\(\int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [104]

Optimal result

Integrand size = 36, antiderivative size = 255 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {(39 i A-89 B) \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(39 i A-89 B) \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}-\frac {(151 i A-361 B) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d} \]

[Out]

-1/8*(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)+1/5*(39*I*A-89*B)*(a+I*a*
tan(d*x+c))^(1/2)/a^3/d-1/20*(39*I*A-89*B)*tan(d*x+c)^2/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+1/5*(I*A-B)*tan(d*x+c)^
4/d/(a+I*a*tan(d*x+c))^(5/2)+1/30*(11*A+21*I*B)*tan(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^(3/2)-1/60*(151*I*A-361*B)
*(a+I*a*tan(d*x+c))^(3/2)/a^4/d

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3676, 3673, 3608, 3561, 212} \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {(-361 B+151 i A) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}+\frac {(-89 B+39 i A) \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}-\frac {(-89 B+39 i A) \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(-B+i A) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

[In]

Int[(Tan[c + d*x]^4*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

-1/4*((I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(5/2)*d) + ((I*A - B)*Tan[c
+ d*x]^4)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((11*A + (21*I)*B)*Tan[c + d*x]^3)/(30*a*d*(a + I*a*Tan[c + d*x
])^(3/2)) - (((39*I)*A - 89*B)*Tan[c + d*x]^2)/(20*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + (((39*I)*A - 89*B)*Sqrt
[a + I*a*Tan[c + d*x]])/(5*a^3*d) - (((151*I)*A - 361*B)*(a + I*a*Tan[c + d*x])^(3/2))/(60*a^4*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3561

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3608

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] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 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 3676

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*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f
*m)), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d
*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan ^3(c+d x) \left (4 a (i A-B)+\frac {1}{2} a (3 A+13 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2} \\ & = \frac {(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\tan ^2(c+d x) \left (-\frac {3}{2} a^2 (11 A+21 i B)+\frac {3}{4} a^2 (17 i A-47 B) \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4} \\ & = \frac {(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {(39 i A-89 B) \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{2} a^3 (39 i A-89 B)-\frac {3}{8} a^3 (151 A+361 i B) \tan (c+d x)\right ) \, dx}{15 a^6} \\ & = \frac {(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {(39 i A-89 B) \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(151 i A-361 B) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}-\frac {\int \sqrt {a+i a \tan (c+d x)} \left (\frac {3}{8} a^3 (151 A+361 i B)-\frac {3}{2} a^3 (39 i A-89 B) \tan (c+d x)\right ) \, dx}{15 a^6} \\ & = \frac {(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {(39 i A-89 B) \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(39 i A-89 B) \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}-\frac {(151 i A-361 B) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}+\frac {(A-i B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = \frac {(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {(39 i A-89 B) \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(39 i A-89 B) \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}-\frac {(151 i A-361 B) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}-\frac {(i A+B) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 a^2 d} \\ & = -\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {(39 i A-89 B) \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(39 i A-89 B) \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}-\frac {(151 i A-361 B) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.11 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.64 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {15 (i A+B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) (-i+\tan (c+d x))^2-2 \left (151 i A-361 B-5 (77 A+179 i B) \tan (c+d x)+60 (-5 i A+11 B) \tan ^2(c+d x)+20 (3 A+5 i B) \tan ^3(c+d x)+20 B \tan ^4(c+d x)\right )}{60 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[(Tan[c + d*x]^4*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(15*(I*A + B)*Hypergeometric2F1[-1/2, 1, 1/2, (1 + I*Tan[c + d*x])/2]*(-I + Tan[c + d*x])^2 - 2*((151*I)*A - 3
61*B - 5*(77*A + (179*I)*B)*Tan[c + d*x] + 60*((-5*I)*A + 11*B)*Tan[c + d*x]^2 + 20*(3*A + (5*I)*B)*Tan[c + d*
x]^3 + 20*B*Tan[c + d*x]^4))/(60*a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+3 i a B \sqrt {a +i a \tan \left (d x +c \right )}+a A \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{2} \left (31 i B +17 A \right )}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{3} \left (9 i B +7 A \right )}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{4} \left (i B +A \right )}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16}\right )}{d \,a^{4}}\)	\(181\)
default	\(\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+3 i a B \sqrt {a +i a \tan \left (d x +c \right )}+a A \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{2} \left (31 i B +17 A \right )}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{3} \left (9 i B +7 A \right )}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{4} \left (i B +A \right )}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16}\right )}{d \,a^{4}}\)	\(181\)
parts	\(\frac {2 i A \left (\sqrt {a +i a \tan \left (d x +c \right )}-\frac {\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16}+\frac {17 a}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {7 a^{2}}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{3}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d \,a^{3}}+\frac {2 B \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-3 a \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16}-\frac {31 a^{2}}{8 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {3 a^{3}}{4 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{4}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d \,a^{4}}\)	\(244\)

