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

Optimal. Leaf size=129 \[ \frac {4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {4 a^2 (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \]

[Out]

4*(-1)^(1/4)*a^2*(I*A+B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d+4*a^2*(I*A+B)*tan(d*x+c)^(1/2)/d-2/15*a^2*(5*A-
7*I*B)*tan(d*x+c)^(3/2)/d+2/5*I*B*tan(d*x+c)^(3/2)*(a^2+I*a^2*tan(d*x+c))/d

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Rubi [A]  time = 0.26, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3594, 3592, 3528, 3533, 205} \[ -\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {4 a^2 (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-1)^(1/4)*a^2*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (4*a^2*(I*A + B)*Sqrt[Tan[c + d*x]])/d
- (2*a^2*(5*A - (7*I)*B)*Tan[c + d*x]^(3/2))/(15*d) + (((2*I)/5)*B*Tan[c + d*x]^(3/2)*(a^2 + I*a^2*Tan[c + d*x
]))/d

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 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 3592

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(B*d*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

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]

Rubi steps

\begin {align*} \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac {2}{5} \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) \left (\frac {1}{2} a (5 A-3 i B)+\frac {1}{2} a (5 i A+7 B) \tan (c+d x)\right ) \, dx\\ &=-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac {2}{5} \int \sqrt {\tan (c+d x)} \left (5 a^2 (A-i B)+5 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac {4 a^2 (i A+B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac {2}{5} \int \frac {-5 a^2 (i A+B)+5 a^2 (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {4 a^2 (i A+B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac {\left (20 a^4 (i A+B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-5 a^2 (i A+B)-5 a^2 (A-i B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {4 \sqrt [4]{-1} a^2 (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {4 a^2 (i A+B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\\ \end {align*}

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Mathematica [B]  time = 5.62, size = 272, normalized size = 2.11 \[ \frac {\cos ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left (\frac {1}{15} (\cos (2 c)-i \sin (2 c)) \sqrt {\tan (c+d x)} \sec ^2(c+d x) (-5 (A-2 i B) \sin (2 (c+d x))+(33 B+30 i A) \cos (2 (c+d x))+30 i A+27 B)-\frac {4 i e^{-2 i c} (A-i B) \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^3*(((-4*I)*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[S
qrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])/(E^((2*I)*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 +
E^((2*I)*(c + d*x)))]) + (Sec[c + d*x]^2*(Cos[2*c] - I*Sin[2*c])*((30*I)*A + 27*B + ((30*I)*A + 33*B)*Cos[2*(c
+ d*x)] - 5*(A - (2*I)*B)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]])/15)*(a + I*a*Tan[c + d*x])^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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fricas [B]  time = 0.71, size = 436, normalized size = 3.38 \[ -\frac {15 \, \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 15 \, \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - {\left ({\left (280 i \, A + 344 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (480 i \, A + 432 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (200 i \, A + 184 \, B\right )} a^{2}\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{60 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(15*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*l
og((4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c)
+ d)*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)/((2*I*A + 2*B)*a^2)) -
15*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log((4*
(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*s
qrt((-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)/((2*I*A + 2*B)*a^2)) - ((280
*I*A + 344*B)*a^2*e^(4*I*d*x + 4*I*c) + (480*I*A + 432*B)*a^2*e^(2*I*d*x + 2*I*c) + (200*I*A + 184*B)*a^2)*sqr
t((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \sqrt {\tan \left (d x + c\right )}\,{d x} \]

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))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.10, size = 537, normalized size = 4.16 \[ -\frac {2 a^{2} B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5 d}+\frac {4 i a^{2} A \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}-\frac {2 a^{2} A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}-\frac {i a^{2} A \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{2 d}+\frac {4 a^{2} B \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}-\frac {i a^{2} A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {i a^{2} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {i a^{2} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{2} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{2} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{2} B \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{2 d}-\frac {i a^{2} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {i a^{2} B \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{2 d}+\frac {4 i a^{2} B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} A \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{2 d}+\frac {a^{2} A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {a^{2} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d} \]

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

[Out]

