3.1115 \(\int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=131 \[ \frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3528, 3537, 63, 208} \[ \frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \]

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*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + ((2*I)*a*(c - I*d)^2*Sqrt[c + d
*Tan[e + f*x]])/f + (2*a*(I*c + d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (((2*I)/5)*a*(c + d*Tan[e + f*x])^(5/2)
)/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 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 3528

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

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]

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx &=\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\int (c+d \tan (e+f x))^{3/2} (a (c-i d)+a (i c+d) \tan (e+f x)) \, dx\\ &=\frac {2 a (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\int \left (a (c-i d)^2+i a (c-i d)^2 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)} \, dx\\ &=\frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 a (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\int \frac {a (c-i d)^3-a (i c+d)^3 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 a (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {\left (i a^2 (c-i d)^6\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a^2 (i c+d)^6+a (c-i d)^3 x\right ) \sqrt {c-\frac {d x}{a (i c+d)^3}}} \, dx,x,-a (i c+d)^3 \tan (e+f x)\right )}{f}\\ &=\frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 a (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {\left (2 a^3 (c-i d)^9\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {a^2 c (c-i d)^3 (i c+d)^3}{d}+a^2 (i c+d)^6-\frac {a^2 (c-i d)^3 (i c+d)^3 x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac {2 i a (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 a (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}\\ \end {align*}

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Mathematica [A]  time = 3.52, size = 208, normalized size = 1.59 \[ \frac {\cos (e+f x) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x)) \left (\frac {1}{15} (\sin (e)+i \cos (e)) \sec ^2(e+f x) \sqrt {c+d \tan (e+f x)} \left (\left (23 c^2-35 i c d-18 d^2\right ) \cos (2 (e+f x))+23 c^2+d (11 c-5 i d) \sin (2 (e+f x))-35 i c d-12 d^2\right )-2 i e^{-i e} (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right )\right )}{f} \]

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)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c - (I*d*(-
1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))]/Sqrt[c - I*d]])/E^(I*e) + (Sec[e + f*x]^2*(I*Cos[e] + Sin
[e])*(23*c^2 - (35*I)*c*d - 12*d^2 + (23*c^2 - (35*I)*c*d - 18*d^2)*Cos[2*(e + f*x)] + (11*c - (5*I)*d)*d*Sin[
2*(e + f*x)])*Sqrt[c + d*Tan[e + f*x]])/15))/f

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fricas [B]  time = 0.69, size = 736, normalized size = 5.62 \[ -\frac {15 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {-\frac {4 \, a^{2} c^{5} - 20 i \, a^{2} c^{4} d - 40 \, a^{2} c^{3} d^{2} + 40 i \, a^{2} c^{2} d^{3} + 20 \, a^{2} c d^{4} - 4 i \, a^{2} d^{5}}{f^{2}}} \log \left (\frac {{\left (2 \, a c^{3} - 4 i \, a c^{2} d - 2 \, a c d^{2} - {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, 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 \, a^{2} c^{5} - 20 i \, a^{2} c^{4} d - 40 \, a^{2} c^{3} d^{2} + 40 i \, a^{2} c^{2} d^{3} + 20 \, a^{2} c d^{4} - 4 i \, a^{2} d^{5}}{f^{2}}} + {\left (2 \, a c^{3} - 6 i \, a c^{2} d - 6 \, a c d^{2} + 2 i \, a d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a c^{2} - 2 i \, a c d - a d^{2}}\right ) - 15 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {-\frac {4 \, a^{2} c^{5} - 20 i \, a^{2} c^{4} d - 40 \, a^{2} c^{3} d^{2} + 40 i \, a^{2} c^{2} d^{3} + 20 \, a^{2} c d^{4} - 4 i \, a^{2} d^{5}}{f^{2}}} \log \left (\frac {{\left (2 \, a c^{3} - 4 i \, a c^{2} d - 2 \, a c d^{2} - {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, 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 \, a^{2} c^{5} - 20 i \, a^{2} c^{4} d - 40 \, a^{2} c^{3} d^{2} + 40 i \, a^{2} c^{2} d^{3} + 20 \, a^{2} c d^{4} - 4 i \, a^{2} d^{5}}{f^{2}}} + {\left (2 \, a c^{3} - 6 i \, a c^{2} d - 6 \, a c d^{2} + 2 i \, a d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a c^{2} - 2 i \, a c d - a d^{2}}\right ) - {\left (184 i \, a c^{2} + 192 \, a c d - 104 i \, a d^{2} + {\left (184 i \, a c^{2} + 368 \, a c d - 184 i \, a d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (368 i \, a c^{2} + 560 \, a c d - 192 i \, a 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}}}{60 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

