3.10.48 \(\int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx\) [948]

Optimal. Leaf size=146 \[ \frac {a^5 x}{c^4}-\frac {i a^5 \log (\cos (e+f x))}{c^4 f}-\frac {4 i a^5}{f (c-i c \tan (e+f x))^4}-\frac {12 i a^5}{f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {32 i a^5 c^5}{3 f \left (c^3-i c^3 \tan (e+f x)\right )^3}+\frac {8 i a^5}{f \left (c^4-i c^4 \tan (e+f x)\right )} \]

[Out]

a^5*x/c^4-I*a^5*ln(cos(f*x+e))/c^4/f-4*I*a^5/f/(c-I*c*tan(f*x+e))^4-12*I*a^5/f/(c^2-I*c^2*tan(f*x+e))^2+32/3*I
*a^5*c^5/f/(c^3-I*c^3*tan(f*x+e))^3+8*I*a^5/f/(c^4-I*c^4*tan(f*x+e))

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Rubi [A]
time = 0.10, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 45} \begin {gather*} \frac {8 i a^5}{f \left (c^4-i c^4 \tan (e+f x)\right )}-\frac {i a^5 \log (\cos (e+f x))}{c^4 f}+\frac {a^5 x}{c^4}-\frac {12 i a^5}{f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {32 i a^5 c^5}{3 f \left (c^3-i c^3 \tan (e+f x)\right )^3}-\frac {4 i a^5}{f (c-i c \tan (e+f x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^5/(c - I*c*Tan[e + f*x])^4,x]

[Out]

(a^5*x)/c^4 - (I*a^5*Log[Cos[e + f*x]])/(c^4*f) - ((4*I)*a^5)/(f*(c - I*c*Tan[e + f*x])^4) - ((12*I)*a^5)/(f*(
c^2 - I*c^2*Tan[e + f*x])^2) + (((32*I)/3)*a^5*c^5)/(f*(c^3 - I*c^3*Tan[e + f*x])^3) + ((8*I)*a^5)/(f*(c^4 - I
*c^4*Tan[e + f*x]))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

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

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx &=\left (a^5 c^5\right ) \int \frac {\sec ^{10}(e+f x)}{(c-i c \tan (e+f x))^9} \, dx\\ &=\frac {\left (i a^5\right ) \text {Subst}\left (\int \frac {(c-x)^4}{(c+x)^5} \, dx,x,-i c \tan (e+f x)\right )}{c^4 f}\\ &=\frac {\left (i a^5\right ) \text {Subst}\left (\int \left (\frac {16 c^4}{(c+x)^5}-\frac {32 c^3}{(c+x)^4}+\frac {24 c^2}{(c+x)^3}-\frac {8 c}{(c+x)^2}+\frac {1}{c+x}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^4 f}\\ &=\frac {a^5 x}{c^4}-\frac {i a^5 \log (\cos (e+f x))}{c^4 f}-\frac {4 i a^5}{f (c-i c \tan (e+f x))^4}+\frac {32 i a^5}{3 c f (c-i c \tan (e+f x))^3}-\frac {12 i a^5}{f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {8 i a^5}{f \left (c^4-i c^4 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 1.48, size = 151, normalized size = 1.03 \begin {gather*} \frac {a^5 \left (-6 i+16 i \cos (2 (e+f x))+3 \cos (4 (e+f x)) \left (-i+4 f x-2 i \log \left (\cos ^2(e+f x)\right )\right )+8 \sin (2 (e+f x))+3 \sin (4 (e+f x))-12 i f x \sin (4 (e+f x))-6 \log \left (\cos ^2(e+f x)\right ) \sin (4 (e+f x))\right ) (\cos (4 e+9 f x)+i \sin (4 e+9 f x))}{12 c^4 f (\cos (f x)+i \sin (f x))^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^5/(c - I*c*Tan[e + f*x])^4,x]

[Out]

(a^5*(-6*I + (16*I)*Cos[2*(e + f*x)] + 3*Cos[4*(e + f*x)]*(-I + 4*f*x - (2*I)*Log[Cos[e + f*x]^2]) + 8*Sin[2*(
e + f*x)] + 3*Sin[4*(e + f*x)] - (12*I)*f*x*Sin[4*(e + f*x)] - 6*Log[Cos[e + f*x]^2]*Sin[4*(e + f*x)])*(Cos[4*
e + 9*f*x] + I*Sin[4*e + 9*f*x]))/(12*c^4*f*(Cos[f*x] + I*Sin[f*x])^5)