-2/5/d*a^2*B*tan(d*x+c)^(5/2)+4*I/d*a^2*A*tan(d*x+c)^(1/2)-2/3/d*a^2*A*tan(d*x+c)^(3/2)-1/2*I/d*a^2*A*2^(1/2)*
ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+4/d*a^2*B*tan(d*x+c)^(1/2)
-I/d*a^2*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))-I/d*a^2*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))-I/
d*a^2*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))-1/d*a^2*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))-1/d*a^
2*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))-1/2/d*a^2*B*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))
/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))-I/d*a^2*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))-1/2*I/d*a^2*B*
2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+4/3*I/d*a^2*B*tan(
d*x+c)^(3/2)+1/2/d*a^2*A*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*
x+c)))+1/d*a^2*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+1/d*a^2*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/
2))

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maxima [A]  time = 0.59, size = 194, normalized size = 1.50 \[ -\frac {12 \, B a^{2} \tan \left (d x + c\right )^{\frac {5}{2}} + 20 \, {\left (A - 2 i \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac {3}{2}} - 4 \, {\left (30 i \, A + 30 \, B\right )} a^{2} \sqrt {\tan \left (d x + c\right )} + 15 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2}}{30 \, d} \]

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))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/30*(12*B*a^2*tan(d*x + c)^(5/2) + 20*(A - 2*I*B)*a^2*tan(d*x + c)^(3/2) - 4*(30*I*A + 30*B)*a^2*sqrt(tan(d*
x + c)) + 15*(2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(
2)*((I - 1)*A + (I + 1)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*(-(I + 1)*A + (I -
1)*B)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(-sqrt(2)*sqrt(
tan(d*x + c)) + tan(d*x + c) + 1))*a^2)/d

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mupad [B]  time = 7.73, size = 256, normalized size = 1.98 \[ \frac {A\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,4{}\mathrm {i}}{d}-\frac {2\,A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}+\frac {4\,B\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,4{}\mathrm {i}}{3\,d}-\frac {2\,B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}+\frac {\sqrt {2}\,A\,a^2\,\ln \left (4\,A\,a^2\,d+\sqrt {2}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d}-\frac {\sqrt {4{}\mathrm {i}}\,A\,a^2\,\ln \left (4\,A\,a^2\,d+2\,\sqrt {4{}\mathrm {i}}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^2\,\ln \left (-B\,a^2\,d\,4{}\mathrm {i}+\sqrt {2}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (1-\mathrm {i}\right )}{d}-\frac {\sqrt {-4{}\mathrm {i}}\,B\,a^2\,\ln \left (-B\,a^2\,d\,4{}\mathrm {i}+2\,\sqrt {-4{}\mathrm {i}}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a^2*tan(c + d*x)^(1/2)*4i)/d - (2*A*a^2*tan(c + d*x)^(3/2))/(3*d) + (4*B*a^2*tan(c + d*x)^(1/2))/d + (B*a^2
*tan(c + d*x)^(3/2)*4i)/(3*d) - (2*B*a^2*tan(c + d*x)^(5/2))/(5*d) + (2^(1/2)*A*a^2*log(4*A*a^2*d - 2^(1/2)*A*
a^2*d*tan(c + d*x)^(1/2)*(2 + 2i))*(1 + 1i))/d - (4i^(1/2)*A*a^2*log(4*A*a^2*d + 2*4i^(1/2)*A*a^2*d*tan(c + d*
x)^(1/2)))/d + (2^(1/2)*B*a^2*log(- B*a^2*d*4i - 2^(1/2)*B*a^2*d*tan(c + d*x)^(1/2)*(2 - 2i))*(1 - 1i))/d - ((
-4i)^(1/2)*B*a^2*log(2*(-4i)^(1/2)*B*a^2*d*tan(c + d*x)^(1/2) - B*a^2*d*4i))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \left (- A \sqrt {\tan {\left (c + d x \right )}}\right )\, dx + \int A \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx + \int \left (- B \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{\frac {7}{2}}{\left (c + d x \right )}\, dx + \int \left (- 2 i A \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int \left (- 2 i B \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx\right ) \]

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

[Out]

-a**2*(Integral(-A*sqrt(tan(c + d*x)), x) + Integral(A*tan(c + d*x)**(5/2), x) + Integral(-B*tan(c + d*x)**(3/
2), x) + Integral(B*tan(c + d*x)**(7/2), x) + Integral(-2*I*A*tan(c + d*x)**(3/2), x) + Integral(-2*I*B*tan(c
+ d*x)**(5/2), x))

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