3.12.33 \(\int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1133]

Optimal. Leaf size=109 \[ -\frac {2 i a \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}-\frac {2 a}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {2 i a}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}} \]

[Out]

-2*I*a*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(5/2)/f+2*I*a/(c-I*d)^2/f/(c+d*tan(f*x+e))^(1/2)-
2/3*a/(I*c+d)/f/(c+d*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.19, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3610, 3618, 65, 214} \begin {gather*} \frac {2 i a}{f (c-i d)^2 \sqrt {c+d \tan (e+f x)}}-\frac {2 a}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}-\frac {2 i a \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (c-i d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

((-2*I)*a*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(5/2)*f) - (2*a)/(3*(I*c + d)*f*(c + d*T
an[e + f*x])^(3/2)) + ((2*I)*a)/((c - I*d)^2*f*Sqrt[c + d*Tan[e + f*x]])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3610

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

Rule 3618

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

Rubi steps

\begin {align*} \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {2 a}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {\int \frac {a (c+i d)+a (i c-d) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx}{c^2+d^2}\\ &=-\frac {2 a}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {2 i a}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\int \frac {a (c+i d)^2+i a (c+i d)^2 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{\left (c^2+d^2\right )^2}\\ &=-\frac {2 a}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {2 i a}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\left (i a^2 (c+i d)^4\right ) \text {Subst}\left (\int \frac {1}{\left (-a^2 (c+i d)^4+a (c+i d)^2 x\right ) \sqrt {c-\frac {i d x}{a (c+i d)^2}}} \, dx,x,i a (c+i d)^2 \tan (e+f x)\right )}{\left (c^2+d^2\right )^2 f}\\ &=-\frac {2 a}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {2 i a}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}}-\frac {\left (2 a^3 (c+i d)^6\right ) \text {Subst}\left (\int \frac {1}{-a^2 (c+i d)^4-\frac {i a^2 c (c+i d)^4}{d}+\frac {i a^2 (c+i d)^4 x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{d \left (c^2+d^2\right )^2 f}\\ &=-\frac {2 i a \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}-\frac {2 a}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {2 i a}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 4.03, size = 198, normalized size = 1.82 \begin {gather*} \frac {\cos (e+f x) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x)) \left (-\frac {2 i e^{-i e} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2}}+\frac {2 \cos (e+f x) (\cos (e)-i \sin (e)) ((4 i c+d) \cos (e+f x)+3 i d \sin (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 (c-i d)^2 (c \cos (e+f x)+d \sin (e+f x))^2}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

(Cos[e + f*x]*(Cos[f*x] - I*Sin[f*x])*(a + I*a*Tan[e + f*x])*(((-2*I)*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e
+ f*x))))/(1 + E^((2*I)*(e + f*x)))]/Sqrt[c - I*d]])/((c - I*d)^(5/2)*E^(I*e)) + (2*Cos[e + f*x]*(Cos[e] - I*S
in[e])*(((4*I)*c + d)*Cos[e + f*x] + (3*I)*d*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*(c - I*d)^2*(c*Cos[e +
f*x] + d*Sin[e + f*x])^2)))/f

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 884 vs. \(2 (91 ) = 182\).
time = 0.32, size = 885, normalized size = 8.12

method	result	size
derivativedivides	\(\frac {a \left (-\frac {2 \left (-i c +d \right )}{3 \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2 \left (-i c^{2}+i d^{2}+2 c d \right )}{\left (c^{2}+d^{2}\right )^{2} \sqrt {c +d \tan \left (f x +e \right )}}+\frac {\frac {\frac {\left (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}-3 i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-3 c^{2} d +d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}+3 i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}+3 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}+\frac {\left (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}-3 i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-3 c^{2} d +d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}+3 i c \,d^{2}+2 c d \sqrt {c^{2}+d^{2}}+3 c^{2} d -d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}+3 i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}+3 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}-\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}+3 i c \,d^{2}+2 c d \sqrt {c^{2}+d^{2}}+3 c^{2} d -d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}}{\left (c^{2}+d^{2}\right )^{2}}\right )}{f}\)	\(885\)
default	\(\frac {a \left (-\frac {2 \left (-i c +d \right )}{3 \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2 \left (-i c^{2}+i d^{2}+2 c d \right )}{\left (c^{2}+d^{2}\right )^{2} \sqrt {c +d \tan \left (f x +e \right )}}+\frac {\frac {\frac {\left (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}-3 i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-3 c^{2} d +d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}+3 i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}+3 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}+\frac {\left (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}-3 i c \,d^{2}-2 c d \sqrt {c^{2}+d^{2}}-3 c^{2} d +d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}+3 i c \,d^{2}+2 c d \sqrt {c^{2}+d^{2}}+3 c^{2} d -d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}+3 i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}+3 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}-\frac {\left (-i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}-i c^{3}+3 i c \,d^{2}+2 c d \sqrt {c^{2}+d^{2}}+3 c^{2} d -d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}}{\left (c^{2}+d^{2}\right )^{2}}\right )}{f}\)	\(885\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((-(2*c*d^4)/((c^2-d^2)^2>0)',
see `assume