[In]

int(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2*I/d/a^4*(-1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)+3*I*a*B*(a+I*a*tan(d*x+c))^(1/2)+a*A*(a+I*a*tan(d*x+c))^(1/2)+1/8
*a^2*(31*I*B+17*A)/(a+I*a*tan(d*x+c))^(1/2)-1/12*a^3*(9*I*B+7*A)/(a+I*a*tan(d*x+c))^(3/2)+1/10*a^4*(A+I*B)/(a+
I*a*tan(d*x+c))^(5/2)-1/16*a^(3/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(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. 457 vs. \(2 (198) = 396\).

Time = 0.26 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.79 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} - {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt {2} {\left ({\left (463 i \, A - 983 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, {\left (-219 i \, A + 509 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 12 \, {\left (-14 i \, A + 29 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-23 i \, A + 33 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{120 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \]

[In]

integrate(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/120*(15*sqrt(1/2)*(a^3*d*e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d
^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(A^
2 - 2*I*A*B - B^2)/(a^5*d^2)) + (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 15*sqrt(1/2)*(a^3*
d*e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))*log(4*(sqrt(2)*sqrt(
1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^
2)) - (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) + sqrt(2)*((463*I*A - 983*B)*e^(8*I*d*x + 8*I*
c) - 3*(-219*I*A + 509*B)*e^(6*I*d*x + 6*I*c) - 12*(-14*I*A + 29*B)*e^(4*I*d*x + 4*I*c) + (-23*I*A + 33*B)*e^(
2*I*d*x + 2*I*c) + 3*I*A - 3*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(a^3*d*e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d
*x + 5*I*c))

Sympy [F]

\[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(tan(d*x+c)**4*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral((A + B*tan(c + d*x))*tan(c + d*x)**4/(I*a*(tan(c + d*x) - I))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.73 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \, {\left (15 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 160 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} B a + 480 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + 3 i \, B\right )} a^{2} + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (17 \, A + 31 i \, B\right )} a^{3} - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (7 \, A + 9 i \, B\right )} a^{4} + 12 \, {\left (A + i \, B\right )} a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\right )}}{240 \, a^{5} d} \]

[In]

integrate(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/240*I*(15*sqrt(2)*(A - I*B)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + s
qrt(I*a*tan(d*x + c) + a))) - 160*I*(I*a*tan(d*x + c) + a)^(3/2)*B*a + 480*sqrt(I*a*tan(d*x + c) + a)*(A + 3*I
*B)*a^2 + 4*(15*(I*a*tan(d*x + c) + a)^2*(17*A + 31*I*B)*a^3 - 10*(I*a*tan(d*x + c) + a)*(7*A + 9*I*B)*a^4 + 1
2*(A + I*B)*a^5)/(I*a*tan(d*x + c) + a)^(5/2))/(a^5*d)

Giac [F]

\[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^4/(I*a*tan(d*x + c) + a)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 8.55 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\frac {A\,1{}\mathrm {i}}{5\,d}-\frac {A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{6\,a\,d}+\frac {A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,17{}\mathrm {i}}{4\,a^2\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a^3\,d}-\frac {6\,B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^3\,d}+\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^4\,d}-\frac {\frac {B\,a^2}{5}+\frac {31\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4}-\frac {3\,B\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2}}{a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d}+\frac {\sqrt {2}\,B\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{5/2}\,d} \]

[In]

int((tan(c + d*x)^4*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

((A*1i)/(5*d) - (A*(a + a*tan(c + d*x)*1i)*7i)/(6*a*d) + (A*(a + a*tan(c + d*x)*1i)^2*17i)/(4*a^2*d))/(a + a*t
an(c + d*x)*1i)^(5/2) + (A*(a + a*tan(c + d*x)*1i)^(1/2)*2i)/(a^3*d) - (6*B*(a + a*tan(c + d*x)*1i)^(1/2))/(a^
3*d) + (2*B*(a + a*tan(c + d*x)*1i)^(3/2))/(3*a^4*d) - ((B*a^2)/5 + (31*B*(a + a*tan(c + d*x)*1i)^2)/4 - (3*B*
a*(a + a*tan(c + d*x)*1i))/2)/(a^2*d*(a + a*tan(c + d*x)*1i)^(5/2)) + (2^(1/2)*A*atan((2^(1/2)*(a + a*tan(c +
d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*1i)/(8*(-a)^(5/2)*d) + (2^(1/2)*B*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*
1i)/(2*a^(1/2)))*1i)/(8*a^(5/2)*d)