3.2.28 \(\int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [128]

Optimal. Leaf size=171 \[ \frac {8 \sqrt [4]{-1} a^3 (i A+B) \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {8 a^3 (i A+B) \sqrt {\tan (c+d x)}}{d}-\frac {8 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{105 d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d} \]

[Out]

8*(-1)^(1/4)*a^3*(I*A+B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d+8*a^3*(I*A+B)*tan(d*x+c)^(1/2)/d-8/105*a^3*(21*
A-23*I*B)*tan(d*x+c)^(3/2)/d+2/7*I*a*B*tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^2/d-2/35*(7*A-11*I*B)*tan(d*x+c)^(3
/2)*(a^3+I*a^3*tan(d*x+c))/d

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Rubi [A]
time = 0.29, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3675, 3673, 3609, 3614, 211} \begin {gather*} \frac {8 \sqrt [4]{-1} a^3 (B+i A) \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {8 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{105 d}-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac {8 a^3 (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(-1)^(1/4)*a^3*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (8*a^3*(I*A + B)*Sqrt[Tan[c + d*x]])/d
- (8*a^3*(21*A - (23*I)*B)*Tan[c + d*x]^(3/2))/(105*d) + (((2*I)/7)*a*B*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*
x])^2)/d - (2*(7*A - (11*I)*B)*Tan[c + d*x]^(3/2)*(a^3 + I*a^3*Tan[c + d*x]))/(35*d)

Rule 211

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

Rule 3609

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 3614

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 3673

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 3675

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))^3 (A+B \tan (c+d x)) \, dx &=\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}+\frac {2}{7} \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 \left (\frac {1}{2} a (7 A-3 i B)+\frac {1}{2} a (7 i A+11 B) \tan (c+d x)\right ) \, dx\\ &=\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac {4}{35} \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) \left (2 a^2 (7 A-6 i B)+a^2 (21 i A+23 B) \tan (c+d x)\right ) \, dx\\ &=-\frac {8 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{105 d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac {4}{35} \int \sqrt {\tan (c+d x)} \left (35 a^3 (A-i B)+35 a^3 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac {8 a^3 (i A+B) \sqrt {\tan (c+d x)}}{d}-\frac {8 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{105 d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac {4}{35} \int \frac {-35 a^3 (i A+B)+35 a^3 (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {8 a^3 (i A+B) \sqrt {\tan (c+d x)}}{d}-\frac {8 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{105 d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac {\left (280 a^6 (i A+B)^2\right ) \text {Subst}\left (\int \frac {1}{-35 a^3 (i A+B)-35 a^3 (A-i B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {8 \sqrt [4]{-1} a^3 (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {8 a^3 (i A+B) \sqrt {\tan (c+d x)}}{d}-\frac {8 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{105 d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(452\) vs. \(2(171)=342\).
time = 8.13, size = 452, normalized size = 2.64 \begin {gather*} -\frac {8 i (A-i B) e^{-3 i c} \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 ) \cos ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{d \sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {\cos ^4(c+d x) \left (\sec (c) \sec ^2(c+d x) (7 A \cos (c)-21 i B \cos (c)+5 B \sin (c)) \left (-\frac {2}{35} i \cos (3 c)-\frac {2}{35} \sin (3 c)\right )+\sec (c) (441 i A \cos (c)+483 B \cos (c)-105 A \sin (c)+155 i B \sin (c)) \left (\frac {2}{105} \cos (3 c)-\frac {2}{105} i \sin (3 c)\right )-i B \sec (c) \sec ^3(c+d x) \left (\frac {2}{7} \cos (3 c)-\frac {2}{7} i \sin (3 c)\right ) \sin (d x)+\sec (c) \sec (c+d x) \left (-\frac {2}{21} \cos (3 c)+\frac {2}{21} i \sin (3 c)\right ) (21 A \sin (d x)-31 i B \sin (d x))\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-8*I)*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[Sqrt[(-1 + E^((2*I
)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*E^(
(3*I)*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x]
+ B*Sin[c + d*x])) + (Cos[c + d*x]^4*(Sec[c]*Sec[c + d*x]^2*(7*A*Cos[c] - (21*I)*B*Cos[c] + 5*B*Sin[c])*(((-2*
I)/35)*Cos[3*c] - (2*Sin[3*c])/35) + Sec[c]*((441*I)*A*Cos[c] + 483*B*Cos[c] - 105*A*Sin[c] + (155*I)*B*Sin[c]
)*((2*Cos[3*c])/105 - ((2*I)/105)*Sin[3*c]) - I*B*Sec[c]*Sec[c + d*x]^3*((2*Cos[3*c])/7 - ((2*I)/7)*Sin[3*c])*
Sin[d*x] + Sec[c]*Sec[c + d*x]*((-2*Cos[3*c])/21 + ((2*I)/21)*Sin[3*c])*(21*A*Sin[d*x] - (31*I)*B*Sin[d*x]))*S
qrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x]
+ B*Sin[c + d*x]))

