3.1155 \(\int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=200 \[ -\frac {\sqrt [4]{-1} a^{5/2} (c+5 i d) \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f \sqrt {c-i d}}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f} \]

[Out]

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

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Rubi [A]  time = 0.67, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3556, 3601, 3544, 208, 3599, 63, 217, 206} \[ -\frac {\sqrt [4]{-1} a^{5/2} (c+5 i d) \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f \sqrt {c-i d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 206

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

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]

Rule 217

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 3544

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 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3599

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 3601

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 {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx &=-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {a \int \frac {\sqrt {a+i a \tan (e+f x)} \left (\frac {1}{2} a (i c+3 d)+\frac {1}{2} a (c+5 i d) \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{d}\\ &=-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\left (4 a^2\right ) \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx+\frac {(a (i c-5 d)) \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 d}\\ &=-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}-\frac {\left (8 i a^4\right ) \operatorname {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 {\left (a^3 (i c-5 d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 d f}\\ &=-\frac {4 i \sqrt {2} a^{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 {c-i d} f}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\left (a^2 (c+5 i d)\right ) \operatorname {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 )}{d f}\\ &=-\frac {4 i \sqrt {2} a^{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 {c-i d} f}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\left (a^2 (c+5 i d)\right ) \operatorname {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 )}{d f}\\ &=-\frac {\sqrt [4]{-1} a^{5/2} (c+5 i d) \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 )}{d^{3/2} f}-\frac {4 i \sqrt {2} a^{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 {c-i d} f}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\\ \end {align*}

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Mathematica [B]  time = 7.22, size = 602, normalized size = 3.01 \[ \frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \cos ^2(e+f x) (a+i a \tan (e+f x))^{5/2} \left (\frac {(-\cos (2 e)+i \sin (2 e)) \cos (e+f x) \left ((8+8 i) d^{3/2} \log \left (2 \left (i \sqrt {c-i d} \sin (e+f x)+\sqrt {c-i d} \cos (e+f x)+\sqrt {i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt {c+d \tan (e+f x)}\right )\right )+\sqrt {c-i d} (c+5 i d) \log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left (-(1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c \left (e^{i (e+f x)}+i\right )+d e^{i (e+f x)}-i d\right )}{\sqrt {d} (5 d-i c) \left (e^{i (e+f x)}+i\right )}\right )-\sqrt {c-i d} (c+5 i d) \log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left ((1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{\sqrt {d} (5 d-i c) \left (e^{i (e+f x)}-i\right )}\right )\right )}{\sqrt {c-i d} \sqrt {i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}+(1+i) \sqrt {d} \sin (2 e) \sqrt {c+d \tan (e+f x)}+(-1+i) \sqrt {d} \cos (2 e) \sqrt {c+d \tan (e+f x)}\right )}{d^{3/2} f (\cos (f x)+i \sin (f x))^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/2 + I/2)*Cos[e + f*x]^2*(a + I*a*Tan[e + f*x])^(5/2)*((Cos[e + f*x]*(Sqrt[c - I*d]*(c + (5*I)*d)*Log[((2 +
2*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)))]))/(Sqrt[d]*((-I)*c + 5*d)*(I
+ E^(I*(e + f*x))))] - Sqrt[c - I*d]*(c + (5*I)*d)*Log[((2 + 2*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)))]))/(Sqrt[d]*((-I)*c + 5*d)*(-I + E^(I*(e + f*x))))] + (8 + 8*I)*d^(3/2)*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[2*e] + I*Sin[2*e]))/(Sqrt[c - I*d]*Sqrt[1 + Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]])
- (1 - I)*Sqrt[d]*Cos[2*e]*Sqrt[c + d*Tan[e + f*x]] + (1 + I)*Sqrt[d]*Sin[2*e]*Sqrt[c + d*Tan[e + f*x]]))/(d^(
3/2)*f*(Cos[f*x] + I*Sin[f*x])^2)

