3.52 \(\int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=148 \[ \frac {A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {x (-7 B+i A)}{8 a^3}-\frac {i B \log (\cos (c+d x))}{a^3 d}+\frac {(-B+i A) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2} \]

[Out]

1/8*(I*A-7*B)*x/a^3-I*B*ln(cos(d*x+c))/a^3/d+1/6*(I*A-B)*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^3+1/8*(A+3*I*B)*tan
(d*x+c)^2/a/d/(a+I*a*tan(d*x+c))^2+1/8*(A+7*I*B)/d/(a^3+I*a^3*tan(d*x+c))

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Rubi [A]  time = 0.37, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3595, 3589, 3475, 12, 3526, 8} \[ \frac {A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {x (-7 B+i A)}{8 a^3}-\frac {i B \log (\cos (c+d x))}{a^3 d}+\frac {(-B+i A) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((I*A - 7*B)*x)/(8*a^3) - (I*B*Log[Cos[c + d*x]])/(a^3*d) + ((I*A - B)*Tan[c + d*x]^3)/(6*d*(a + I*a*Tan[c + d
*x])^3) + ((A + (3*I)*B)*Tan[c + d*x]^2)/(8*a*d*(a + I*a*Tan[c + d*x])^2) + (A + (7*I)*B)/(8*d*(a^3 + I*a^3*Ta
n[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

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

Rule 3475

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

Rule 3526

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

Rule 3589

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

Rule 3595

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*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*a*f*
m), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx &=\frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^2(c+d x) (3 a (i A-B)+6 i a B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\tan (c+d x) \left (-6 a^2 (A+3 i B)-24 a^2 B \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}-\frac {i \int -\frac {6 a^3 (i A-7 B) \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{24 a^5}+\frac {(i B) \int \tan (c+d x) \, dx}{a^3}\\ &=-\frac {i B \log (\cos (c+d x))}{a^3 d}+\frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}-\frac {(A+7 i B) \int \frac {\tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=-\frac {i B \log (\cos (c+d x))}{a^3 d}+\frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(i A-7 B) \int 1 \, dx}{8 a^3}\\ &=\frac {(i A-7 B) x}{8 a^3}-\frac {i B \log (\cos (c+d x))}{a^3 d}+\frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.49, size = 178, normalized size = 1.20 \[ \frac {\sec ^3(c+d x) ((-51 B+9 i A) \cos (c+d x)-2 \cos (3 (c+d x)) (6 A d x-i A-48 B \log (\cos (c+d x))+42 i B d x+B)-27 A \sin (c+d x)+2 A \sin (3 (c+d x))-12 i A d x \sin (3 (c+d x))-81 i B \sin (c+d x)+2 i B \sin (3 (c+d x))+84 B d x \sin (3 (c+d x))+96 i B \sin (3 (c+d x)) \log (\cos (c+d x)))}{96 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*(((9*I)*A - 51*B)*Cos[c + d*x] - 2*Cos[3*(c + d*x)]*((-I)*A + B + 6*A*d*x + (42*I)*B*d*x - 48*
B*Log[Cos[c + d*x]]) - 27*A*Sin[c + d*x] - (81*I)*B*Sin[c + d*x] + 2*A*Sin[3*(c + d*x)] + (2*I)*B*Sin[3*(c + d
*x)] - (12*I)*A*d*x*Sin[3*(c + d*x)] + 84*B*d*x*Sin[3*(c + d*x)] + (96*I)*B*Log[Cos[c + d*x]]*Sin[3*(c + d*x)]
))/(96*a^3*d*(-I + Tan[c + d*x])^3)

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fricas [A]  time = 1.10, size = 103, normalized size = 0.70 \[ \frac {{\left ({\left (12 i \, A - 180 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} - 96 i \, B e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 6 \, {\left (3 \, A + 11 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 \, {\left (3 \, A + 5 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, A + 2 i \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*((12*I*A - 180*B)*d*x*e^(6*I*d*x + 6*I*c) - 96*I*B*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 6*(
3*A + 11*I*B)*e^(4*I*d*x + 4*I*c) - 3*(3*A + 5*I*B)*e^(2*I*d*x + 2*I*c) + 2*A + 2*I*B)*e^(-6*I*d*x - 6*I*c)/(a
^3*d)