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Maple [A]
time = 0.04, size = 276, normalized size = 1.61

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {2 i B \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}-\frac {2 i A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-\frac {6 B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {8 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-2 A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+8 i A \left (\sqrt {\tan }\left (d x +c \right )\right )+8 B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(276\)
default	\(\frac {a^{3} \left (-\frac {2 i B \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}-\frac {2 i A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-\frac {6 B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {8 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-2 A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+8 i A \left (\sqrt {\tan }\left (d x +c \right )\right )+8 B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(276\)

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*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*a^3*(-2/7*I*B*tan(d*x+c)^(7/2)-2/5*I*A*tan(d*x+c)^(5/2)-6/5*B*tan(d*x+c)^(5/2)+8/3*I*B*tan(d*x+c)^(3/2)-2*
A*tan(d*x+c)^(3/2)+8*I*A*tan(d*x+c)^(1/2)+8*B*tan(d*x+c)^(1/2)+1/4*(-4*I*A-4*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)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-
1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(-4*I*B+4*A)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*t
an(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))))

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Maxima [A]
time = 0.50, size = 216, normalized size = 1.26 \begin {gather*} -\frac {30 i \, B a^{3} \tan \left (d x + c\right )^{\frac {7}{2}} + 42 \, {\left (i \, A + 3 \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac {5}{2}} + 70 \, {\left (3 \, A - 4 i \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac {3}{2}} + 840 \, {\left (-i \, A - B\right )} a^{3} \sqrt {\tan \left (d x + c\right )} + 105 \, {\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^{3}}{105 \, d} \end {gather*}

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

[Out]

-1/105*(30*I*B*a^3*tan(d*x + c)^(7/2) + 42*(I*A + 3*B)*a^3*tan(d*x + c)^(5/2) + 70*(3*A - 4*I*B)*a^3*tan(d*x +
c)^(3/2) + 840*(-I*A - B)*a^3*sqrt(tan(d*x + c)) + 105*(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(ta
n(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^3)/d

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (137) = 274\).
time = 3.29, size = 498, normalized size = 2.91 \begin {gather*} -\frac {2 \, {\left (105 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{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 (-i \, A - B\right )} a^{3}}\right ) - 105 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{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 (-i \, A - B\right )} a^{3}}\right ) + 2 \, {\left ({\left (-273 i \, A - 319 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (-336 i \, A - 323 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-567 i \, A - 551 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left (-42 i \, A - 41 \, B\right )} a^{3}\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 )}}{105 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

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

[Out]

-2/105*(105*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(
2*I*d*x + 2*I*c) + d)*log(-2*((A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/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)/((-
I*A - B)*a^3)) - 105*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c)
+ 3*d*e^(2*I*d*x + 2*I*c) + d)*log(-2*((A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/
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)/((-I*A - B)*a^3)) + 2*((-273*I*A - 319*B)*a^3*e^(6*I*d*x + 6*I*c) + 2*(-336*I*A - 323*B)*a^3*e^(4*I*d*x
+ 4*I*c) + (-567*I*A - 551*B)*a^3*e^(2*I*d*x + 2*I*c) + 4*(-42*I*A - 41*B)*a^3)*sqrt((-I*e^(2*I*d*x + 2*I*c) +
I)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d
)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i a^{3} \left (\int \left (- 3 A \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int A \tan ^{\frac {7}{2}}{\left (c + d x \right )}\, dx + \int \left (- 3 B \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{\frac {9}{2}}{\left (c + d x \right )}\, dx + \int i A \sqrt {\tan {\left (c + d x \right )}}\, dx + \int \left (- 3 i A \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx + \int i B \tan ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int \left (- 3 i B \tan ^{\frac {7}{2}}{\left (c + d x \right )}\right )\, dx\right ) \end {gather*}

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

[Out]

-I*a**3*(Integral(-3*A*tan(c + d*x)**(3/2), x) + Integral(A*tan(c + d*x)**(7/2), x) + Integral(-3*B*tan(c + d*
x)**(5/2), x) + Integral(B*tan(c + d*x)**(9/2), x) + Integral(I*A*sqrt(tan(c + d*x)), x) + Integral(-3*I*A*tan
(c + d*x)**(5/2), x) + Integral(I*B*tan(c + d*x)**(3/2), x) + Integral(-3*I*B*tan(c + d*x)**(7/2), x))

