3.827 \(\int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx\)

Optimal. Leaf size=208 \[ -\frac {2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{9009 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac {2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac {(-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}} \]

[Out]

-1/13*(I*A+B)*(a+I*a*tan(f*x+e))^(7/2)/f/(c-I*c*tan(f*x+e))^(13/2)-1/143*(3*I*A-10*B)*(a+I*a*tan(f*x+e))^(7/2)
/c/f/(c-I*c*tan(f*x+e))^(11/2)-2/1287*(3*I*A-10*B)*(a+I*a*tan(f*x+e))^(7/2)/c^2/f/(c-I*c*tan(f*x+e))^(9/2)-2/9
009*(3*I*A-10*B)*(a+I*a*tan(f*x+e))^(7/2)/c^3/f/(c-I*c*tan(f*x+e))^(7/2)

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Rubi [A]  time = 0.29, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ -\frac {2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{9009 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac {2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac {(-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}} \]

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])^(13/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(13*f*(c - I*c*Tan[e + f*x])^(13/2)) - (((3*I)*A - 10*B)*(a + I*a*Ta
n[e + f*x])^(7/2))/(143*c*f*(c - I*c*Tan[e + f*x])^(11/2)) - (2*((3*I)*A - 10*B)*(a + I*a*Tan[e + f*x])^(7/2))
/(1287*c^2*f*(c - I*c*Tan[e + f*x])^(9/2)) - (2*((3*I)*A - 10*B)*(a + I*a*Tan[e + f*x])^(7/2))/(9009*c^3*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))^{13/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2} (A+B x)}{(c-i c x)^{15/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}+\frac {(a (3 A+10 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 )}{13 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac {(3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}+\frac {(2 a (3 A+10 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 )}{143 c f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac {(3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac {2 (3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}+\frac {(2 a (3 A+10 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 )}{1287 c^2 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac {(3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac {2 (3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac {2 (3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{9009 c^3 f (c-i c \tan (e+f x))^{7/2}}\\ \end {align*}

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Mathematica [B]  time = 17.07, size = 495, normalized size = 2.38 \[ \frac {\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt {\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((B-i A) \cos (6 f x) \left (\frac {\cos (3 e)}{112 c^7}+\frac {i \sin (3 e)}{112 c^7}\right )+(A+i B) \sin (6 f x) \left (\frac {\cos (3 e)}{112 c^7}+\frac {i \sin (3 e)}{112 c^7}\right )+(8 B-15 i A) \cos (8 f x) \left (\frac {\cos (5 e)}{504 c^7}+\frac {i \sin (5 e)}{504 c^7}\right )+(B-30 i A) \cos (10 f x) \left (\frac {\cos (7 e)}{792 c^7}+\frac {i \sin (7 e)}{792 c^7}\right )+(25 A-12 i B) \cos (12 f x) \left (\frac {\sin (9 e)}{1144 c^7}-\frac {i \cos (9 e)}{1144 c^7}\right )+(A-i B) \cos (14 f x) \left (\frac {\sin (11 e)}{208 c^7}-\frac {i \cos (11 e)}{208 c^7}\right )+(15 A+8 i B) \sin (8 f x) \left (\frac {\cos (5 e)}{504 c^7}+\frac {i \sin (5 e)}{504 c^7}\right )+(30 A+i B) \sin (10 f x) \left (\frac {\cos (7 e)}{792 c^7}+\frac {i \sin (7 e)}{792 c^7}\right )+(25 A-12 i B) \sin (12 f x) \left (\frac {\cos (9 e)}{1144 c^7}+\frac {i \sin (9 e)}{1144 c^7}\right )+(A-i B) \sin (14 f x) \left (\frac {\cos (11 e)}{208 c^7}+\frac {i \sin (11 e)}{208 c^7}\right )\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

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])^(13/2),x]

[Out]

(Cos[e + f*x]^4*(((-I)*A + B)*Cos[6*f*x]*(Cos[3*e]/(112*c^7) + ((I/112)*Sin[3*e])/c^7) + ((-15*I)*A + 8*B)*Cos
[8*f*x]*(Cos[5*e]/(504*c^7) + ((I/504)*Sin[5*e])/c^7) + ((-30*I)*A + B)*Cos[10*f*x]*(Cos[7*e]/(792*c^7) + ((I/
792)*Sin[7*e])/c^7) + (25*A - (12*I)*B)*Cos[12*f*x]*(((-1/1144*I)*Cos[9*e])/c^7 + Sin[9*e]/(1144*c^7)) + (A -
I*B)*Cos[14*f*x]*(((-1/208*I)*Cos[11*e])/c^7 + Sin[11*e]/(208*c^7)) + (A + I*B)*(Cos[3*e]/(112*c^7) + ((I/112)
*Sin[3*e])/c^7)*Sin[6*f*x] + (15*A + (8*I)*B)*(Cos[5*e]/(504*c^7) + ((I/504)*Sin[5*e])/c^7)*Sin[8*f*x] + (30*A
+ I*B)*(Cos[7*e]/(792*c^7) + ((I/792)*Sin[7*e])/c^7)*Sin[10*f*x] + (25*A - (12*I)*B)*(Cos[9*e]/(1144*c^7) + (
(I/1144)*Sin[9*e])/c^7)*Sin[12*f*x] + (A - I*B)*(Cos[11*e]/(208*c^7) + ((I/208)*Sin[11*e])/c^7)*Sin[14*f*x])*S
qrt[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*(C
os[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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fricas [A]  time = 1.10, size = 143, normalized size = 0.69 \[ \frac {{\left ({\left (-693 i \, A - 693 \, B\right )} a^{3} e^{\left (15 i \, f x + 15 i \, e\right )} + {\left (-3150 i \, A - 1512 \, B\right )} a^{3} e^{\left (13 i \, f x + 13 i \, e\right )} + {\left (-5460 i \, A + 182 \, B\right )} a^{3} e^{\left (11 i \, f x + 11 i \, e\right )} + {\left (-4290 i \, A + 2288 \, B\right )} a^{3} e^{\left (9 i \, f x + 9 i \, e\right )} + {\left (-1287 i \, A + 1287 \, B\right )} a^{3} e^{\left (7 i \, f x + 7 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{72072 \, c^{7} f} \]

