3.1114 \(\int (a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=166 \[ -\frac {2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\frac {4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 a^2 (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 i a^2 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}-\frac {4 i a^2 (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \]

[Out]

-4*I*a^2*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f+4*I*a^2*(c-I*d)^2*(c+d*tan(f*x+e))^(1/2
)/f+4/3*a^2*(I*c+d)*(c+d*tan(f*x+e))^(3/2)/f+4/5*I*a^2*(c+d*tan(f*x+e))^(5/2)/f-2/7*a^2*(c+d*tan(f*x+e))^(7/2)
/d/f

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Rubi [A]  time = 0.44, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3543, 3528, 3537, 63, 208} \[ -\frac {2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\frac {4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 a^2 (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 i a^2 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}-\frac {4 i a^2 (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*a^2*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + ((4*I)*a^2*(c - I*d)^2*Sqrt[c
+ d*Tan[e + f*x]])/f + (4*a^2*(I*c + d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (((4*I)/5)*a^2*(c + d*Tan[e + f*x
])^(5/2))/f - (2*a^2*(c + d*Tan[e + f*x])^(7/2))/(7*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 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]

Rule 3543

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

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx &=-\frac {2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\int \left (2 a^2+2 i a^2 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2} \, dx\\ &=\frac {4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\int (c+d \tan (e+f x))^{3/2} \left (2 a^2 (c-i d)+2 a^2 (i c+d) \tan (e+f x)\right ) \, dx\\ &=\frac {4 a^2 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\int \left (2 a^2 (c-i d)^2+2 i a^2 (c-i d)^2 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)} \, dx\\ &=\frac {4 i a^2 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^2 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\int \frac {2 a^2 (c-i d)^3-2 a^2 (i c+d)^3 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {4 i a^2 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^2 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\frac {\left (4 i a^4 (c-i d)^6\right ) \operatorname {Subst}\left (\int \frac {1}{\left (4 a^4 (i c+d)^6+2 a^2 (c-i d)^3 x\right ) \sqrt {c-\frac {d x}{2 a^2 (i c+d)^3}}} \, dx,x,-2 a^2 (i c+d)^3 \tan (e+f x)\right )}{f}\\ &=\frac {4 i a^2 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^2 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}-\frac {\left (16 a^6 (c-i d)^9\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {4 a^4 c (c-i d)^3 (i c+d)^3}{d}+4 a^4 (i c+d)^6-\frac {4 a^4 (c-i d)^3 (i c+d)^3 x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac {4 i a^2 (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {4 i a^2 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^2 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\\ \end {align*}

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Mathematica [A]  time = 8.27, size = 271, normalized size = 1.63 \[ \frac {a^2 (\cos (e+f x)+i \sin (e+f x))^2 \left (-\frac {(\cos (2 e)-i \sin (2 e)) \sec ^2(e+f x) \sqrt {c+d \tan (e+f x)} \left (15 c^3+d \left (45 c^2-154 i c d-55 d^2\right ) \tan (e+f x)-322 i c^2 d+\cos (2 (e+f x)) \left (15 c^3+d \left (45 c^2-154 i c d-85 d^2\right ) \tan (e+f x)-322 i c^2 d-535 c d^2+252 i d^3\right )-445 c d^2+168 i d^3\right )}{105 d}-4 i e^{-2 i e} (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right )\right )}{f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[e + f*x] + I*Sin[e + f*x])^2*(((-4*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^((2*I)*e) - (Sec[e + f*x]^2*(Cos[2*e] - I*Sin[2*e])*Sqrt[c +
d*Tan[e + f*x]]*(15*c^3 - (322*I)*c^2*d - 445*c*d^2 + (168*I)*d^3 + d*(45*c^2 - (154*I)*c*d - 55*d^2)*Tan[e +
f*x] + Cos[2*(e + f*x)]*(15*c^3 - (322*I)*c^2*d - 535*c*d^2 + (252*I)*d^3 + d*(45*c^2 - (154*I)*c*d - 85*d^2)*
Tan[e + f*x])))/(105*d)))/(f*(Cos[f*x] + I*Sin[f*x])^2)

