3.12.52 \(\int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx\) [1152]

Optimal. Leaf size=250 \[ \frac {\sqrt [4]{-1} (5 i c-d) d^{3/2} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a} f}-\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} f}-\frac {(c+2 i d) d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{a f}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}} \]

[Out]

(-1)^(1/4)*(5*I*c-d)*d^(3/2)*arctanh((-1)^(3/4)*d^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c+d*tan(f*x+e))^(1/2
))/f/a^(1/2)-1/2*I*(c-I*d)^(5/2)*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a*tan(f*x+e
))^(1/2))/f*2^(1/2)/a^(1/2)-(c+2*I*d)*d*(a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/a/f+(I*c-d)*(c+d*tan(f
*x+e))^(3/2)/f/(a+I*a*tan(f*x+e))^(1/2)

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Rubi [A]
time = 0.66, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {3639, 3678, 3682, 3625, 214, 3680, 65, 223, 212} \begin {gather*} \frac {\sqrt [4]{-1} d^{3/2} (-d+5 i c) \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a} f}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}-\frac {d (c+2 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{a f}-\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(1/4)*((5*I)*c - d)*d^(3/2)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + d*
Tan[e + f*x]])])/(Sqrt[a]*f) - (I*(c - I*d)^(5/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c -
I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[2]*Sqrt[a]*f) - ((c + (2*I)*d)*d*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c +
d*Tan[e + f*x]])/(a*f) + ((I*c - d)*(c + d*Tan[e + f*x])^(3/2))/(f*Sqrt[a + I*a*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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3625

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

Rule 3639

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

Rule 3678

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

Rule 3680

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

Rule 3682

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

Rubi steps

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(549\) vs. \(2(250)=500\).
time = 8.32, size = 549, normalized size = 2.20 \begin {gather*} \frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (\cos (f x)+i \sin (f x)) \left (-\frac {\left (d^{3/2} (-5 i c+d) \left (\log \left (\frac {(1+i) e^{\frac {i e}{2}} \left (-i d+d e^{i (e+f x)}+i c \left (i+e^{i (e+f x)}\right )-(1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )}{d^{5/2} (-5 i c+d) \left (i+e^{i (e+f x)}\right )}\right )-\log \left (\frac {(1+i) e^{\frac {i e}{2}} \left (c+i d+i c e^{i (e+f x)}+d e^{i (e+f x)}+(1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )}{d^{5/2} (-5 i c+d) \left (-i+e^{i (e+f x)}\right )}\right )\right )+(1+i) (c-i d)^{5/2} \log \left (2 \left (\sqrt {c-i d} \cos (e+f x)+i \sqrt {c-i d} \sin (e+f x)+\sqrt {1+\cos (2 (e+f x))+i \sin (2 (e+f x))} \sqrt {c+d \tan (e+f x)}\right )\right )\right ) (\cos (e)+i \sin (e))}{\sqrt {1+\cos (2 (e+f x))+i \sin (2 (e+f x))}}+(1+i) (\cos (f x)-i \sin (f x)) \sqrt {c+d \tan (e+f x)} \left (c^2+2 i c d-2 d^2-i d^2 \tan (e+f x)\right )\right )}{f \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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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. 2646 vs. \(2 (200 ) = 400\).
time = 0.56, size = 2647, normalized size = 10.59

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2647\)
default	\(\text {Expression too large to display}\)	\(2647\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(I*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*ln((3*a*c
+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1
/2))/(tan(f*x+e)+I))*c^2*d*tan(f*x+e)^2+2*I*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*
x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d^4*tan(f*x+e)^2-4*c^3*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^
(1/2)*(I*a*d)^(1/2)+8*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*d^3+16*(a*(c+d*tan(f*x+e))*(
1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c*d^2-12*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*d^3*ta
n(f*x+e)+4*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(
I*a*d)^(1/2))*a*d^4*tan(f*x+e)-12*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/
2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^3-2*I*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*t
an(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d^4+2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*ln((3*a*c+I
*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2
))/(tan(f*x+e)+I))*c^3+2*I*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan
(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^3*tan(f*x+
e)-I*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*
(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^2*d+I*2^(1/2)*(-a*(I*d-c))^(1/2)
*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+
e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*d^3*tan(f*x+e)^2-2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*ln((3*a
*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^
(1/2))/(tan(f*x+e)+I))*c*d^2*tan(f*x+e)^2+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*
c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)
+I))*c^2*d*tan(f*x+e)-I*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*
x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*d^3-24*I*ln(1/2
*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c
*d^3*tan(f*x+e)+20*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c*d^2*tan(f*x+e)+2*2^(1/2)*(-a*
(I*d-c))^(1/2)*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a
*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*d^3*tan(f*x+e)+2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(
1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*t
an(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c*d^2-2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I
*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I)
)*c^3*tan(f*x+e)^2-10*I*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)
^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^2*d^2*tan(f*x+e)^2-4*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2
)*d^3*tan(f*x+e)^2+10*I*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)
^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^2*d^2-4*I*c^3*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*tan(f*x
+e)-12*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c^2*d+12*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2
*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^3*tan(f*x+e)^2-20*ln(1/2*
(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^
2*d^2*tan(f*x+e)-4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c*d^2*tan(f*x+e)^2+12*(a*(c+d*tan
(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c^2*d*tan(f*x+e)+2*I*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*l
n((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x
+e)))^(1/2))/(tan(f*x+e)+I))*c*d^2*tan(f*x+e))/a/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(I*c-d)/(-tan(f*x
+e)+I)^2/(I*a*d)^(1/2)