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 843 vs. \(2 (88) = 176\).
time = 1.02, size = 843, normalized size = 7.73 \begin {gather*} \frac {3 \, {\left ({\left (c^{4} - 4 i \, c^{3} d - 6 \, c^{2} d^{2} + 4 i \, c d^{3} + d^{4}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} - 2 i \, c^{3} d - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} + 2 \, c^{2} d^{2} + d^{4}\right )} f\right )} \sqrt {-\frac {4 i \, a^{2}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} \log \left (\frac {{\left (2 \, a c + {\left ({\left (i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i \, a^{2}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} + 2 \, {\left (a c - i \, a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) - 3 \, {\left ({\left (c^{4} - 4 i \, c^{3} d - 6 \, c^{2} d^{2} + 4 i \, c d^{3} + d^{4}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} - 2 i \, c^{3} d - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} + 2 \, c^{2} d^{2} + d^{4}\right )} f\right )} \sqrt {-\frac {4 i \, a^{2}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} \log \left (\frac {{\left (2 \, a c + {\left ({\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i \, a^{2}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} + 2 \, {\left (a c - i \, a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) - 16 \, {\left (-2 i \, a c + a d + 2 \, {\left (-i \, a c - a d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-4 i \, a c - a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left ({\left (c^{4} - 4 i \, c^{3} d - 6 \, c^{2} d^{2} + 4 i \, c d^{3} + d^{4}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} - 2 i \, c^{3} d - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} + 2 \, c^{2} d^{2} + d^{4}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/12*(3*((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f*x + 4*I*e) + 2*(c^4 - 2*I*c^3*d - 2*I*c*d^
3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c^4 + 2*c^2*d^2 + d^4)*f)*sqrt(-4*I*a^2/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 1
0*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log((2*a*c + ((I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (
I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)
)*sqrt(-4*I*a^2/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2)) + 2*(a*c - I*a*d)*e^(2*
I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a) - 3*((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f*x + 4*
I*e) + 2*(c^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c^4 + 2*c^2*d^2 + d^4)*f)*sqrt(-4*I*a^2/
((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log((2*a*c + ((-I*c^3 - 3*c^2*d + 3*I*c
*d^2 + d^3)*f*e^(2*I*f*x + 2*I*e) + (-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e
) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-4*I*a^2/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^
4 + d^5)*f^2)) + 2*(a*c - I*a*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a) - 16*(-2*I*a*c + a*d + 2*(-I*a*c
- a*d)*e^(4*I*f*x + 4*I*e) + (-4*I*a*c - a*d)*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c +
I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f*x + 4*I*e) + 2*(c
^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c^4 + 2*c^2*d^2 + d^4)*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i a \left (\int \left (- \frac {i}{c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} + 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} + d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}\right )\, dx + \int \frac {\tan {\left (e + f x \right )}}{c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} + 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} + d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))**(5/2),x)

[Out]

I*a*(Integral(-I/(c**2*sqrt(c + d*tan(e + f*x)) + 2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x) + d**2*sqrt(c +
d*tan(e + f*x))*tan(e + f*x)**2), x) + Integral(tan(e + f*x)/(c**2*sqrt(c + d*tan(e + f*x)) + 2*c*d*sqrt(c + d
*tan(e + f*x))*tan(e + f*x) + d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2), x))

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (88) = 176\).
time = 0.81, size = 228, normalized size = 2.09 \begin {gather*} -2 \, a {\left (\frac {3 \, d \tan \left (f x + e\right ) + 4 \, c - i \, d}{-3 \, {\left (-i \, c^{2} f - 2 \, c d f + i \, d^{2} f\right )} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} - \frac {2 i \, \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{{\left (c^{2} f - 2 i \, c d f - d^{2} f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

-2*a*((3*d*tan(f*x + e) + 4*c - I*d)/((3*I*c^2*f + 6*c*d*f - 3*I*d^2*f)*(d*tan(f*x + e) + c)^(3/2)) - 2*I*arct
an(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2))
- I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/((c^2*f - 2*I*c*d*f -
d^2*f)*sqrt(-2*c + 2*sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)))

