3.10.7 \(\int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3 \, dx\) [907]

Optimal. Leaf size=61 \[ -\frac {i a^2 c^3 \sec ^4(e+f x)}{4 f}+\frac {a^2 c^3 \tan (e+f x)}{f}+\frac {a^2 c^3 \tan ^3(e+f x)}{3 f} \]

[Out]

-1/4*I*a^2*c^3*sec(f*x+e)^4/f+a^2*c^3*tan(f*x+e)/f+1/3*a^2*c^3*tan(f*x+e)^3/f

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 61, 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, 3567, 3852} \begin {gather*} \frac {a^2 c^3 \tan ^3(e+f x)}{3 f}+\frac {a^2 c^3 \tan (e+f x)}{f}-\frac {i a^2 c^3 \sec ^4(e+f x)}{4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^3,x]

[Out]

((-1/4*I)*a^2*c^3*Sec[e + f*x]^4)/f + (a^2*c^3*Tan[e + f*x])/f + (a^2*c^3*Tan[e + f*x]^3)/(3*f)

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

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]))

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3 \, dx &=\left (a^2 c^2\right ) \int \sec ^4(e+f x) (c-i c \tan (e+f x)) \, dx\\ &=-\frac {i a^2 c^3 \sec ^4(e+f x)}{4 f}+\left (a^2 c^3\right ) \int \sec ^4(e+f x) \, dx\\ &=-\frac {i a^2 c^3 \sec ^4(e+f x)}{4 f}-\frac {\left (a^2 c^3\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{f}\\ &=-\frac {i a^2 c^3 \sec ^4(e+f x)}{4 f}+\frac {a^2 c^3 \tan (e+f x)}{f}+\frac {a^2 c^3 \tan ^3(e+f x)}{3 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.03, size = 52, normalized size = 0.85 \begin {gather*} \frac {a^2 c^3 \sec (e) \sec ^4(e+f x) (-3 i \cos (e)-3 \sin (e)+4 \sin (e+2 f x)+\sin (3 e+4 f x))}{12 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^3,x]

[Out]

(a^2*c^3*Sec[e]*Sec[e + f*x]^4*((-3*I)*Cos[e] - 3*Sin[e] + 4*Sin[e + 2*f*x] + Sin[3*e + 4*f*x]))/(12*f)

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 54, normalized size = 0.89

method	result	size
risch	\(\frac {4 i a^{2} c^{3} \left (4 \,{\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}\)	\(39\)
derivativedivides	\(\frac {i a^{2} c^{3} \left (-i \tan \left (f x +e \right )-\frac {\left (\tan ^{4}\left (f x +e \right )\right )}{4}-\frac {i \left (\tan ^{3}\left (f x +e \right )\right )}{3}-\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}\right )}{f}\)	\(54\)
default	\(\frac {i a^{2} c^{3} \left (-i \tan \left (f x +e \right )-\frac {\left (\tan ^{4}\left (f x +e \right )\right )}{4}-\frac {i \left (\tan ^{3}\left (f x +e \right )\right )}{3}-\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}\right )}{f}\)	\(54\)
norman	\(\frac {a^{2} c^{3} \tan \left (f x +e \right )}{f}+\frac {a^{2} c^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{3 f}-\frac {i a^{2} c^{3} \left (\tan ^{2}\left (f x +e \right )\right )}{2 f}-\frac {i a^{2} c^{3} \left (\tan ^{4}\left (f x +e \right )\right )}{4 f}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

I/f*a^2*c^3*(-I*tan(f*x+e)-1/4*tan(f*x+e)^4-1/3*I*tan(f*x+e)^3-1/2*tan(f*x+e)^2)

________________________________________________________________________________________

Maxima [A]
time = 0.53, size = 72, normalized size = 1.18 \begin {gather*} -\frac {3 i \, a^{2} c^{3} \tan \left (f x + e\right )^{4} - 4 \, a^{2} c^{3} \tan \left (f x + e\right )^{3} + 6 i \, a^{2} c^{3} \tan \left (f x + e\right )^{2} - 12 \, a^{2} c^{3} \tan \left (f x + e\right )}{12 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 1.20, size = 84, normalized size = 1.38 \begin {gather*} -\frac {4 \, {\left (-4 i \, a^{2} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} c^{3}\right )}}{3 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (53) = 106\).
time = 0.28, size = 112, normalized size = 1.84 \begin {gather*} \frac {16 i a^{2} c^{3} e^{2 i e} e^{2 i f x} + 4 i a^{2} c^{3}}{3 f e^{8 i e} e^{8 i f x} + 12 f e^{6 i e} e^{6 i f x} + 18 f e^{4 i e} e^{4 i f x} + 12 f e^{2 i e} e^{2 i f x} + 3 f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**2*(c-I*c*tan(f*x+e))**3,x)

[Out]

(16*I*a**2*c**3*exp(2*I*e)*exp(2*I*f*x) + 4*I*a**2*c**3)/(3*f*exp(8*I*e)*exp(8*I*f*x) + 12*f*exp(6*I*e)*exp(6*
I*f*x) + 18*f*exp(4*I*e)*exp(4*I*f*x) + 12*f*exp(2*I*e)*exp(2*I*f*x) + 3*f)

________________________________________________________________________________________

Giac [A]
time = 0.62, size = 84, normalized size = 1.38 \begin {gather*} -\frac {4 \, {\left (-4 i \, a^{2} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} c^{3}\right )}}{3 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 4.75, size = 80, normalized size = 1.31 \begin {gather*} \frac {a^2\,c^3\,\sin \left (e+f\,x\right )\,\left (12\,{\cos \left (e+f\,x\right )}^3-{\cos \left (e+f\,x\right )}^2\,\sin \left (e+f\,x\right )\,6{}\mathrm {i}+4\,\cos \left (e+f\,x\right )\,{\sin \left (e+f\,x\right )}^2-{\sin \left (e+f\,x\right )}^3\,3{}\mathrm {i}\right )}{12\,f\,{\cos \left (e+f\,x\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^2*(c - c*tan(e + f*x)*1i)^3,x)

[Out]

(a^2*c^3*sin(e + f*x)*(4*cos(e + f*x)*sin(e + f*x)^2 - cos(e + f*x)^2*sin(e + f*x)*6i + 12*cos(e + f*x)^3 - si
n(e + f*x)^3*3i))/(12*f*cos(e + f*x)^4)

________________________________________________________________________________________