3.170 \(\int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=252 \[ -\frac{(-1)^{3/4} a^{5/2} (46 A-45 i B) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}-\frac{a^2 (2 A-3 i B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{a^2 (19 B+18 i A) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

-((-1)^(3/4)*a^(5/2)*(46*A - (45*I)*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x
]]])/(8*d) - ((4 + 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*
x]]])/d + (a^2*((18*I)*A + 19*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(8*d) - (a^2*(2*A - (3*I)*B)*T
an[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(4*d) + ((I/3)*a*B*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/
2))/d

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Rubi [A]  time = 0.922251, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.237, Rules used = {3594, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{(-1)^{3/4} a^{5/2} (46 A-45 i B) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}-\frac{a^2 (2 A-3 i B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{a^2 (19 B+18 i A) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-1)^(3/4)*a^(5/2)*(46*A - (45*I)*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x
]]])/(8*d) - ((4 + 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*
x]]])/d + (a^2*((18*I)*A + 19*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(8*d) - (a^2*(2*A - (3*I)*B)*T
an[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(4*d) + ((I/3)*a*B*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/
2))/d

Rule 3594

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*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1]

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 \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac{i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{1}{3} \int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} \left (\frac{3}{2} a (2 A-i B)+\frac{3}{2} a (2 i A+3 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{a^2 (2 A-3 i B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{1}{6} \int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (\frac{3}{4} a^2 (14 A-13 i B)+\frac{3}{4} a^2 (18 i A+19 B) \tan (c+d x)\right ) \, dx\\ &=\frac{a^2 (18 i A+19 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{a^2 (2 A-3 i B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{8} a^3 (18 i A+19 B)+\frac{3}{8} a^3 (46 A-45 i B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{6 a}\\ &=\frac{a^2 (18 i A+19 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{a^2 (2 A-3 i B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\left (4 a^2 (i A+B)\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx+\frac{1}{16} (a (46 i A+45 B)) \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{a^2 (18 i A+19 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{a^2 (2 A-3 i B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{\left (8 a^4 (A-i B)\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{\left (a^3 (46 i A+45 B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{16 d}\\ &=-\frac{(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{a^2 (18 i A+19 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{a^2 (2 A-3 i B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{\left (a^3 (46 i A+45 B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{8 d}\\ &=-\frac{(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{a^2 (18 i A+19 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{a^2 (2 A-3 i B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{\left (a^3 (46 i A+45 B)\right ) \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 )}{8 d}\\ &=-\frac{\sqrt [4]{-1} a^{5/2} (46 i A+45 B) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}-\frac{(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{a^2 (18 i A+19 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{a^2 (2 A-3 i B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\\ \end{align*}

Mathematica [B]  time = 8.98103, size = 537, normalized size = 2.13 \[ \frac{\cos ^3(c+d x) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \left ((6 A-13 i B) \sec (c+d x) \left (-\frac{1}{12} \sin (3 c+d x)-\frac{1}{12} i \cos (3 c+d x)\right )+(91 B+66 i A) \left (\frac{1}{24} \cos (2 c)-\frac{1}{24} i \sin (2 c)\right )+\left (-\frac{1}{3} B \cos (2 c)+\frac{1}{3} i B \sin (2 c)\right ) \sec ^2(c+d x)\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac{e^{-2 i c} \sqrt{e^{i d x}} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} (a+i a \tan (c+d x))^{5/2} \left (\sqrt{2} (45 B+46 i A) \left (\log \left (-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )-256 i (A-i B) \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )\right ) (A+B \tan (c+d x))}{32 \sqrt{2} d \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec ^{\frac{7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[E^(I*d*x)]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*((-256*I)*(A - I*B)*Log[E^(
I*(c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] + Sqrt[2]*((46*I)*A + 45*B)*(Log[1 - 3*E^((2*I)*(c + d*x)) - 2*
Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] - Log[1 - 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*E^(I*(c +
d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]]))*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(32*Sqrt[2]*d*E^((2
*I)*c)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sec[c + d*x]^(7/2)*(Cos[
d*x] + I*Sin[d*x])^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^3*(((66*I)*A + 91*B)*(Cos[2*c]/24
- (I/24)*Sin[2*c]) + Sec[c + d*x]^2*(-(B*Cos[2*c])/3 + (I/3)*B*Sin[2*c]) + (6*A - (13*I)*B)*Sec[c + d*x]*((-I/
12)*Cos[3*c + d*x] - Sin[3*c + d*x]/12))*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))
/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B]  time = 0.042, size = 653, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(16*B*(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)^2-52*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+5
4*I*A*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))*(-I*a)^(1/2
)*a-108*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+24*A*(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)-48*I*(I*a)^(1/2)*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))*a+57*B*(-I*a)^(1/2)*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-114*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)
*(-I*a)^(1/2)*(I*a)^(1/2)+96*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))*(-I*a)^(1/2)*a-48*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))*a*(I*a)^(1/2)-96*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))/(I*a)^(1/2)/(-I*a)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2
)

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

[Out]

Timed out

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Fricas [B]  time = 1.93646, size = 2603, normalized size = 10.33 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*sqrt(2)*((66*I*A + 91*B)*a^2*e^(4*I*d*x + 4*I*c) + (108*I*A + 98*B)*a^2*e^(2*I*d*x + 2*I*c) + (42*I*A
+ 39*B)*a^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^
(I*d*x + I*c) + 3*sqrt((2116*I*A^2 + 4140*A*B - 2025*I*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x +
2*I*c) + d)*log((sqrt(2)*((46*I*A + 45*B)*a^2*e^(2*I*d*x + 2*I*c) + (46*I*A + 45*B)*a^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^(I*d*x + I*c) + 2*sqrt((2116*I*A^
2 + 4140*A*B - 2025*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((46*I*A + 45*B)*a^2)) - 3*sqr
t((2116*I*A^2 + 4140*A*B - 2025*I*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqr
t(2)*((46*I*A + 45*B)*a^2*e^(2*I*d*x + 2*I*c) + (46*I*A + 45*B)*a^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^(I*d*x + I*c) - 2*sqrt((2116*I*A^2 + 4140*A*B - 2025*I
*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((46*I*A + 45*B)*a^2)) - 24*sqrt((32*I*A^2 + 64*A*B
- 32*I*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^
(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a^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^(I*d*x + I*c) + sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*
e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^2)) + 24*sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*
I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a^2)*s
qrt(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^(I*d*x + I*c)
- sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^2)
))/(d*e^(4*I*d*x + 4*I*c) + 2*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(tan(d*x+c)**(1/2)*(a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.49186, size = 347, normalized size = 1.38 \begin{align*} -\frac{-i \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a{\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 )} - 2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} -{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B{\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 )}}{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )} a - 2 \, a^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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