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Maple [A]
time = 0.25, size = 79, normalized size = 0.54

method	result	size
derivativedivides	\(\frac {a^{5} \left (-\frac {4 i}{\left (\tan \left (f x +e \right )+i\right )^{4}}-\frac {8}{\tan \left (f x +e \right )+i}+\frac {32}{3 \left (\tan \left (f x +e \right )+i\right )^{3}}+i \ln \left (\tan \left (f x +e \right )+i\right )+\frac {12 i}{\left (\tan \left (f x +e \right )+i\right )^{2}}\right )}{f \,c^{4}}\)	\(79\)
default	\(\frac {a^{5} \left (-\frac {4 i}{\left (\tan \left (f x +e \right )+i\right )^{4}}-\frac {8}{\tan \left (f x +e \right )+i}+\frac {32}{3 \left (\tan \left (f x +e \right )+i\right )^{3}}+i \ln \left (\tan \left (f x +e \right )+i\right )+\frac {12 i}{\left (\tan \left (f x +e \right )+i\right )^{2}}\right )}{f \,c^{4}}\)	\(79\)
risch	\(-\frac {i a^{5} {\mathrm e}^{8 i \left (f x +e \right )}}{4 c^{4} f}+\frac {i a^{5} {\mathrm e}^{6 i \left (f x +e \right )}}{3 c^{4} f}-\frac {i a^{5} {\mathrm e}^{4 i \left (f x +e \right )}}{2 c^{4} f}+\frac {i a^{5} {\mathrm e}^{2 i \left (f x +e \right )}}{c^{4} f}-\frac {2 a^{5} e}{c^{4} f}-\frac {i a^{5} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{c^{4} f}\)	\(122\)
norman	\(\frac {\frac {a^{5} x}{c}+\frac {a^{5} x \left (\tan ^{8}\left (f x +e \right )\right )}{c}+\frac {8 i a^{5}}{3 c f}+\frac {4 a^{5} x \left (\tan ^{2}\left (f x +e \right )\right )}{c}+\frac {6 a^{5} x \left (\tan ^{4}\left (f x +e \right )\right )}{c}+\frac {4 a^{5} x \left (\tan ^{6}\left (f x +e \right )\right )}{c}-\frac {40 a^{5} \left (\tan ^{3}\left (f x +e \right )\right )}{3 c f}+\frac {32 a^{5} \left (\tan ^{5}\left (f x +e \right )\right )}{3 c f}-\frac {8 a^{5} \left (\tan ^{7}\left (f x +e \right )\right )}{c f}+\frac {20 i a^{5} \left (\tan ^{6}\left (f x +e \right )\right )}{c f}+\frac {44 i a^{5} \left (\tan ^{2}\left (f x +e \right )\right )}{3 c f}}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{4} c^{3}}+\frac {i a^{5} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 c^{4} f}\)	\(226\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

1/f*a^5/c^4*(-4*I/(tan(f*x+e)+I)^4-8/(tan(f*x+e)+I)+32/3/(tan(f*x+e)+I)^3+I*ln(tan(f*x+e)+I)+12*I/(tan(f*x+e)+
I)^2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,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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Fricas [A]
time = 1.61, size = 87, normalized size = 0.60 \begin {gather*} \frac {-3 i \, a^{5} e^{\left (8 i \, f x + 8 i \, e\right )} + 4 i \, a^{5} e^{\left (6 i \, f x + 6 i \, e\right )} - 6 i \, a^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - 12 i \, a^{5} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{12 \, c^{4} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

1/12*(-3*I*a^5*e^(8*I*f*x + 8*I*e) + 4*I*a^5*e^(6*I*f*x + 6*I*e) - 6*I*a^5*e^(4*I*f*x + 4*I*e) + 12*I*a^5*e^(2
*I*f*x + 2*I*e) - 12*I*a^5*log(e^(2*I*f*x + 2*I*e) + 1))/(c^4*f)