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Giac [A]
time = 1.00, size = 160, normalized size = 0.94 \begin {gather*} -\frac {\left (4 i - 4\right ) \, \sqrt {2} {\left (A a^{3} - i \, B a^{3}\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} - \frac {2 \, {\left (15 i \, B a^{3} d^{6} \tan \left (d x + c\right )^{\frac {7}{2}} + 21 i \, A a^{3} d^{6} \tan \left (d x + c\right )^{\frac {5}{2}} + 63 \, B a^{3} d^{6} \tan \left (d x + c\right )^{\frac {5}{2}} + 105 \, A a^{3} d^{6} \tan \left (d x + c\right )^{\frac {3}{2}} - 140 i \, B a^{3} d^{6} \tan \left (d x + c\right )^{\frac {3}{2}} - 420 i \, A a^{3} d^{6} \sqrt {\tan \left (d x + c\right )} - 420 \, B a^{3} d^{6} \sqrt {\tan \left (d x + c\right )}\right )}}{105 \, d^{7}} \end {gather*}

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

[Out]

-(4*I - 4)*sqrt(2)*(A*a^3 - I*B*a^3)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 2/105*(15*I*B*a^3*d
^6*tan(d*x + c)^(7/2) + 21*I*A*a^3*d^6*tan(d*x + c)^(5/2) + 63*B*a^3*d^6*tan(d*x + c)^(5/2) + 105*A*a^3*d^6*ta
n(d*x + c)^(3/2) - 140*I*B*a^3*d^6*tan(d*x + c)^(3/2) - 420*I*A*a^3*d^6*sqrt(tan(d*x + c)) - 420*B*a^3*d^6*sqr
t(tan(d*x + c)))/d^7

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Mupad [B]
time = 8.92, size = 292, normalized size = 1.71 \begin {gather*} \frac {A\,a^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,8{}\mathrm {i}}{d}-\frac {2\,A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{d}-\frac {A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,2{}\mathrm {i}}{5\,d}+\frac {8\,B\,a^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,8{}\mathrm {i}}{3\,d}-\frac {6\,B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}-\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,2{}\mathrm {i}}{7\,d}+\frac {\sqrt {2}\,A\,a^3\,\ln \left (8\,A\,a^3\,d+\sqrt {2}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4-4{}\mathrm {i}\right )\right )\,\left (2+2{}\mathrm {i}\right )}{d}-\frac {\sqrt {16{}\mathrm {i}}\,A\,a^3\,\ln \left (8\,A\,a^3\,d+2\,\sqrt {16{}\mathrm {i}}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^3\,\ln \left (-B\,a^3\,d\,8{}\mathrm {i}+\sqrt {2}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4+4{}\mathrm {i}\right )\right )\,\left (2-2{}\mathrm {i}\right )}{d}-\frac {\sqrt {-16{}\mathrm {i}}\,B\,a^3\,\ln \left (-B\,a^3\,d\,8{}\mathrm {i}+2\,\sqrt {-16{}\mathrm {i}}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \end {gather*}

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)^3,x)

[Out]

(A*a^3*tan(c + d*x)^(1/2)*8i)/d - (2*A*a^3*tan(c + d*x)^(3/2))/d - (A*a^3*tan(c + d*x)^(5/2)*2i)/(5*d) + (8*B*
a^3*tan(c + d*x)^(1/2))/d + (B*a^3*tan(c + d*x)^(3/2)*8i)/(3*d) - (6*B*a^3*tan(c + d*x)^(5/2))/(5*d) - (B*a^3*
tan(c + d*x)^(7/2)*2i)/(7*d) + (2^(1/2)*A*a^3*log(8*A*a^3*d - 2^(1/2)*A*a^3*d*tan(c + d*x)^(1/2)*(4 + 4i))*(2
+ 2i))/d - (16i^(1/2)*A*a^3*log(8*A*a^3*d + 2*16i^(1/2)*A*a^3*d*tan(c + d*x)^(1/2)))/d + (2^(1/2)*B*a^3*log(-
B*a^3*d*8i - 2^(1/2)*B*a^3*d*tan(c + d*x)^(1/2)*(4 - 4i))*(2 - 2i))/d - ((-16i)^(1/2)*B*a^3*log(2*(-16i)^(1/2)
*B*a^3*d*tan(c + d*x)^(1/2) - B*a^3*d*8i))/d

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