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fricas [B]  time = 0.83, size = 924, normalized size = 5.57 \[ -\frac {105 \, {\left (d f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, d f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {16 \, a^{4} c^{5} - 80 i \, a^{4} c^{4} d - 160 \, a^{4} c^{3} d^{2} + 160 i \, a^{4} c^{2} d^{3} + 80 \, a^{4} c d^{4} - 16 i \, a^{4} d^{5}}{f^{2}}} \log \left (\frac {{\left (4 \, a^{2} c^{3} - 8 i \, a^{2} c^{2} d - 4 \, a^{2} 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 {16 \, a^{4} c^{5} - 80 i \, a^{4} c^{4} d - 160 \, a^{4} c^{3} d^{2} + 160 i \, a^{4} c^{2} d^{3} + 80 \, a^{4} c d^{4} - 16 i \, a^{4} d^{5}}{f^{2}}} + {\left (4 \, a^{2} c^{3} - 12 i \, a^{2} c^{2} d - 12 \, a^{2} c d^{2} + 4 i \, a^{2} d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, {\left (a^{2} c^{2} - 2 i \, a^{2} c d - a^{2} d^{2}\right )}}\right ) - 105 \, {\left (d f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, d f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {16 \, a^{4} c^{5} - 80 i \, a^{4} c^{4} d - 160 \, a^{4} c^{3} d^{2} + 160 i \, a^{4} c^{2} d^{3} + 80 \, a^{4} c d^{4} - 16 i \, a^{4} d^{5}}{f^{2}}} \log \left (\frac {{\left (4 \, a^{2} c^{3} - 8 i \, a^{2} c^{2} d - 4 \, a^{2} 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 {16 \, a^{4} c^{5} - 80 i \, a^{4} c^{4} d - 160 \, a^{4} c^{3} d^{2} + 160 i \, a^{4} c^{2} d^{3} + 80 \, a^{4} c d^{4} - 16 i \, a^{4} d^{5}}{f^{2}}} + {\left (4 \, a^{2} c^{3} - 12 i \, a^{2} c^{2} d - 12 \, a^{2} c d^{2} + 4 i \, a^{2} d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, {\left (a^{2} c^{2} - 2 i \, a^{2} c d - a^{2} d^{2}\right )}}\right ) + {\left (120 \, a^{2} c^{3} - 2216 i \, a^{2} c^{2} d - 3048 \, a^{2} c d^{2} + 1336 i \, a^{2} d^{3} + {\left (120 \, a^{2} c^{3} - 2936 i \, a^{2} c^{2} d - 5512 \, a^{2} c d^{2} + 2696 i \, a^{2} d^{3}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (360 \, a^{2} c^{3} - 8088 i \, a^{2} c^{2} d - 12632 \, a^{2} c d^{2} + 4904 i \, a^{2} d^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (360 \, a^{2} c^{3} - 7368 i \, a^{2} c^{2} d - 10168 \, a^{2} c d^{2} + 4504 i \, a^{2} d^{3}\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}}}{420 \, {\left (d f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, d f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*(105*(d*f*e^(6*I*f*x + 6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(16*
a^4*c^5 - 80*I*a^4*c^4*d - 160*a^4*c^3*d^2 + 160*I*a^4*c^2*d^3 + 80*a^4*c*d^4 - 16*I*a^4*d^5)/f^2)*log(1/2*(4*
a^2*c^3 - 8*I*a^2*c^2*d - 4*a^2*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(-(16*a^4*c^5 - 80*I*a^4*c^4*d - 160*a^4*c^3*d^2 + 160*I*a^4*c^2*d^3 +
80*a^4*c*d^4 - 16*I*a^4*d^5)/f^2) + (4*a^2*c^3 - 12*I*a^2*c^2*d - 12*a^2*c*d^2 + 4*I*a^2*d^3)*e^(2*I*f*x + 2*
I*e))*e^(-2*I*f*x - 2*I*e)/(a^2*c^2 - 2*I*a^2*c*d - a^2*d^2)) - 105*(d*f*e^(6*I*f*x + 6*I*e) + 3*d*f*e^(4*I*f*
x + 4*I*e) + 3*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(16*a^4*c^5 - 80*I*a^4*c^4*d - 160*a^4*c^3*d^2 + 160*I*a^4
*c^2*d^3 + 80*a^4*c*d^4 - 16*I*a^4*d^5)/f^2)*log(1/2*(4*a^2*c^3 - 8*I*a^2*c^2*d - 4*a^2*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(-(16*a^4*c^5
- 80*I*a^4*c^4*d - 160*a^4*c^3*d^2 + 160*I*a^4*c^2*d^3 + 80*a^4*c*d^4 - 16*I*a^4*d^5)/f^2) + (4*a^2*c^3 - 12*
I*a^2*c^2*d - 12*a^2*c*d^2 + 4*I*a^2*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a^2*c^2 - 2*I*a^2*c*d - a
^2*d^2)) + (120*a^2*c^3 - 2216*I*a^2*c^2*d - 3048*a^2*c*d^2 + 1336*I*a^2*d^3 + (120*a^2*c^3 - 2936*I*a^2*c^2*d
- 5512*a^2*c*d^2 + 2696*I*a^2*d^3)*e^(6*I*f*x + 6*I*e) + (360*a^2*c^3 - 8088*I*a^2*c^2*d - 12632*a^2*c*d^2 +
4904*I*a^2*d^3)*e^(4*I*f*x + 4*I*e) + (360*a^2*c^3 - 7368*I*a^2*c^2*d - 10168*a^2*c*d^2 + 4504*I*a^2*d^3)*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)))/(d*f*e^(6*I*f*x +
6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*d*f*e^(2*I*f*x + 2*I*e) + d*f)