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Sympy [A]
time = 0.45, size = 209, normalized size = 1.43 \begin {gather*} - \frac {i a^{5} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{4} f} + \begin {cases} \frac {- 6 i a^{5} c^{12} f^{3} e^{8 i e} e^{8 i f x} + 8 i a^{5} c^{12} f^{3} e^{6 i e} e^{6 i f x} - 12 i a^{5} c^{12} f^{3} e^{4 i e} e^{4 i f x} + 24 i a^{5} c^{12} f^{3} e^{2 i e} e^{2 i f x}}{24 c^{16} f^{4}} & \text {for}\: c^{16} f^{4} \neq 0 \\\frac {x \left (2 a^{5} e^{8 i e} - 2 a^{5} e^{6 i e} + 2 a^{5} e^{4 i e} - 2 a^{5} e^{2 i e}\right )}{c^{4}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**5/(c-I*c*tan(f*x+e))**4,x)

[Out]

-I*a**5*log(exp(2*I*f*x) + exp(-2*I*e))/(c**4*f) + Piecewise(((-6*I*a**5*c**12*f**3*exp(8*I*e)*exp(8*I*f*x) +
8*I*a**5*c**12*f**3*exp(6*I*e)*exp(6*I*f*x) - 12*I*a**5*c**12*f**3*exp(4*I*e)*exp(4*I*f*x) + 24*I*a**5*c**12*f
**3*exp(2*I*e)*exp(2*I*f*x))/(24*c**16*f**4), Ne(c**16*f**4, 0)), (x*(2*a**5*exp(8*I*e) - 2*a**5*exp(6*I*e) +
2*a**5*exp(4*I*e) - 2*a**5*exp(2*I*e))/c**4, True))

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Giac [A]
time = 1.12, size = 227, normalized size = 1.55 \begin {gather*} -\frac {\frac {420 i \, a^{5} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{4}} - \frac {840 i \, a^{5} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c^{4}} + \frac {420 i \, a^{5} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c^{4}} + \frac {2283 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 18264 \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 70644 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 136808 \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 191170 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 136808 \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 70644 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 18264 \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2283 i \, a^{5}}{c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{8}}}{420 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-1/420*(420*I*a^5*log(tan(1/2*f*x + 1/2*e) + 1)/c^4 - 840*I*a^5*log(tan(1/2*f*x + 1/2*e) + I)/c^4 + 420*I*a^5*
log(tan(1/2*f*x + 1/2*e) - 1)/c^4 + (2283*I*a^5*tan(1/2*f*x + 1/2*e)^8 - 18264*a^5*tan(1/2*f*x + 1/2*e)^7 - 70
644*I*a^5*tan(1/2*f*x + 1/2*e)^6 + 136808*a^5*tan(1/2*f*x + 1/2*e)^5 + 191170*I*a^5*tan(1/2*f*x + 1/2*e)^4 - 1
36808*a^5*tan(1/2*f*x + 1/2*e)^3 - 70644*I*a^5*tan(1/2*f*x + 1/2*e)^2 + 18264*a^5*tan(1/2*f*x + 1/2*e) + 2283*
I*a^5)/(c^4*(tan(1/2*f*x + 1/2*e) + I)^8))/f

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Mupad [B]
time = 6.07, size = 146, normalized size = 1.00 \begin {gather*} \frac {a^5\,\left (\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )-6\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^4-12\,{\mathrm {tan}\left (e+f\,x\right )}^2+\frac {8}{3}-\frac {\mathrm {tan}\left (e+f\,x\right )\,32{}\mathrm {i}}{3}-\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}+\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3\,4{}\mathrm {i}+{\mathrm {tan}\left (e+f\,x\right )}^3\,8{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^4\,f\,{\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^5/(c - c*tan(e + f*x)*1i)^4,x)

[Out]

(a^5*(log(tan(e + f*x) + 1i) - (tan(e + f*x)*32i)/3 - log(tan(e + f*x) + 1i)*tan(e + f*x)*4i - 6*log(tan(e + f
*x) + 1i)*tan(e + f*x)^2 + log(tan(e + f*x) + 1i)*tan(e + f*x)^3*4i + log(tan(e + f*x) + 1i)*tan(e + f*x)^4 -
12*tan(e + f*x)^2 + tan(e + f*x)^3*8i + 8/3)*1i)/(c^4*f*(tan(e + f*x)*1i - 1)^4)

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