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fricas [B]  time = 0.55, size = 781, normalized size = 3.90 \[ -\frac {2 \, \sqrt {2} a^{2} \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}} e^{\left (i \, f x + i \, e\right )} + d f \sqrt {\frac {i \, a^{5} c^{2} - 10 \, a^{5} c d - 25 i \, a^{5} d^{2}}{d^{3} f^{2}}} \log \left (\frac {{\left (2 \, d^{2} f \sqrt {\frac {i \, a^{5} c^{2} - 10 \, a^{5} c d - 25 i \, a^{5} d^{2}}{d^{3} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left (-i \, a^{2} c + 5 \, a^{2} d + {\left (-i \, a^{2} c + 5 \, a^{2} 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}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{-i \, a^{2} c + 5 \, a^{2} d}\right ) - d f \sqrt {\frac {i \, a^{5} c^{2} - 10 \, a^{5} c d - 25 i \, a^{5} d^{2}}{d^{3} f^{2}}} \log \left (-\frac {{\left (2 \, d^{2} f \sqrt {\frac {i \, a^{5} c^{2} - 10 \, a^{5} c d - 25 i \, a^{5} d^{2}}{d^{3} f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left (-i \, a^{2} c + 5 \, a^{2} d + {\left (-i \, a^{2} c + 5 \, a^{2} 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}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{-i \, a^{2} c + 5 \, a^{2} d}\right ) - \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c + d\right )} f^{2}}} d f \log \left (\frac {{\left (\sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c + d\right )} f^{2}}} {\left (i \, c + d\right )} f e^{\left (i \, f x + i \, e\right )} + 4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\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^{2}}\right ) + \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c + d\right )} f^{2}}} d f \log \left (\frac {{\left (\sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c + d\right )} f^{2}}} {\left (-i \, c - d\right )} f e^{\left (i \, f x + i \, e\right )} + 4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\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^{2}}\right )}{2 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*sqrt(2)*a^2*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) + d*f*sqrt((I*a^5*c^2 - 10*a^5*c*d - 25*I*a^5*d^2)/(d^3*f^2))*log((2*d^2*f*s
qrt((I*a^5*c^2 - 10*a^5*c*d - 25*I*a^5*d^2)/(d^3*f^2))*e^(I*f*x + I*e) + sqrt(2)*(-I*a^2*c + 5*a^2*d + (-I*a^2
*c + 5*a^2*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))*s
qrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(-I*a^2*c + 5*a^2*d)) - d*f*sqrt((I*a^5*c^2 - 10*a^5*c*d -
25*I*a^5*d^2)/(d^3*f^2))*log(-(2*d^2*f*sqrt((I*a^5*c^2 - 10*a^5*c*d - 25*I*a^5*d^2)/(d^3*f^2))*e^(I*f*x + I*e)
- sqrt(2)*(-I*a^2*c + 5*a^2*d + (-I*a^2*c + 5*a^2*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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(-I*a^2*c + 5*a^2*d
)) - sqrt(-32*I*a^5/((I*c + d)*f^2))*d*f*log(1/4*(sqrt(-32*I*a^5/((I*c + d)*f^2))*(I*c + d)*f*e^(I*f*x + I*e)
+ 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*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^2) + sqrt(-32*I*a^5/((I*c + d)*f^2))*d*f*log(1/4
*(sqrt(-32*I*a^5/((I*c + d)*f^2))*(-I*c - d)*f*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sqr
t(((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^2))/(d*f)

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giac [A]  time = 2.10, size = 214, normalized size = 1.07 \[ -\frac {{\left (2 \, {\left (d \tan \left (f x + e\right ) + c\right )} a^{2} - 2 \, a^{2} c - 2 i \, a^{2} d\right )} \sqrt {2 \, a d^{2} + 2 \, \sqrt {{\left (d \tan \left (f x + e\right ) + c\right )}^{2} - 2 \, {\left (d \tan \left (f x + e\right ) + c\right )} c + c^{2} + d^{2}} a d} {\left (\frac {i \, {\left (d \tan \left (f x + e\right ) + c\right )} a d - i \, a c d}{a d^{2} + \sqrt {{\left (d \tan \left (f x + e\right ) + c\right )}^{2} a^{2} d^{2} - 2 \, {\left (d \tan \left (f x + e\right ) + c\right )} a^{2} c d^{2} + a^{2} c^{2} d^{2} + a^{2} d^{4}}} + 1\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{4 \, {\left ({\left (d \tan \left (f x + e\right ) + c\right )} d^{2} - c d^{2} + i \, d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*(d*tan(f*x + e) + c)*a^2 - 2*a^2*c - 2*I*a^2*d)*sqrt(2*a*d^2 + 2*sqrt((d*tan(f*x + e) + c)^2 - 2*(d*ta
n(f*x + e) + c)*c + c^2 + d^2)*a*d)*((I*(d*tan(f*x + e) + c)*a*d - I*a*c*d)/(a*d^2 + sqrt((d*tan(f*x + e) + c)
^2*a^2*d^2 - 2*(d*tan(f*x + e) + c)*a^2*c*d^2 + a^2*c^2*d^2 + a^2*d^4)) + 1)*log(abs(d*tan(f*x + e) + c))/((d*
tan(f*x + e) + c)*d^2 - c*d^2 + I*d^3)

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maple [B]  time = 0.50, size = 1295, normalized size = 6.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),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((d^2-2*c*d-c^2)>0)', see `assu
me?` for more details)Is (d^2-2*c*d-c^2)    *(d^2+2*c*d-c^2)    positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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