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giac [B]  time = 2.02, size = 375, normalized size = 2.26 \[ \frac {2 \, {\left (8 i \, a^{2} c^{3} + 24 \, a^{2} c^{2} d - 24 i \, a^{2} c d^{2} - 8 \, a^{2} d^{3}\right )} \arctan \left (\frac {4 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {30 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} d^{6} f^{6} - 84 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} d^{7} f^{6} - 140 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} c d^{7} f^{6} - 420 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{2} c^{2} d^{7} f^{6} - 140 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} d^{8} f^{6} - 840 \, \sqrt {d \tan \left (f x + e\right ) + c} a^{2} c d^{8} f^{6} + 420 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{2} d^{9} f^{6}}{105 \, d^{7} f^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(8*I*a^2*c^3 + 24*a^2*c^2*d - 24*I*a^2*c*d^2 - 8*a^2*d^3)*arctan(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))))/(sqrt(-8*c + 8*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1))
- 1/105*(30*(d*tan(f*x + e) + c)^(7/2)*a^2*d^6*f^6 - 84*I*(d*tan(f*x + e) + c)^(5/2)*a^2*d^7*f^6 - 140*I*(d*t
an(f*x + e) + c)^(3/2)*a^2*c*d^7*f^6 - 420*I*sqrt(d*tan(f*x + e) + c)*a^2*c^2*d^7*f^6 - 140*(d*tan(f*x + e) +
c)^(3/2)*a^2*d^8*f^6 - 840*sqrt(d*tan(f*x + e) + c)*a^2*c*d^8*f^6 + 420*I*sqrt(d*tan(f*x + e) + c)*a^2*d^9*f^6
)/(d^7*f^7)