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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((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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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. 1196 vs. \(2 (198) = 396\).
time = 1.83, size = 1196, normalized size = 4.78 \begin {gather*} -\frac {{\left (\sqrt {2} a f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (-\frac {i \, \sqrt {2} a f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left (c^{2} - 2 i \, c d - d^{2} + {\left (c^{2} - 2 i \, c d - d^{2}\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}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{c^{2} - 2 i \, c d - d^{2}}\right ) - \sqrt {2} a f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (-\frac {-i \, \sqrt {2} a f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left (c^{2} - 2 i \, c d - d^{2} + {\left (c^{2} - 2 i \, c d - d^{2}\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}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{c^{2} - 2 i \, c d - d^{2}}\right ) + a f \sqrt {\frac {-25 i \, c^{2} d^{3} + 10 \, c d^{4} + i \, d^{5}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (-\frac {4 \, {\left (2 \, \sqrt {2} {\left ({\left (5 i \, c d^{3} - d^{4}\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (5 i \, c d^{3} - d^{4}\right )} e^{\left (i \, f x + 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}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left ({\left (a c d - 3 i \, a d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (a c d + i \, a d^{2}\right )} f\right )} \sqrt {\frac {-25 i \, c^{2} d^{3} + 10 \, c d^{4} + i \, d^{5}}{a f^{2}}}\right )}}{5 \, c^{4} - 4 i \, c^{3} d + 6 \, c^{2} d^{2} - 4 i \, c d^{3} + d^{4} + {\left (5 \, c^{4} - 4 i \, c^{3} d + 6 \, c^{2} d^{2} - 4 i \, c d^{3} + d^{4}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}\right ) - a f \sqrt {\frac {-25 i \, c^{2} d^{3} + 10 \, c d^{4} + i \, d^{5}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (-\frac {4 \, {\left (2 \, \sqrt {2} {\left ({\left (5 i \, c d^{3} - d^{4}\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (5 i \, c d^{3} - d^{4}\right )} e^{\left (i \, f x + 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}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left ({\left (a c d - 3 i \, a d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (a c d + i \, a d^{2}\right )} f\right )} \sqrt {\frac {-25 i \, c^{2} d^{3} + 10 \, c d^{4} + i \, d^{5}}{a f^{2}}}\right )}}{5 \, c^{4} - 4 i \, c^{3} d + 6 \, c^{2} d^{2} - 4 i \, c d^{3} + d^{4} + {\left (5 \, c^{4} - 4 i \, c^{3} d + 6 \, c^{2} d^{2} - 4 i \, c d^{3} + d^{4}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}\right ) + 2 \, \sqrt {2} {\left (-i \, c^{2} + 2 \, c d + i \, d^{2} + {\left (-i \, c^{2} + 2 \, c d + 3 i \, d^{2}\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}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/4*(sqrt(2)*a*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a*f^2))*e^(I*f*x + I*
e)*log(-(I*sqrt(2)*a*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a*f^2))*e^(I*f*x
+ I*e) - sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c^2 - 2*I*c*d - d^2)*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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))/(c^2 - 2*I*c*d - d^2)) - sq
rt(2)*a*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a*f^2))*e^(I*f*x + I*e)*log(-
(-I*sqrt(2)*a*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a*f^2))*e^(I*f*x + I*e)
- sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c^2 - 2*I*c*d - d^2)*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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))/(c^2 - 2*I*c*d - d^2)) + a*f*sqrt(
(-25*I*c^2*d^3 + 10*c*d^4 + I*d^5)/(a*f^2))*e^(I*f*x + I*e)*log(-4*(2*sqrt(2)*((5*I*c*d^3 - d^4)*e^(3*I*f*x +
3*I*e) + (5*I*c*d^3 - d^4)*e^(I*f*x + 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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)) + ((a*c*d - 3*I*a*d^2)*f*e^(2*I*f*x + 2*I*e) + (a*c*d + I*a*d^2)*f)*
sqrt((-25*I*c^2*d^3 + 10*c*d^4 + I*d^5)/(a*f^2)))/(5*c^4 - 4*I*c^3*d + 6*c^2*d^2 - 4*I*c*d^3 + d^4 + (5*c^4 -
4*I*c^3*d + 6*c^2*d^2 - 4*I*c*d^3 + d^4)*e^(2*I*f*x + 2*I*e))) - a*f*sqrt((-25*I*c^2*d^3 + 10*c*d^4 + I*d^5)/(
a*f^2))*e^(I*f*x + I*e)*log(-4*(2*sqrt(2)*((5*I*c*d^3 - d^4)*e^(3*I*f*x + 3*I*e) + (5*I*c*d^3 - d^4)*e^(I*f*x
+ 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))*sqrt(a/(e^(2*I*f*x + 2*I*e)
+ 1)) - ((a*c*d - 3*I*a*d^2)*f*e^(2*I*f*x + 2*I*e) + (a*c*d + I*a*d^2)*f)*sqrt((-25*I*c^2*d^3 + 10*c*d^4 + I*d
^5)/(a*f^2)))/(5*c^4 - 4*I*c^3*d + 6*c^2*d^2 - 4*I*c*d^3 + d^4 + (5*c^4 - 4*I*c^3*d + 6*c^2*d^2 - 4*I*c*d^3 +
d^4)*e^(2*I*f*x + 2*I*e))) + 2*sqrt(2)*(-I*c^2 + 2*c*d + I*d^2 + (-I*c^2 + 2*c*d + 3*I*d^2)*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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))
)*e^(-I*f*x - I*e)/(a*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regu
lar value [

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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