3.1133 \(\int \frac{a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=109 \[ \frac{2 i a}{f (c-i d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 a}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}-\frac{2 i a \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]
[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]])
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Rubi [A] time = 0.264716, antiderivative size = 109, normalized size of antiderivative = 1.,
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 =
{3529, 3537, 63, 208} \[ \frac{2 i a}{f (c-i d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 a}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}-\frac{2 i a \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]
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 3529
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 3537
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^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]
Rule 63
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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
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 ) \operatorname{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 ) \operatorname{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*}
Mathematica [A] time = 3.94207, size = 198, normalized size = 1.82 \[ \frac{\cos (e+f x) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x)) \left (\frac{2 (\cos (e)-i \sin (e)) \cos (e+f x) \sqrt{c+d \tan (e+f x)} ((d+4 i c) \cos (e+f x)+3 i d \sin (e+f x))}{3 (c-i d)^2 (c \cos (e+f x)+d \sin (e+f x))^2}-\frac{2 i e^{-i e} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )}{(c-i d)^{5/2}}\right )}{f} \]
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] time = 0.03, size = 2597, normalized size = 23.8 \begin{align*} \text{result too large to display} \end{align*}
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)
[Out]
-3*I/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e)
)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c*d^2-2/3/f*a/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)*d+3*I/f*a/(c^2+d^2)^(5/
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))*c*d^2-3/2*I/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(
c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c*d^2+3/2*I/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2
*c)^(1/2)*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))*c*d^2+3/f*a/
(c^2+d^2)^(5/2)/(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))*c^2*d+I/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^
(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^3-2/f*a/(c^2+d^2)^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))
*c*d-1/f*a/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d
*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c*d-3/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1
/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2*d-3/2/f*a/(c^2+d^2)^(5/2)/(2*(c^2+
d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*
c^2*d+3/2/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*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))*c^2*d-I/f*a/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^
2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2+1/2*I/f*a/(c^2+d^2)^2/(2*(c^2
+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))
*c^2-1/2*I/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*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))*c^3-1/2*I/f*a/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*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))*c^2+1/2*I/f*a/(c^2+d^2)^2/(2*(c^2+d^2
)^(1/2)+2*c)^(1/2)*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))*d^2
-I/f*a/(c^2+d^2)^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))*d^2-1/2*I/f*a/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^
(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*d^2-I/f*a/(c^2+d^2)^(5/2)/(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))*c
^3+1/2*I/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(
1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c^3+2*I/f*a/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*c^2-4/f*a/(c^2+d^2)^2/(c+d
*tan(f*x+e))^(1/2)*c*d+1/2/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2
+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*d^3-2*I/f*a/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*d^2-1/2/
f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*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))*d^3+1/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)
+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*d^3-1/f*a/(c^2+d^2)^(5/2)/(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))*
d^3+2/3*I/f*a/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)*c+1/f*a/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*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))*c*d+2/f*a/(c^2+d^2)^2/(2*(c^2+d^2)^
(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)
)*c*d+I/f*a/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))
^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*d^2+I/f*a/(c^2+d^2)^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))*c^2
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
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
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Fricas [B] time = 3.23654, size = 2099, normalized size = 19.26 \begin{align*} \text{result too large to display} \end{align*}
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]
(((3*c^4 - 12*I*c^3*d - 18*c^2*d^2 + 12*I*c*d^3 + 3*d^4)*f*e^(4*I*f*x + 4*I*e) + (6*c^4 - 12*I*c^3*d - 12*I*c*
d^3 - 6*d^4)*f*e^(2*I*f*x + 2*I*e) + 3*(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 - 2*I*a*
d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a) - ((3*c^4 - 12*I*c^3*d - 18*c^2*d^2 + 12*I*c*d^3 + 3*d^4)*f*e^
(4*I*f*x + 4*I*e) + (6*c^4 - 12*I*c^3*d - 12*I*c*d^3 - 6*d^4)*f*e^(2*I*f*x + 2*I*e) + 3*(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 - 2*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)))/((12*c^4 - 48*I*c^3*d - 72*c^2*d^2 + 48*I*c*d^3 + 12*d^4
)*f*e^(4*I*f*x + 4*I*e) + (24*c^4 - 48*I*c^3*d - 48*I*c*d^3 - 24*d^4)*f*e^(2*I*f*x + 2*I*e) + 12*(c^4 + 2*c^2*
d^2 + d^4)*f)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \frac{i \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 + \int \frac{1}{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{align*}
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]
a*(Integral(I*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) + Integral(1/(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] time = 1.45792, size = 308, normalized size = 2.83 \begin{align*} -2 \, a{\left (\frac{3 \, d \tan \left (f x + e\right ) + 4 \, c - i \, d}{{\left (3 i \, c^{2} f + 6 \, c d f - 3 i \, d^{2} f\right )}{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}} - \frac{4 i \, \arctan \left (\frac{4 \,{\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{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} - i \, \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d - \sqrt{c^{2} + d^{2}} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}}\right )}{{\left (c^{2} f - 2 i \, c d f - d^{2} f\right )} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}{\left (-\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}}\right )} \end{align*}
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)) - 4*I*arct
an(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-8*c + 8*sqrt(c^2 + d^2))
- I*sqrt(-8*c + 8*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-8*c + 8*sqrt(c^2 + d^2))))/((c^2*f - 2*I*c*d*f -
d^2*f)*sqrt(-8*c + 8*sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)))