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]

-1/60*(15*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(4*a^2*c^5 - 20*I*a^2*c^4*d - 40*a^2*c^3
*d^2 + 40*I*a^2*c^2*d^3 + 20*a^2*c*d^4 - 4*I*a^2*d^5)/f^2)*log((2*a*c^3 - 4*I*a*c^2*d - 2*a*c*d^2 - (I*f*e^(2*
I*f*x + 2*I*e) + I*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*a^2*c
^5 - 20*I*a^2*c^4*d - 40*a^2*c^3*d^2 + 40*I*a^2*c^2*d^3 + 20*a^2*c*d^4 - 4*I*a^2*d^5)/f^2) + (2*a*c^3 - 6*I*a*
c^2*d - 6*a*c*d^2 + 2*I*a*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*c^2 - 2*I*a*c*d - a*d^2)) - 15*(f*
e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(4*a^2*c^5 - 20*I*a^2*c^4*d - 40*a^2*c^3*d^2 + 40*I*a
^2*c^2*d^3 + 20*a^2*c*d^4 - 4*I*a^2*d^5)/f^2)*log((2*a*c^3 - 4*I*a*c^2*d - 2*a*c*d^2 - (-I*f*e^(2*I*f*x + 2*I*
e) - I*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*a^2*c^5 - 20*I*a^
2*c^4*d - 40*a^2*c^3*d^2 + 40*I*a^2*c^2*d^3 + 20*a^2*c*d^4 - 4*I*a^2*d^5)/f^2) + (2*a*c^3 - 6*I*a*c^2*d - 6*a*
c*d^2 + 2*I*a*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*c^2 - 2*I*a*c*d - a*d^2)) - (184*I*a*c^2 + 192
*a*c*d - 104*I*a*d^2 + (184*I*a*c^2 + 368*a*c*d - 184*I*a*d^2)*e^(4*I*f*x + 4*I*e) + (368*I*a*c^2 + 560*a*c*d
- 192*I*a*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)))
/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

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giac [B]  time = 0.91, size = 315, normalized size = 2.40 \[ -\frac {2 \, {\left (-4 i \, a c^{3} - 12 \, a c^{2} d + 12 i \, a c d^{2} + 4 \, a d^{3}\right )} \arctan \left (\frac {4 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {-6 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a f^{4} - 10 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a c f^{4} - 30 i \, \sqrt {d \tan \left (f x + e\right ) + c} a c^{2} f^{4} - 10 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a d f^{4} - 60 \, \sqrt {d \tan \left (f x + e\right ) + c} a c d f^{4} + 30 i \, \sqrt {d \tan \left (f x + e\right ) + c} a d^{2} f^{4}}{15 \, f^{5}} \]

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*(-4*I*a*c^3 - 12*a*c^2*d + 12*I*a*c*d^2 + 4*a*d^3)*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*s
qrt(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))))/(sqrt(-8*c + 8*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) - 1/1
5*(-6*I*(d*tan(f*x + e) + c)^(5/2)*a*f^4 - 10*I*(d*tan(f*x + e) + c)^(3/2)*a*c*f^4 - 30*I*sqrt(d*tan(f*x + e)
+ c)*a*c^2*f^4 - 10*(d*tan(f*x + e) + c)^(3/2)*a*d*f^4 - 60*sqrt(d*tan(f*x + e) + c)*a*c*d*f^4 + 30*I*sqrt(d*t
an(f*x + e) + c)*a*d^2*f^4)/f^5

