\(\int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) [125]

Optimal result

Integrand size = 36, antiderivative size = 127 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4 \sqrt [4]{-1} a^2 (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (7 i A+5 B)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (A-i B)}{d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3674, 3672, 3610, 3614, 211} \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4 \sqrt [4]{-1} a^2 (B+i A) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (5 B+7 i A)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (A-i B)}{d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[In]

Int[((a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

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

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 3610

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

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 3672

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

Rule 3674

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

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

Mathematica [A] (verified)

Time = 2.65 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.80 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 a^2 \left (3 A+5 (2 i A+B) \tan (c+d x)-30 (A-i B) \tan ^2(c+d x)+(15+15 i) \sqrt {2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {\tan (c+d x)}}{\sqrt {2}}\right ) \tan ^{\frac {5}{2}}(c+d x)\right )}{15 d \tan ^{\frac {5}{2}}(c+d x)} \]

[In]

Integrate[((a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

(-2*a^2*(3*A + 5*((2*I)*A + B)*Tan[c + d*x] - 30*(A - I*B)*Tan[c + d*x]^2 + (15 + 15*I)*Sqrt[2]*(A - I*B)*ArcT
anh[((1 + I)*Sqrt[Tan[c + d*x]])/Sqrt[2]]*Tan[c + d*x]^(5/2)))/(15*d*Tan[c + d*x]^(5/2))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (107 ) = 214\).

Time = 0.03 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.89

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {2 A}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 \left (2 i B -2 A \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 \left (2 i A +B \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {\left (-2 i A -2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(240\)
default	\(\frac {a^{2} \left (-\frac {2 A}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 \left (2 i B -2 A \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 \left (2 i A +B \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {\left (-2 i A -2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(240\)
parts	\(\frac {\left (2 i A \,a^{2}+B \,a^{2}\right ) \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\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}+\frac {\left (2 i B \,a^{2}-A \,a^{2}\right ) \left (-\frac {\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 {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {A \,a^{2} \left (\frac {\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 {2}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}-\frac {B \,a^{2} \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 d}\)	\(435\)

[In]

int((a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (103) = 206\).

Time = 0.27 (sec) , antiderivative size = 502, normalized size of antiderivative = 3.95 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{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^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + 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 (-i \, A - B\right )} a^{2}}\right ) - 15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{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^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + 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 (-i \, A - B\right )} a^{2}}\right ) + 2 \, {\left ({\left (-43 i \, A - 35 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (11 i \, A + 25 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (31 i \, A + 35 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-23 i \, A - 25 \, 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}}}{15 \, {\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 )}} \]

[In]

integrate((a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/15*(15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/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^2*e^(2*I*d*x + 2*I*c) + sqrt(-(-I*A^2 - 2*A*B + 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)/((-I*
A - B)*a^2)) - 15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/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^2*e^(2*I*d*x + 2*I*c) - sqrt(-(-I*A^2 - 2*A*B + 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)/((-I*A - B)*a^2)) + 2*((-43*I*A - 35*B)*a^2*e^(6*I*d*x + 6*I*c) + (11*I*A + 25*B)*a^2*e^(4*I*d*x + 4*I*c) +
(31*I*A + 35*B)*a^2*e^(2*I*d*x + 2*I*c) + (-23*I*A - 25*B)*a^2)*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)

Sympy [F]

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

[In]

integrate((a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c))/tan(d*x+c)**(7/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.54 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {15 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2} - \frac {4 \, {\left (30 \, {\left (A - i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} + 5 \, {\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - 3 \, A a^{2}\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{30 \, d} \]

[In]

integrate((a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

-1/30*(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 - 4*(30*(A - I*B)*a^2*tan(d*x + c)^2 + 5*(-2*I*A - B)*a^2*tan(d*x + c) - 3
*A*a^2)/tan(d*x + c)^(5/2))/d

Giac [A] (verification not implemented)

none

Time = 1.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {\left (2 i - 2\right ) \, \sqrt {2} {\left (A a^{2} - i \, B a^{2}\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} + \frac {2 \, {\left (30 \, A a^{2} \tan \left (d x + c\right )^{2} - 30 i \, B a^{2} \tan \left (d x + c\right )^{2} - 10 i \, A a^{2} \tan \left (d x + c\right ) - 5 \, B a^{2} \tan \left (d x + c\right ) - 3 \, A a^{2}\right )}}{15 \, d \tan \left (d x + c\right )^{\frac {5}{2}}} \]

[In]

integrate((a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

-(2*I - 2)*sqrt(2)*(A*a^2 - I*B*a^2)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d + 2/15*(30*A*a^2*tan(
d*x + c)^2 - 30*I*B*a^2*tan(d*x + c)^2 - 10*I*A*a^2*tan(d*x + c) - 5*B*a^2*tan(d*x + c) - 3*A*a^2)/(d*tan(d*x
+ c)^(5/2))

Mupad [B] (verification not implemented)

Time = 9.50 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.03 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {\frac {2\,A\,a^2}{5\,d}-\frac {4\,A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d}+\frac {A\,a^2\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{3\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}-\frac {\frac {2\,B\,a^2}{3\,d}+\frac {B\,a^2\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}+\frac {\sqrt {2}\,A\,a^2\,\ln \left (4\,A\,a^2\,d+\sqrt {2}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d}-\frac {\sqrt {4{}\mathrm {i}}\,A\,a^2\,\ln \left (4\,A\,a^2\,d+2\,\sqrt {4{}\mathrm {i}}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^2\,\ln \left (-B\,a^2\,d\,4{}\mathrm {i}+\sqrt {2}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (1-\mathrm {i}\right )}{d}-\frac {\sqrt {-4{}\mathrm {i}}\,B\,a^2\,\ln \left (-B\,a^2\,d\,4{}\mathrm {i}+2\,\sqrt {-4{}\mathrm {i}}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^2)/tan(c + d*x)^(7/2),x)

[Out]

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