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Mupad [B]
time = 25.07, size = 2500, normalized size = 22.94 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)/(c + d*tan(e + f*x))^(5/2),x)

[Out]

((a*c*2i)/(3*f*(c^2 + d^2)) + (a*(c^2 - d^2)*(c + d*tan(e + f*x))*2i)/(f*(c^2 + d^2)^2))/(c + d*tan(e + f*x))^
(3/2) + (log(((((320*a^4*c^2*d^8*f^4 - 16*a^4*d^10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*d^4*f^4 - 400*a^4
*c^8*d^2*f^4)^(1/2) - 4*a^2*c^5*f^2 - 20*a^2*c*d^4*f^2 + 40*a^2*c^3*d^2*f^2)/(c^10*f^4 + d^10*f^4 + 5*c^2*d^8*
f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2)*((c + d*tan(e + f*x))^(1/2)*(320*a^2*c^4*d^14*f^
3 - 16*a^2*d^18*f^3 + 1024*a^2*c^6*d^12*f^3 + 1440*a^2*c^8*d^10*f^3 + 1024*a^2*c^10*d^8*f^3 + 320*a^2*c^12*d^6
*f^3 - 16*a^2*c^16*d^2*f^3) - ((((320*a^4*c^2*d^8*f^4 - 16*a^4*d^10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*
d^4*f^4 - 400*a^4*c^8*d^2*f^4)^(1/2) - 4*a^2*c^5*f^2 - 20*a^2*c*d^4*f^2 + 40*a^2*c^3*d^2*f^2)/(c^10*f^4 + d^10
*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2)*(((((320*a^4*c^2*d^8*f^4 - 16*a
^4*d^10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*d^4*f^4 - 400*a^4*c^8*d^2*f^4)^(1/2) - 4*a^2*c^5*f^2 - 20*a^
2*c*d^4*f^2 + 40*a^2*c^3*d^2*f^2)/(c^10*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c
^8*d^2*f^4))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(64*c*d^22*f^5 + 640*c^3*d^20*f^5 + 2880*c^5*d^18*f^5 + 7680*c^7
*d^16*f^5 + 13440*c^9*d^14*f^5 + 16128*c^11*d^12*f^5 + 13440*c^13*d^10*f^5 + 7680*c^15*d^8*f^5 + 2880*c^17*d^6
*f^5 + 640*c^19*d^4*f^5 + 64*c^21*d^2*f^5))/4 - 32*a*d^21*f^4 - 160*a*c^2*d^19*f^4 - 128*a*c^4*d^17*f^4 + 896*
a*c^6*d^15*f^4 + 3136*a*c^8*d^13*f^4 + 4928*a*c^10*d^11*f^4 + 4480*a*c^12*d^9*f^4 + 2432*a*c^14*d^7*f^4 + 736*
a*c^16*d^5*f^4 + 96*a*c^18*d^3*f^4))/4))/4 + 16*a^3*c*d^15*f^2 + 96*a^3*c^3*d^13*f^2 + 240*a^3*c^5*d^11*f^2 +
320*a^3*c^7*d^9*f^2 + 240*a^3*c^9*d^7*f^2 + 96*a^3*c^11*d^5*f^2 + 16*a^3*c^13*d^3*f^2)*(((320*a^4*c^2*d^8*f^4
- 16*a^4*d^10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*d^4*f^4 - 400*a^4*c^8*d^2*f^4)^(1/2) - 4*a^2*c^5*f^2 -
20*a^2*c*d^4*f^2 + 40*a^2*c^3*d^2*f^2)/(c^10*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4
+ 5*c^8*d^2*f^4))^(1/2))/4 + (log(((-((320*a^4*c^2*d^8*f^4 - 16*a^4*d^10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^
4*c^6*d^4*f^4 - 400*a^4*c^8*d^2*f^4)^(1/2) + 4*a^2*c^5*f^2 + 20*a^2*c*d^4*f^2 - 40*a^2*c^3*d^2*f^2)/(c^10*f^4
+ d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2)*((c + d*tan(e + f*x))^(1/
2)*(320*a^2*c^4*d^14*f^3 - 16*a^2*d^18*f^3 + 1024*a^2*c^6*d^12*f^3 + 1440*a^2*c^8*d^10*f^3 + 1024*a^2*c^10*d^8