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maple [B]  time = 0.21, size = 2785, normalized size = 21.26 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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mupad [B]  time = 28.13, size = 3899, normalized size = 29.76 \[ \text {result too large to display} \]

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]

log(((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)
/f^4)^(1/2)*(((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3
*d^2*f^2)/f^4)^(1/2)*(64*a*c^3*d^3 + 64*a*c*d^5 - 32*c*d^2*f*(((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/
2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) - (16*a
^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^2))/2 - (8*a^3*d^3*(3*c^2 - d^2)*(c
^2 + d^2)^3)/f^3)*((20*a^4*c^2*d^8*f^4 - a^4*d^10*f^4 - 110*a^4*c^4*d^6*f^4 + 100*a^4*c^6*d^4*f^4 - 25*a^4*c^8
*d^2*f^4)^(1/2)/(4*f^4) - (a^2*c^5)/(4*f^2) - (5*a^2*c*d^4)/(4*f^2) + (5*a^2*c^3*d^2)/(2*f^2))^(1/2) - log((((
(-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(
1/2)*(((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^
2)/f^4)^(1/2)*(64*a*c^3*d^3 + 64*a*c*d^5 + 32*c*d^2*f*(((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^
2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) + (16*a^2*d^2*
(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^2))/2 - (8*a^3*d^3*(3*c^2 - d^2)*(c^2 + d^
2)^3)/f^3)*(((20*a^4*c^2*d^8*f^4 - a^4*d^10*f^4 - 110*a^4*c^4*d^6*f^4 + 100*a^4*c^6*d^4*f^4 - 25*a^4*c^8*d^2*f
^4)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/(4*f^4))^(1/2) - log(((-((-a^4*d^2*f^4*(5*c^4
+ d^4 - 10*c^2*d^2)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(((-((-a^4*d^2*f
^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(64*a*
c^3*d^3 + 64*a*c*d^5 + 32*c*d^2*f*(-((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c
*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) + (16*a^2*d^2*(c + d*tan(e + f*x)
)^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^2))/2 - (8*a^3*d^3*(3*c^2 - d^2)*(c^2 + d^2)^3)/f^3)*(-((20*a
^4*c^2*d^8*f^4 - a^4*d^10*f^4 - 110*a^4*c^4*d^6*f^4 + 100*a^4*c^6*d^4*f^4 - 25*a^4*c^8*d^2*f^4)^(1/2) + a^2*c^
5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/(4*f^4))^(1/2) + log(((-((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2
)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(((-((-a^4*d^2*f^4*(5*c^4 + d^4 -
10*c^2*d^2)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(64*a*c^3*d^3 + 64*a*c*d
^5 - 32*c*d^2*f*(-((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*
c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) - (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6
+ 15*c^2*d^4 - 15*c^4*d^2))/f^2))/2 - (8*a^3*d^3*(3*c^2 - d^2)*(c^2 + d^2)^3)/f^3)*((5*a^2*c^3*d^2)/(2*f^2) -
(a^2*c^5)/(4*f^2) - (5*a^2*c*d^4)/(4*f^2) - (20*a^4*c^2*d^8*f^4 - a^4*d^10*f^4 - 110*a^4*c^4*d^6*f^4 + 100*a^
4*c^6*d^4*f^4 - 25*a^4*c^8*d^2*f^4)^(1/2)/(4*f^4))^(1/2) - log(((((-((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^
2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(a*c^4*d^2*32i - a*d^6*32i + 32*c*d^
2*f*(-((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)
/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^
4 - 15*c^4*d^2))/f^2)*(-((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 1
