3.162 \(\int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=204 \[ -\frac{(-1)^{3/4} a^{3/2} (12 A-11 i B) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{(2+2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{a (5 B+4 i A) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d} \]

[Out]

-((-1)^(3/4)*a^(3/2)*(12*A - (11*I)*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x
]]])/(4*d) - ((2 + 2*I)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*
x]]])/d + (a*((4*I)*A + 5*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(4*d) + ((I/2)*a*B*Tan[c + d*x]^(3
/2)*Sqrt[a + I*a*Tan[c + d*x]])/d

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Rubi [A]  time = 0.699549, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{(-1)^{3/4} a^{3/2} (12 A-11 i B) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{(2+2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{a (5 B+4 i A) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

-((-1)^(3/4)*a^(3/2)*(12*A - (11*I)*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x
]]])/(4*d) - ((2 + 2*I)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*
x]]])/d + (a*((4*I)*A + 5*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(4*d) + ((I/2)*a*B*Tan[c + d*x]^(3
/2)*Sqrt[a + I*a*Tan[c + d*x]])/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))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac{i a B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{1}{2} \int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (4 A-3 i B)+\frac{1}{2} a (4 i A+5 B) \tan (c+d x)\right ) \, dx\\ &=\frac{a (4 i A+5 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \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{1}{4} a^2 (4 i A+5 B)+\frac{1}{4} a^2 (12 A-11 i B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{2 a}\\ &=\frac{a (4 i A+5 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}-(2 a (i A+B)) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx+\frac{1}{8} (12 i A+11 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 (4 i A+5 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}-\frac{\left (4 a^3 (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^2 (12 i A+11 B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=-\frac{(2+2 i) a^{3/2} (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 (4 i A+5 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{\left (a^2 (12 i A+11 B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 d}\\ &=-\frac{(2+2 i) a^{3/2} (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 (4 i A+5 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{\left (a^2 (12 i A+11 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 )}{4 d}\\ &=-\frac{\sqrt [4]{-1} a^{3/2} (12 i A+11 B) \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{(2+2 i) a^{3/2} (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 (4 i A+5 B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}\\ \end{align*}

Mathematica [A]  time = 5.95936, size = 389, normalized size = 1.91 \[ \frac{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\frac{\sqrt{2} e^{-i (c+d x)} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (\sqrt{2} (11 B+12 i A) \left (\log \left (-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )-64 i (A-i B) \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )\right )}{\sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}}}+8 (\sin (c)+i \cos (c)) \sqrt{\tan (c+d x)} \sqrt{\sec (c+d x)} (\cos (d x)-i \sin (d x)) (4 A+2 B \tan (c+d x)-5 i B)\right )}{32 d \sec ^{\frac{5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x])*((Sqrt[2]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2
*I)*(c + d*x)))]*((-64*I)*(A - I*B)*Log[E^(I*(c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] + Sqrt[2]*((12*I)*A
+ 11*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))]])))/(E^(I*(c + d*x))*Sqrt[-1 +
E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]) + 8*Sqrt[Sec[c + d*x]]*(I*Cos[c] + Sin[
c])*(Cos[d*x] - I*Sin[d*x])*Sqrt[Tan[c + d*x]]*(4*A - (5*I)*B + 2*B*Tan[c + d*x])))/(32*d*Sec[c + d*x]^(5/2)*(
A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B]  time = 0.048, size = 565, normalized size = 2.8 \begin{align*} -{\frac{a}{8\,d}\sqrt{\tan \left ( dx+c \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( -4\,iB\sqrt{ia}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\tan \left ( dx+c \right ) +4\,iA\sqrt{-ia}\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-8\,iA\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia}-4\,i\sqrt{ia}\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 ) a+5\,B\sqrt{-ia}\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-10\,B\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia}+8\,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}-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 ) a\sqrt{ia}-8\,\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(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)

[Out]

-1/8/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(-4*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)+4*I*A*(-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-8*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)-4*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+5*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-10*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)+8*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)-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))*a*(I*a)^(1/2)-8*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sqrt{\tan \left (d x + c\right )}\,{d x} \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))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(3/2)*sqrt(tan(d*x + c)), x)

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Fricas [B]  time = 1.92043, size = 2290, normalized size = 11.23 \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))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/8*(2*sqrt(2)*((4*I*A + 7*B)*a*e^(2*I*d*x + 2*I*c) + (4*I*A + 3*B)*a)*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((144*I*A^2 + 264*A*B - 121*I*B^
2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((12*I*A + 11*B)*a*e^(2*I*d*x + 2*I*c) + (12*I*A + 11*B)*
a)*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((144*I*A^2 + 264*A*B - 121*I*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((12*I*A +
11*B)*a)) - sqrt((144*I*A^2 + 264*A*B - 121*I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((12*I*A
+ 11*B)*a*e^(2*I*d*x + 2*I*c) + (12*I*A + 11*B)*a)*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((144*I*A^2 + 264*A*B - 121*I*B^2)*a^3/d^2)*d*e^(
2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((12*I*A + 11*B)*a)) - 4*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*(d*
e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a)*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((8*I*A^2
+ 16*A*B - 8*I*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)) + 4*sqrt((8*I*A^2
+ 16*A*B - 8*I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) +
(2*I*A + 2*B)*a)*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((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2
*I*A + 2*B)*a)))/(d*e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

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))**(3/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.36409, size = 348, normalized size = 1.71 \begin{align*} \frac{\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 )}^{2} 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 )} +{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} -{\left (2 \, a \tan \left (d x + c\right ) - 2 i \, a\right )} 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 (a \tan \left (d x + c\right ) - i \, a\right )} a + 2 i \, 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))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

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