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

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

[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)

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Rubi [A]  time = 0.57449, antiderivative size = 176, normalized size of antiderivative = 1., 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 = {3560, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{7 (-1)^{3/4} \sqrt{a} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{i \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{(1+i) \sqrt{a} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

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 3560

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 3597

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 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]

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 205

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

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 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 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 203

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

Mathematica [A]  time = 3.76396, size = 267, normalized size = 1.52 \[ \frac{i e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \sqrt{a+i a \tan (c+d x)} \left (\frac{\sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \left (1-3 e^{2 i (c+d x)}\right ) \sqrt{\tan (c+d x)}}{\left (1+e^{2 i (c+d x)}\right )^2}+4 \sqrt{2} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-7 \sqrt{\tan (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{4 \sqrt{2} d \sqrt{-1+e^{2 i (c+d x)}}} \]

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]  time = 0.127, size = 472, normalized size = 2.7 \begin{align*}{\frac{1}{8\,d \left ( -\tan \left ( dx+c \right ) +i \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{\tan \left ( dx+c \right ) } \left ( 6\,i\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia}\tan \left ( dx+c \right ) +4\,i\sqrt{2}\ln \left ( -{\frac{1}{\tan \left ( dx+c \right ) +i} \left ( -2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+ia-3\,a\tan \left ( dx+c \right ) \right ) } \right ) \sqrt{ia}a+7\,i\ln \left ({\frac{1}{2} \left ( 2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a \right ){\frac{1}{\sqrt{ia}}}} \right ) a\sqrt{-ia}\tan \left ( dx+c \right ) -4\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}-4\,\sqrt{2}\ln \left ( -{\frac{-2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+ia-3\,a\tan \left ( dx+c \right ) }{\tan \left ( dx+c \right ) +i}} \right ) \sqrt{ia}\tan \left ( dx+c \right ) a+2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia}+7\,\ln \left ( 1/2\,{\frac{2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a}{\sqrt{ia}}} \right ) a\sqrt{-ia} \right ){\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{-ia}}}{\frac{1}{\sqrt{ia}}}} \end{align*}

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)

[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)+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+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*(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*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)*tan(d*x+c)*a+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., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}

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]  time = 2.4461, size = 1713, normalized size = 9.73 \begin{align*} \text{result too large to display} \end{align*}

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^(2*I*d*x + 2*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)*e^(I*d*x + I*c) + 8*I*d*sqrt(49/16*I*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*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)*e^(I*d*x + I*c) - 8*I*d*sqrt(49/16
*I*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*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)*e^(I*d*x + I*c) + I*d*sqrt(2*I*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*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)*e^(I*d*x + I*c) - I*d*sqrt(2*I*a/d^2)*e^(2*I*d*x + 2
*I*c))*e^(-2*I*d*x - 2*I*c)))/(d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.20472, size = 216, normalized size = 1.23 \begin{align*} \frac{{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 2 \,{\left (2 \, a \tan \left (d x + c\right ) - 2 i \, a\right )} a - 2 i \, a^{2}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (\frac{-i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + i \, a^{2}}{\sqrt{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + a^{4}}} + 1\right )} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - 2 \, a^{3}\right )}} \end{align*}

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]

1/2*(-2*I*(I*a*tan(d*x + c) + a)^2 - 2*(2*a*tan(d*x + c) - 2*I*a)*a - 2*I*a^2)*sqrt(-2*(I*a*tan(d*x + c) + a)*
a + 2*a^2)*((-I*(I*a*tan(d*x + c) + a)*a + I*a^2)/sqrt((I*a*tan(d*x + c) + a)^2*a^2 - 2*(I*a*tan(d*x + c) + a)
*a^3 + a^4) + 1)*log(sqrt(I*a*tan(d*x + c) + a))/((I*a*tan(d*x + c) + a)*a^2 - 2*a^3)