cos[18*f*x]*(((-1/1088*I)*Cos[15*e])/c
^9 + Sin[15*e]/(1088*c^9)) + (A + I*B)*(Cos[3*e]/(448*c^9) + ((I/448)*Sin[3*e])/c^9)*Sin[6*f*x] + (22*A + (15*
I)*B)*(Cos[5*e]/(2016*c^9) + ((I/2016)*Sin[5*e])/c^9)*Sin[8*f*x] + (145*A + (51*I)*B)*(Cos[7*e]/(6336*c^9) + (
(I/6336)*Sin[7*e])/c^9)*Sin[10*f*x] + (60*A + I*B)*(Cos[9*e]/(2288*c^9) + ((I/2288)*Sin[9*e])/c^9)*Sin[12*f*x]
+ (215*A - (69*I)*B)*(Cos[11*e]/(12480*c^9) + ((I/12480)*Sin[11*e])/c^9)*Sin[14*f*x] + (50*A - (33*I)*B)*(Cos
[13*e]/(8160*c^9) + ((I/8160)*Sin[13*e])/c^9)*Sin[16*f*x] + (A - I*B)*(Cos[15*e]/(1088*c^9) + ((I/1088)*Sin[15
*e])/c^9)*Sin[18*f*x])*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a*Tan[e + f*x])^(7/2)*(A
+ B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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fricas [A]  time = 0.80, size = 183, normalized size = 0.58 \[ \frac {{\left ({\left (-45045 i \, A - 45045 \, B\right )} a^{3} e^{\left (19 i \, f x + 19 i \, e\right )} + {\left (-300300 i \, A - 198198 \, B\right )} a^{3} e^{\left (17 i \, f x + 17 i \, e\right )} + {\left (-844305 i \, A - 270963 \, B\right )} a^{3} e^{\left (15 i \, f x + 15 i \, e\right )} + {\left (-1285200 i \, A + 21420 \, B\right )} a^{3} e^{\left (13 i \, f x + 13 i \, e\right )} + {\left (-1121575 i \, A + 394485 \, B\right )} a^{3} e^{\left (11 i \, f x + 11 i \, e\right )} + {\left (-534820 i \, A + 364650 \, B\right )} a^{3} e^{\left (9 i \, f x + 9 i \, e\right )} + {\left (-109395 i \, A + 109395 \, B\right )} a^{3} e^{\left (7 i \, f x + 7 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{24504480 \, c^{9} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(17/2),x, algorithm="fricas")

[Out]

1/24504480*((-45045*I*A - 45045*B)*a^3*e^(19*I*f*x + 19*I*e) + (-300300*I*A - 198198*B)*a^3*e^(17*I*f*x + 17*I
*e) + (-844305*I*A - 270963*B)*a^3*e^(15*I*f*x + 15*I*e) + (-1285200*I*A + 21420*B)*a^3*e^(13*I*f*x + 13*I*e)
+ (-1121575*I*A + 394485*B)*a^3*e^(11*I*f*x + 11*I*e) + (-534820*I*A + 364650*B)*a^3*e^(9*I*f*x + 9*I*e) + (-1
09395*I*A + 109395*B)*a^3*e^(7*I*f*x + 7*I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) +
1))/(c^9*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {17}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(17/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(7/2)/(-I*c*tan(f*x + e) + c)^(17/2), x)

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maple [A]  time = 0.44, size = 230, normalized size = 0.73 \[ \frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (109881 i B \left (\tan ^{2}\left (f x +e \right )\right )+11175 i A \left (\tan ^{3}\left (f x +e \right )\right )-96 B \left (\tan ^{7}\left (f x +e \right )\right )+40 i A \left (\tan ^{7}\left (f x +e \right )\right )-400 A \left (\tan ^{6}\left (f x +e \right )\right )-960 i B \left (\tan ^{6}\left (f x +e \right )\right )+4464 B \left (\tan ^{5}\left (f x +e \right )\right )+103165 i A \tan \left (f x +e \right )+5400 A \left (\tan ^{4}\left (f x +e \right )\right )+12960 i B \left (\tan ^{4}\left (f x +e \right )\right )-26820 B \left (\tan ^{3}\left (f x +e \right )\right )-1860 i A \left (\tan ^{5}\left (f x +e \right )\right )-18030 A \left (\tan ^{2}\left (f x +e \right )\right )+5871 i B +58710 B \tan \left (f x +e \right )+66260 A \right )}{765765 f \,c^{9} \left (\tan \left (f x +e \right )+i\right )^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(17/2),x)

[Out]