0*a^2*c^3*d^2*f^2)/f^4)^(1/2))/2 - (8*a^3*c*d^2*(c^2 - 3*d^2)*(c^2*1i + d^2*1i)^3)/f^3)*(-((20*a^4*c^2*d^8*f^4
- a^4*d^10*f^4 - 110*a^4*c^4*d^6*f^4 + 100*a^4*c^6*d^4*f^4 - 25*a^4*c^8*d^2*f^4)^(1/2) + a^2*c^5*f^2 + 5*a^2*
c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/(4*f^4))^(1/2) - log(((((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) -
a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(a*c^4*d^2*32i - a*d^6*32i + 32*c*d^2*f*(((-a^4
*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*
(c + d*tan(e + f*x))^(1/2)))/(2*f) + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d
^2))/f^2)*(((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2
*f^2)/f^4)^(1/2))/2 - (8*a^3*c*d^2*(c^2 - 3*d^2)*(c^2*1i + d^2*1i)^3)/f^3)*(((20*a^4*c^2*d^8*f^4 - a^4*d^10*f^
4 - 110*a^4*c^4*d^6*f^4 + 100*a^4*c^6*d^4*f^4 - 25*a^4*c^8*d^2*f^4)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10
*a^2*c^3*d^2*f^2)/(4*f^4))^(1/2) + log(- ((((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2
- 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(a*d^6*32i - a*c^4*d^2*32i + 32*c*d^2*f*(((-a^4*d^2*f^4*(5*
c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e
+ f*x))^(1/2)))/(2*f) + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^2)*((
(-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(
1/2))/2 - (8*a^3*c*d^2*(c^2 - 3*d^2)*(c^2*1i + d^2*1i)^3)/f^3)*((20*a^4*c^2*d^8*f^4 - a^4*d^10*f^4 - 110*a^4*c
^4*d^6*f^4 + 100*a^4*c^6*d^4*f^4 - 25*a^4*c^8*d^2*f^4)^(1/2)/(4*f^4) - (a^2*c^5)/(4*f^2) - (5*a^2*c*d^4)/(4*f^
2) + (5*a^2*c^3*d^2)/(2*f^2))^(1/2) + log(- ((((-((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + a^2*c^5*
f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(a*d^6*32i - a*c^4*d^2*32i + 32*c*d^2*f*(-((-a^4*d^2*f^
4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*
tan(e + f*x))^(1/2)))/(2*f) + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^
2)*(-((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + a^2*c^5*f^2 + 5*a^2*c*d^4*f^2 - 10*a^2*c^3*d^2*f^2)/
f^4)^(1/2))/2 - (8*a^3*c*d^2*(c^2 - 3*d^2)*(c^2*1i + d^2*1i)^3)/f^3)*((5*a^2*c^3*d^2)/(2*f^2) - (a^2*c^5)/(4*f
^2) - (5*a^2*c*d^4)/(4*f^2) - (20*a^4*c^2*d^8*f^4 - a^4*d^10*f^4 - 110*a^4*c^4*d^6*f^4 + 100*a^4*c^6*d^4*f^4 -
25*a^4*c^8*d^2*f^4)^(1/2)/(4*f^4))^(1/2) + ((a*c^2*4i)/f - (a*(c^2 + d^2)*2i)/f)*(c + d*tan(e + f*x))^(1/2) +
(a*(c + d*tan(e + f*x))^(5/2)*2i)/(5*f) + (a*c*(c + d*tan(e + f*x))^(3/2)*2i)/(3*f) + (2*a*d*(c + d*tan(e + f
*x))^(3/2))/(3*f) + (4*a*c*d*(c + d*tan(e + f*x))^(1/2))/f

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int \left (- i c^{2} \sqrt {c + d \tan {\left (e + f x \right )}}\right )\, dx + \int c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\, dx + \int d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- i d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- 2 i c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx\right ) \]

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)), x) + Integral(c**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) +
Integral(d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**3, x) + Integral(-I*d**2*sqrt(c + d*tan(e + f*x))*tan(e
+ f*x)**2, x) + Integral(2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(-2*I*c*d*sqrt(c + d*tan
(e + f*x))*tan(e + f*x), x))

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