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))^(13/2),x, algorithm="fricas")

[Out]

1/72072*((-693*I*A - 693*B)*a^3*e^(15*I*f*x + 15*I*e) + (-3150*I*A - 1512*B)*a^3*e^(13*I*f*x + 13*I*e) + (-546
0*I*A + 182*B)*a^3*e^(11*I*f*x + 11*I*e) + (-4290*I*A + 2288*B)*a^3*e^(9*I*f*x + 9*I*e) + (-1287*I*A + 1287*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^7*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {13}{2}}}\,{d x} \]

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))^(13/2),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)^(13/2), x)

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maple [A]  time = 0.53, size = 184, normalized size = 0.88 \[ \frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (6 i A \left (\tan ^{5}\left (f x +e \right )\right )-160 i B \left (\tan ^{4}\left (f x +e \right )\right )-20 B \left (\tan ^{5}\left (f x +e \right )\right )-177 i A \left (\tan ^{3}\left (f x +e \right )\right )-48 A \left (\tan ^{4}\left (f x +e \right )\right )-1643 i B \left (\tan ^{2}\left (f x +e \right )\right )+590 B \left (\tan ^{3}\left (f x +e \right )\right )-1569 i A \tan \left (f x +e \right )+408 A \left (\tan ^{2}\left (f x +e \right )\right )-97 i B -776 B \tan \left (f x +e \right )-930 A \right )}{9009 f \,c^{7} \left (\tan \left (f x +e \right )+i\right )^{8}} \]

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))^(13/2),x)

[Out]

1/9009*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^7*(1+tan(f*x+e)^2)*(6*I*A*tan(f*x+e)^
5-160*I*B*tan(f*x+e)^4-20*B*tan(f*x+e)^5-177*I*A*tan(f*x+e)^3-48*A*tan(f*x+e)^4-1643*I*B*tan(f*x+e)^2+590*B*ta
n(f*x+e)^3-1569*I*A*tan(f*x+e)+408*A*tan(f*x+e)^2-97*I*B-776*B*tan(f*x+e)-930*A)/(tan(f*x+e)+I)^8

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maxima [A]  time = 0.97, size = 276, normalized size = 1.33 \[ -\frac {{\left ({\left (693 i \, A + 693 \, B\right )} a^{3} \cos \left (\frac {13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (2457 i \, A + 819 \, B\right )} a^{3} \cos \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (3003 i \, A - 1001 \, B\right )} a^{3} \cos \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (1287 i \, A - 1287 \, B\right )} a^{3} \cos \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 693 \, {\left (A - i \, B\right )} a^{3} \sin \left (\frac {13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 819 \, {\left (3 \, A - i \, B\right )} a^{3} \sin \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 1001 \, {\left (3 \, A + i \, B\right )} a^{3} \sin \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 1287 \, {\left (A + i \, B\right )} a^{3} \sin \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{72072 \, c^{\frac {13}{2}} f} \]

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))^(13/2),x, algorithm="maxima")

[Out]

-1/72072*((693*I*A + 693*B)*a^3*cos(13/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (2457*I*A + 819*B)*a^3
*cos(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (3003*I*A - 1001*B)*a^3*cos(9/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))) + (1287*I*A - 1287*B)*a^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 693*(
A - I*B)*a^3*sin(13/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 819*(3*A - I*B)*a^3*sin(11/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))) - 1001*(3*A + I*B)*a^3*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) -
1287*(A + I*B)*a^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(13/2)*f)

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mupad [B]  time = 13.49, size = 167, normalized size = 0.80 \[ -\frac {\sqrt {a+\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}\,\left (\frac {a^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\left (3\,A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{72\,c^6\,f}+\frac {a^3\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\left (3\,A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{88\,c^6\,f}+\frac {a^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{56\,c^6\,f}+\frac {a^3\,{\mathrm {e}}^{e\,12{}\mathrm {i}+f\,x\,12{}\mathrm {i}}\,\left (A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{104\,c^6\,f}\right )}{\sqrt {c-\frac {c\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}} \]

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)^(13/2),x)

[Out]

-((a + (a*sin(e + f*x)*1i)/cos(e + f*x))^(1/2)*((a^3*exp(e*8i + f*x*8i)*(3*A + B*1i)*1i)/(72*c^6*f) + (a^3*exp
(e*10i + f*x*10i)*(3*A - B*1i)*1i)/(88*c^6*f) + (a^3*exp(e*6i + f*x*6i)*(A + B*1i)*1i)/(56*c^6*f) + (a^3*exp(e
*12i + f*x*12i)*(A - B*1i)*1i)/(104*c^6*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))**(13/2),x)

[Out]

Timed out

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