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maple [B]  time = 0.22, size = 2886, normalized size = 17.39 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*I/f*a^2*d^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+6*I/f*a^2*d^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-I/f*a^2*d^4/(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))+4/5*I*a^2*(
c+d*tan(f*x+e))^(5/2)/f+2/f*a^2*d/(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+2/f*a^2*d^3/(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-2/f*a^2*d/
(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^3-2/f*a^2*d^3/(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*t
an(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c-4/f*a^2*d^3/(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/f*a^2*d/(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+4/f*a^2*d^3/(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+I/f*a^2*d^4/(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))-2/7*a^2*(c+d*tan(f*x+e))^(7/2)/d/f-2/f*a^2*d^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*t
an(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/f*a^2*d^3/(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))+2/f*a^2*d^3
/(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))+8/f*a^2*d*(c+d*tan(f*x+e))^(1/2)*c+4/3*I/f*a^2*(c+d*tan(f*x+e))^(3/2)*c-4*I/f*a^2*d^2*(c+d*tan
(f*x+e))^(1/2)+4*I/f*a^2*c^2*(c+d*tan(f*x+e))^(1/2)+4/3/f*a^2*d*(c+d*tan(f*x+e))^(3/2)+I/f*a^2/(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^4-3*I/f*a^2*d^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*I/f*a^2*d^4/(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))-4/f*a^2*d/(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+2*I/f*a^2*d^4/(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))+3*I/f*a^2*d^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*I/f*a^2/(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-I/f*a^2/(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^4-2*I/f*a^2/(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+I/f*a^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^3-6/f*a^2*d/(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^2-3/f*a^2*d/(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/f*a^2*d^3/(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))+6/f*a^2*d/(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+3/f*a^2*d/(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-I/f*a^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^3-2*I/f*a^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*I/f*a^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

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

[Out]

Timed out

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mupad [B]  time = 19.22, size = 257, normalized size = 1.55 \[ -\left (\frac {2\,a^2\,\left (c-d\,1{}\mathrm {i}\right )}{5\,d\,f}-\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{5\,d\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}-{\left (c-d\,1{}\mathrm {i}\right )}^2\,\left (\frac {2\,a^2\,\left (c-d\,1{}\mathrm {i}\right )}{d\,f}-\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{d\,f}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}-\frac {\left (c-d\,1{}\mathrm {i}\right )\,\left (\frac {2\,a^2\,\left (c-d\,1{}\mathrm {i}\right )}{d\,f}-\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{d\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3}-\frac {2\,a^2\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}}{7\,d\,f}-\frac {\sqrt {4{}\mathrm {i}}\,a^2\,\mathrm {atan}\left (\frac {\sqrt {4{}\mathrm {i}}\,{\left (d+c\,1{}\mathrm {i}\right )}^{5/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{2\,\left (-c^3\,1{}\mathrm {i}-3\,c^2\,d+c\,d^2\,3{}\mathrm {i}+d^3\right )}\right )\,{\left (d+c\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*a^2*(c - d*1i))/(5*d*f) - (2*a^2*(c + d*1i))/(5*d*f))*(c + d*tan(e + f*x))^(5/2) - (c - d*1i)^2*((2*a^2*
(c - d*1i))/(d*f) - (2*a^2*(c + d*1i))/(d*f))*(c + d*tan(e + f*x))^(1/2) - ((c - d*1i)*((2*a^2*(c - d*1i))/(d*
f) - (2*a^2*(c + d*1i))/(d*f))*(c + d*tan(e + f*x))^(3/2))/3 - (2*a^2*(c + d*tan(e + f*x))^(7/2))/(7*d*f) - (4
i^(1/2)*a^2*atan((4i^(1/2)*(c*1i + d)^(5/2)*(c + d*tan(e + f*x))^(1/2)*1i)/(2*(c*d^2*3i - 3*c^2*d - c^3*1i + d
^3)))*(c*1i + d)^(5/2)*2i)/f

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \left (- c^{2} \sqrt {c + d \tan {\left (e + f x \right )}}\right )\, dx + \int c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{4}{\left (e + f x \right )}\, dx + \int \left (- 2 i c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- 2 i d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx + \int 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- 4 i c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*(Integral(-c**2*sqrt(c + d*tan(e + f*x)), x) + Integral(c**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x
) + Integral(-d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(d**2*sqrt(c + d*tan(e + f*x))*tan(e
+ f*x)**4, x) + Integral(-2*I*c**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(-2*I*d**2*sqrt(c + d*
tan(e + f*x))*tan(e + f*x)**3, x) + Integral(-2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(2*c*d
*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**3, x) + Integral(-4*I*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x)
)

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