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giac [A]  time = 1.40, size = 130, normalized size = 0.88 \[ \frac {\frac {6 \, {\left (A + 15 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac {6 \, {\left (A - i \, B\right )} \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {11 \, A \tan \left (d x + c\right )^{3} + 165 i \, B \tan \left (d x + c\right )^{3} + 51 i \, A \tan \left (d x + c\right )^{2} + 291 \, B \tan \left (d x + c\right )^{2} + 75 \, A \tan \left (d x + c\right ) - 171 i \, B \tan \left (d x + c\right ) - 29 i \, A - 29 \, B}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/96*(6*(A + 15*I*B)*log(tan(d*x + c) - I)/a^3 - 6*(A - I*B)*log(-I*tan(d*x + c) + 1)/a^3 - (11*A*tan(d*x + c)
^3 + 165*I*B*tan(d*x + c)^3 + 51*I*A*tan(d*x + c)^2 + 291*B*tan(d*x + c)^2 + 75*A*tan(d*x + c) - 171*I*B*tan(d
*x + c) - 29*I*A - 29*B)/(a^3*(tan(d*x + c) - I)^3))/d

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maple [A]  time = 0.22, size = 203, normalized size = 1.37 \[ -\frac {A \ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}+\frac {i B \ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}+\frac {\ln \left (\tan \left (d x +c \right )-i\right ) A}{16 d \,a^{3}}+\frac {15 i \ln \left (\tan \left (d x +c \right )-i\right ) B}{16 d \,a^{3}}+\frac {17 B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {7 i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}+\frac {5 A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {7 i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x)

[Out]

-1/16/d/a^3*A*ln(tan(d*x+c)+I)+1/16*I/d/a^3*B*ln(tan(d*x+c)+I)+1/16/d/a^3*ln(tan(d*x+c)-I)*A+15/16*I/d/a^3*ln(
tan(d*x+c)-I)*B+17/8/d/a^3/(tan(d*x+c)-I)*B-7/8*I/d/a^3/(tan(d*x+c)-I)*A+5/8/d/a^3/(tan(d*x+c)-I)^2*A+7/8*I/d/
a^3/(tan(d*x+c)-I)^2*B+1/6*I/d/a^3/(tan(d*x+c)-I)^3*A-1/6/d/a^3/(tan(d*x+c)-I)^3*B

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 6.64, size = 146, normalized size = 0.99 \[ \frac {\frac {5\,A}{12\,a^3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {7\,A}{8\,a^3}+\frac {B\,17{}\mathrm {i}}{8\,a^3}\right )+\frac {B\,17{}\mathrm {i}}{12\,a^3}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {27\,B}{8\,a^3}+\frac {A\,9{}\mathrm {i}}{8\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,15{}\mathrm {i}\right )}{16\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(c + d*x)^3*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

((5*A)/(12*a^3) - tan(c + d*x)^2*((7*A)/(8*a^3) + (B*17i)/(8*a^3)) + (B*17i)/(12*a^3) + tan(c + d*x)*((A*9i)/(
8*a^3) - (27*B)/(8*a^3)))/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) - (log(tan(c + d*x)
+ 1i)*(A - B*1i))/(16*a^3*d) + (log(tan(c + d*x) - 1i)*(A + B*15i))/(16*a^3*d)

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sympy [A]  time = 2.42, size = 303, normalized size = 2.05 \[ - \frac {i B \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} + \begin {cases} - \frac {\left (\left (- 512 A a^{6} d^{2} e^{6 i c} - 512 i B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (2304 A a^{6} d^{2} e^{8 i c} + 3840 i B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (- 4608 A a^{6} d^{2} e^{10 i c} - 16896 i B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: 24576 a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {i A - 15 B}{8 a^{3}} - \frac {\left (- i A e^{6 i c} + 3 i A e^{4 i c} - 3 i A e^{2 i c} + i A + 15 B e^{6 i c} - 11 B e^{4 i c} + 5 B e^{2 i c} - B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- i A + 15 B\right )}{8 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**3,x)

[Out]

-I*B*log(exp(2*I*d*x) + exp(-2*I*c))/(a**3*d) + Piecewise((-((-512*A*a**6*d**2*exp(6*I*c) - 512*I*B*a**6*d**2*
exp(6*I*c))*exp(-6*I*d*x) + (2304*A*a**6*d**2*exp(8*I*c) + 3840*I*B*a**6*d**2*exp(8*I*c))*exp(-4*I*d*x) + (-46
08*A*a**6*d**2*exp(10*I*c) - 16896*I*B*a**6*d**2*exp(10*I*c))*exp(-2*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), N
e(24576*a**9*d**3*exp(12*I*c), 0)), (x*(-(I*A - 15*B)/(8*a**3) - (-I*A*exp(6*I*c) + 3*I*A*exp(4*I*c) - 3*I*A*e
xp(2*I*c) + I*A + 15*B*exp(6*I*c) - 11*B*exp(4*I*c) + 5*B*exp(2*I*c) - B)*exp(-6*I*c)/(8*a**3)), True)) - x*(-
I*A + 15*B)/(8*a**3)

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