*f^3 + 320*a^2*c^12*d^6*f^3 - 16*a^2*c^16*d^2*f^3) - ((-((320*a^4*c^2*d^8*f^4 - 16*a^4*d^10*f^4 - 1760*a^4*c^4
*d^6*f^4 + 1600*a^4*c^6*d^4*f^4 - 400*a^4*c^8*d^2*f^4)^(1/2) + 4*a^2*c^5*f^2 + 20*a^2*c*d^4*f^2 - 40*a^2*c^3*d
^2*f^2)/(c^10*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2)*(((-((3
20*a^4*c^2*d^8*f^4 - 16*a^4*d^10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*d^4*f^4 - 400*a^4*c^8*d^2*f^4)^(1/2
) + 4*a^2*c^5*f^2 + 20*a^2*c*d^4*f^2 - 40*a^2*c^3*d^2*f^2)/(c^10*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f
^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(64*c*d^22*f^5 + 640*c^3*d^20*f^5 + 288
0*c^5*d^18*f^5 + 7680*c^7*d^16*f^5 + 13440*c^9*d^14*f^5 + 16128*c^11*d^12*f^5 + 13440*c^13*d^10*f^5 + 7680*c^1
5*d^8*f^5 + 2880*c^17*d^6*f^5 + 640*c^19*d^4*f^5 + 64*c^21*d^2*f^5))/4 - 32*a*d^21*f^4 - 160*a*c^2*d^19*f^4 -
128*a*c^4*d^17*f^4 + 896*a*c^6*d^15*f^4 + 3136*a*c^8*d^13*f^4 + 4928*a*c^10*d^11*f^4 + 4480*a*c^12*d^9*f^4 + 2
432*a*c^14*d^7*f^4 + 736*a*c^16*d^5*f^4 + 96*a*c^18*d^3*f^4))/4))/4 + 16*a^3*c*d^15*f^2 + 96*a^3*c^3*d^13*f^2
+ 240*a^3*c^5*d^11*f^2 + 320*a^3*c^7*d^9*f^2 + 240*a^3*c^9*d^7*f^2 + 96*a^3*c^11*d^5*f^2 + 16*a^3*c^13*d^3*f^2
)*(-((320*a^4*c^2*d^8*f^4 - 16*a^4*d^10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*d^4*f^4 - 400*a^4*c^8*d^2*f^
4)^(1/2) + 4*a^2*c^5*f^2 + 20*a^2*c*d^4*f^2 - 40*a^2*c^3*d^2*f^2)/(c^10*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^
4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2))/4 - log(16*a^3*c*d^15*f^2 - (((320*a^4*c^2*d^8*f^4 - 16*a^
4*d^10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*d^4*f^4 - 400*a^4*c^8*d^2*f^4)^(1/2) - 4*a^2*c^5*f^2 - 20*a^2
*c*d^4*f^2 + 40*a^2*c^3*d^2*f^2)/(16*c^10*f^4 + 16*d^10*f^4 + 80*c^2*d^8*f^4 + 160*c^4*d^6*f^4 + 160*c^6*d^4*f
^4 + 80*c^8*d^2*f^4))^(1/2)*((c + d*tan(e + f*x))^(1/2)*(320*a^2*c^4*d^14*f^3 - 16*a^2*d^18*f^3 + 1024*a^2*c^6
*d^12*f^3 + 1440*a^2*c^8*d^10*f^3 + 1024*a^2*c^10*d^8*f^3 + 320*a^2*c^12*d^6*f^3 - 16*a^2*c^16*d^2*f^3) + (((3
20*a^4*c^2*d^8*f^4 - 16*a^4*d^10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*d^4*f^4 - 400*a^4*c^8*d^2*f^4)^(1/2
) - 4*a^2*c^5*f^2 - 20*a^2*c*d^4*f^2 + 40*a^2*c^3*d^2*f^2)/(16*c^10*f^4 + 16*d^10*f^4 + 80*c^2*d^8*f^4 + 160*c
^4*d^6*f^4 + 160*c^6*d^4*f^4 + 80*c^8*d^2*f^4))^(1/2)*(896*a*c^6*d^15*f^4 - (((320*a^4*c^2*d^8*f^4 - 16*a^4*d^
10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*d^4*f^4 - 400*a^4*c^8*d^2*f^4)^(1/2) - 4*a^2*c^5*f^2 - 20*a^2*c*d
^4*f^2 + 40*a^2*c^3*d^2*f^2)/(16*c^10*f^4 + 16*d^10*f^4 + 80*c^2*d^8*f^4 + 160*c^4*d^6*f^4 + 160*c^6*d^4*f^4 +
80*c^8*d^2*f^4))^(1/2)*(c + d*tan(e + f*x))^(1...

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