3.2.86 \(\int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [186]

Optimal. Leaf size=176 \[ \frac {7 (-1)^{3/4} \sqrt {a} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}+\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.35, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3641, 3678, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {7 (-1)^{3/4} \sqrt {a} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(-1)^(3/4)*Sqrt[a]*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(4*d) + ((1
+ I)*Sqrt[a]*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - ((I/4)*Sqrt[Tan[c +
d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d + (Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(2*d)

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 209

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[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 3641

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[d*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[1/(a*(m + n - 1)), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) - a*c^2*(m + n - 1) + d*(b*d*m
- a*c*(m + 2*n - 2))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[
a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] || 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 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx &=\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {\int \sqrt {\tan (c+d x)} \left (\frac {3 a}{2}+\frac {1}{2} i a \tan (c+d x)\right ) \sqrt {a+i a \tan (c+d x)} \, dx}{2 a}\\ &=-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {i a^2}{4}+\frac {7}{4} a^2 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}+i \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx-\frac {(7 i) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a}\\ &=-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {(7 i a) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {(7 i a) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 d}\\ &=\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {(7 i a) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}\\ &=\frac {7 (-1)^{3/4} \sqrt {a} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}+\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\\ \end {align*}

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Mathematica [A]
time = 4.11, size = 267, normalized size = 1.52 \begin {gather*} \frac {i e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \left (4 \sqrt {2} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+\frac {\sqrt {2} e^{i (c+d x)} \left (1-3 e^{2 i (c+d x)}\right ) \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\tan (c+d x)}}{\left (1+e^{2 i (c+d x)}\right )^2}-7 \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right ) \sqrt {\tan (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{4 \sqrt {2} d \sqrt {-1+e^{2 i (c+d x)}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*(1 + E^((2*I)*(c + d*x)))*(4*Sqrt[2]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*
ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] + (Sqrt[2]*E^(I*(c + d*x))*(1 - 3*E^((2*I)*(c + d*x)))
*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[Tan[c + d*x]])/(1 + E^((2*I)*(c + d*x)))^2 - 7*ArcTanh[(Sqrt[2]*E^(I*(c +
d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]]*Sqrt[Tan[c + d*x]])*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[2]*d*E^(I*(c + d
*x))*Sqrt[-1 + E^((2*I)*(c + d*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. 471 vs. \(2 (137 ) = 274\).
time = 0.35, size = 472, normalized size = 2.68

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (\sqrt {\tan }\left (d x +c \right )\right ) \left (6 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+7 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\, \tan \left (d x +c \right )+4 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a -4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )-4 \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+7 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{8 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \left (-\tan \left (d x +c \right )+i\right ) \sqrt {-i a}}\)	\(472\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (\sqrt {\tan }\left (d x +c \right )\right ) \left (6 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+7 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\, \tan \left (d x +c \right )+4 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a -4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )-4 \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+7 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{8 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \left (-\tan \left (d x +c \right )+i\right ) \sqrt {-i a}}\)	\(472\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(I*a*tan(d*x + c) + a)*tan(d*x + c)^(5/2), x)

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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. 559 vs. \(2 (128) = 256\).
time = 0.37, size = 559, normalized size = 3.18 \begin {gather*} \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + i \, e^{\left (i \, d x + i \, c\right )}\right )} + 2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {49 i \, a}{16 \, d^{2}}} \log \left (\frac {1}{7} \, {\left (7 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + 8 i \, d \sqrt {\frac {49 i \, a}{16 \, d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {49 i \, a}{16 \, d^{2}}} \log \left (\frac {1}{7} \, {\left (7 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - 8 i \, d \sqrt {\frac {49 i \, a}{16 \, d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(-
3*I*e^(3*I*d*x + 3*I*c) + I*e^(I*d*x + I*c)) + 2*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(49/16*I*a/d^2)*log(1/7*(7*sq
rt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d
*x + 2*I*c) + 1) + 8*I*d*sqrt(49/16*I*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 2*(d*e^(2*I*d*x + 2*I*c) + d
)*sqrt(49/16*I*a/d^2)*log(1/7*(7*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(
e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1) - 8*I*d*sqrt(49/16*I*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I
*c)) - 2*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(2*I*a/d^2)*log((sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e
^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1) + I*d*sqrt(2*I*a/d^2)*e^(I*d*x +
I*c))*e^(-I*d*x - I*c)) + 2*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(2*I*a/d^2)*log((sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*
c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1) - I*d*sqrt(2*I
*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)))/(d*e^(2*I*d*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*tan(c + d*x)**(5/2), 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((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 2.38Unable to divide, perhaps due to rounding error%%%{%%{[%%%{%%{poly1[-16*i,0]:[1,0,-2]%
%},[0]%%%},

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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