1/765765*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^9*(1+tan(f*x+e)^2)*(109881*I*B*tan(
f*x+e)^2+11175*I*A*tan(f*x+e)^3-96*B*tan(f*x+e)^7+40*I*A*tan(f*x+e)^7-400*A*tan(f*x+e)^6-960*I*B*tan(f*x+e)^6+
4464*B*tan(f*x+e)^5+103165*I*A*tan(f*x+e)+5400*A*tan(f*x+e)^4+12960*I*B*tan(f*x+e)^4-26820*B*tan(f*x+e)^3-1860
*I*A*tan(f*x+e)^5-18030*A*tan(f*x+e)^2+5871*I*B+58710*B*tan(f*x+e)+66260*A)/(tan(f*x+e)+I)^10

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maxima [A]  time = 1.30, size = 410, normalized size = 1.31 \[ -\frac {{\left ({\left (45045 i \, A + 45045 \, B\right )} a^{3} \cos \left (\frac {17}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (255255 i \, A + 153153 \, B\right )} a^{3} \cos \left (\frac {15}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (589050 i \, A + 117810 \, B\right )} a^{3} \cos \left (\frac {13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (696150 i \, A - 139230 \, B\right )} a^{3} \cos \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (425425 i \, A - 255255 \, B\right )} a^{3} \cos \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (109395 i \, A - 109395 \, B\right )} a^{3} \cos \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 45045 \, {\left (A - i \, B\right )} a^{3} \sin \left (\frac {17}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 51051 \, {\left (5 \, A - 3 i \, B\right )} a^{3} \sin \left (\frac {15}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 117810 \, {\left (5 \, A - i \, B\right )} a^{3} \sin \left (\frac {13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 139230 \, {\left (5 \, A + i \, B\right )} a^{3} \sin \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 85085 \, {\left (5 \, A + 3 i \, B\right )} a^{3} \sin \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 109395 \, {\left (A + i \, B\right )} a^{3} \sin \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{24504480 \, c^{\frac {17}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(17/2),x, algorithm="maxima")

[Out]

-1/24504480*((45045*I*A + 45045*B)*a^3*cos(17/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (255255*I*A + 1
53153*B)*a^3*cos(15/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (589050*I*A + 117810*B)*a^3*cos(13/2*arct
an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (696150*I*A - 139230*B)*a^3*cos(11/2*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e))) + (425425*I*A - 255255*B)*a^3*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (109395*I*
A - 109395*B)*a^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 45045*(A - I*B)*a^3*sin(17/2*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 51051*(5*A - 3*I*B)*a^3*sin(15/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x +
2*e))) - 117810*(5*A - I*B)*a^3*sin(13/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 139230*(5*A + I*B)*a^3
*sin(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 85085*(5*A + 3*I*B)*a^3*sin(9/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))) - 109395*(A + I*B)*a^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(
c^(17/2)*f)

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mupad [B]  time = 14.48, size = 229, normalized size = 0.73 \[ -\frac {\sqrt {a+\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}\,\left (\frac {a^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\left (5\,A+B\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{288\,c^8\,f}+\frac {a^3\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\left (5\,A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{176\,c^8\,f}+\frac {a^3\,{\mathrm {e}}^{e\,12{}\mathrm {i}+f\,x\,12{}\mathrm {i}}\,\left (5\,A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{208\,c^8\,f}+\frac {a^3\,{\mathrm {e}}^{e\,14{}\mathrm {i}+f\,x\,14{}\mathrm {i}}\,\left (5\,A-B\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{480\,c^8\,f}+\frac {a^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{224\,c^8\,f}+\frac {a^3\,{\mathrm {e}}^{e\,16{}\mathrm {i}+f\,x\,16{}\mathrm {i}}\,\left (A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{544\,c^8\,f}\right )}{\sqrt {c-\frac {c\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(7/2))/(c - c*tan(e + f*x)*1i)^(17/2),x)

[Out]

-((a + (a*sin(e + f*x)*1i)/cos(e + f*x))^(1/2)*((a^3*exp(e*8i + f*x*8i)*(5*A + B*3i)*1i)/(288*c^8*f) + (a^3*ex
p(e*10i + f*x*10i)*(5*A + B*1i)*1i)/(176*c^8*f) + (a^3*exp(e*12i + f*x*12i)*(5*A - B*1i)*1i)/(208*c^8*f) + (a^
3*exp(e*14i + f*x*14i)*(5*A - B*3i)*1i)/(480*c^8*f) + (a^3*exp(e*6i + f*x*6i)*(A + B*1i)*1i)/(224*c^8*f) + (a^
3*exp(e*16i + f*x*16i)*(A - B*1i)*1i)/(544*c^8*f)))/(c - (c*sin(e + f*x)*1i)/cos(e + f*x))^(1/2)

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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((a+I*a*tan(f*x+e))**(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(17/2),x)

[Out]

Timed out

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