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

Optimal. Leaf size=248 \[ \frac {(-1)^{3/4} a^{3/2} (23 B+22 i A) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}+\frac {(2+2 i) a^{3/2} (B+i A) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a (7 B+6 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {a (10 A-9 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[Out]

1/8*(-1)^(3/4)*a^(3/2)*(22*I*A+23*B)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+(2
+2*I)*a^(3/2)*(I*A+B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+1/8*a*(10*A-9*I*B)*ta
n(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d+1/12*a*(6*I*A+7*B)*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(3/2)/d+1/3*I
*a*B*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(5/2)/d

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Rubi [A]  time = 0.90, antiderivative size = 248, normalized size of antiderivative = 1.00, 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} \[ \frac {(-1)^{3/4} a^{3/2} (23 B+22 i A) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}+\frac {(2+2 i) a^{3/2} (B+i A) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a (7 B+6 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {a (10 A-9 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(3/4)*a^(3/2)*((22*I)*A + 23*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]
]])/(8*d) + ((2 + 2*I)*a^(3/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x
]]])/d + (a*(10*A - (9*I)*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(8*d) + (a*((6*I)*A + 7*B)*Tan[c +
d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(12*d) + ((I/3)*a*B*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/d

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 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])

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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \tan ^{\frac {3}{2}}(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 {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {1}{3} \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (6 A-5 i B)+\frac {1}{2} a (6 i A+7 B) \tan (c+d x)\right ) \, dx\\ &=\frac {a (6 i A+7 B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} a^2 (6 i A+7 B)+\frac {3}{4} a^2 (10 A-9 i B) \tan (c+d x)\right ) \, dx}{6 a}\\ &=\frac {a (10 A-9 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {a (6 i A+7 B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{8} a^3 (10 A-9 i B)-\frac {3}{8} a^3 (22 i A+23 B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{6 a^2}\\ &=\frac {a (10 A-9 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {a (6 i A+7 B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-(2 a (A-i B)) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx+\frac {1}{16} (22 A-23 i 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 (10 A-9 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {a (6 i A+7 B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\left (a^2 (22 A-23 i 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 {\left (4 a^3 (i A+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 {(2+2 i) a^{3/2} (i A+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 (10 A-9 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {a (6 i A+7 B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\left (a^2 (22 A-23 i 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 {(2+2 i) a^{3/2} (i A+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 (10 A-9 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {a (6 i A+7 B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\left (a^2 (22 A-23 i 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^{3/2} (22 A-23 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 {(2+2 i) a^{3/2} (i A+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 (10 A-9 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {a (6 i A+7 B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\\ \end {align*}

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Mathematica [A]  time = 7.95, size = 420, normalized size = 1.69 \[ \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\frac {\sqrt {2} e^{-i (c+d x)} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (\sqrt {2} (22 A-23 i B) \left (\log \left (-2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )-128 (A-i B) \log \left (\sqrt {-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )\right )}{\sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}}}+\frac {4 (\cos (c)-i \sin (c)) \sqrt {\tan (c+d x)} \sec ^{\frac {5}{2}}(c+d x) (2 (7 B+6 i A) \sin (2 (c+d x))+5 (6 A-7 i B) \cos (2 (c+d x))+30 A-19 i B)}{3 \cos (d x)+3 i \sin (d x)}\right )}{64 d \sec ^{\frac {5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((Sqrt[2]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(-128*(A - I*B)*Log[E^(I*(c + d*x
)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] + Sqrt[2]*(22*A - (23*I)*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)))]) + (4*Sec[c + d*x]^(5/2)*(Cos[c] - I*Sin[c])*(30*A - (19*I)*B + 5*(6*A - (7*I)*B)*Cos[2*(c +
d*x)] + 2*((6*I)*A + 7*B)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]])/(3*Cos[d*x] + (3*I)*Sin[d*x]))*(a + I*a*Tan[c
+ d*x])^(3/2)*(A + B*Tan[c + d*x]))/(64*d*Sec[c + d*x]^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [B]  time = 1.73, size = 895, normalized size = 3.61 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/48*(2*sqrt(2)*(7*(6*A - 7*I*B)*a*e^(5*I*d*x + 5*I*c) + 2*(30*A - 19*I*B)*a*e^(3*I*d*x + 3*I*c) + 3*(6*A - 7*
I*B)*a*e^(I*d*x + I*c))*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*sqrt((-484*I*A^2 - 1012*A*B + 529*I*B^2)*a^3/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)*((22*I*A + 23*B)*a*e^(2*I*d*x + 2*I*c) + (22*I*A + 23*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)) + 2*I*sqrt((-484*I*A^2 - 1012*A*B + 529*I*B^2
)*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((22*I*A + 23*B)*a)) - 3*sqrt((-484*I*A^2 - 1012*A*B + 529*I*B^
2)*a^3/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)*((22*I*A + 23*B)*a*e^(2*I*d*x +
2*I*c) + (22*I*A + 23*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)) - 2*I*sqrt((-484*I*A^2 - 1012*A*B + 529*I*B^2)*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((22
*I*A + 23*B)*a)) - 24*sqrt((-8*I*A^2 - 16*A*B + 8*I*B^2)*a^3/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)*((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)) + I*sqrt((-8*I*A^2 - 16*A*B + 8*I*B^2)*a^3/d^2
)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)) + 24*sqrt((-8*I*A^2 - 16*A*B + 8*I*B^2)*a^3/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)*((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)) - I*sq
rt((-8*I*A^2 - 16*A*B + 8*I*B^2)*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)))/(d*e^(4*I*d*
x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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giac [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(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.38, size = 650, normalized size = 2.62 \[ \frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (16 i B \left (\tan ^{2}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+24 i A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+27 i B \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 ) \sqrt {-i a}\, a -54 i B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+28 B \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 )-24 i \sqrt {i a}\, \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 ) a -30 A \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 ) \sqrt {-i a}\, a +60 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+24 \sqrt {i a}\, \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 ) a -48 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 ) \sqrt {-i a}\, a -48 \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 ) \sqrt {-i a}\, a \right )}{48 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(3/2)*(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*(16*I*B*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)+24*I*A*(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)+27
*I*B*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-54*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+28*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)-24*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-30*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+60*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+
I*tan(d*x+c)))^(1/2)+24*(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-48*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*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)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(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)*tan(d*x + c)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(3/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

int(tan(c + d*x)^(3/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2), x)

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sympy [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(tan(d*x+c)**(3/2)*(a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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