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

Optimal. Leaf size=255 \[ -\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {(-361 B+151 i A) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}+\frac {(-89 B+39 i A) \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}-\frac {(-89 B+39 i A) \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(-B+i A) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

[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

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Rubi [A]  time = 0.78, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3595, 3592, 3527, 3480, 206} \[ -\frac {(-89 B+39 i A) \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(-361 B+151 i A) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}+\frac {(-89 B+39 i A) \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(-B+i A) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(4*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 206

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

Rule 3480

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 3527

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 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 3595

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*} \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) \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) \operatorname {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) \tanh ^{-1}\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*}

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Mathematica [A]  time = 5.96, size = 191, normalized size = 0.75 \[ \frac {\frac {120 (B+i A) e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (1+e^{2 i (c+d x)}\right )^{5/2}}+\sec ^4(c+d x) ((747 B-317 i A) \cos (2 (c+d x))+(493 B-233 i A) \cos (4 (c+d x))+340 A \sin (2 (c+d x))+230 A \sin (4 (c+d x))-84 i A+780 i B \sin (2 (c+d x))+490 i B \sin (4 (c+d x))+174 B)}{120 a^2 d (\tan (c+d x)-i)^2 \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((120*(I*A + B)*E^((5*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])/(1 + E^((2*I)*(c + d*x)))^(5/2) + Sec[c + d*x]^4
*((-84*I)*A + 174*B + ((-317*I)*A + 747*B)*Cos[2*(c + d*x)] + ((-233*I)*A + 493*B)*Cos[4*(c + d*x)] + 340*A*Si
n[2*(c + d*x)] + (780*I)*B*Sin[2*(c + d*x)] + 230*A*Sin[4*(c + d*x)] + (490*I)*B*Sin[4*(c + d*x)]))/(120*a^2*d
*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.60, size = 457, normalized size = 1.79 \[ -\frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} \log \left (\frac {{\left (4 \, \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} \log \left (-\frac {{\left (4 \, \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} - {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} {\left ({\left (463 i \, A - 983 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (657 i \, A - 1527 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (168 i \, A - 348 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-23 i \, A + 33 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{120 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)) + (4*I*A + 4*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)*s
qrt(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)) - (4*I*A + 4*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) + (657*I*A - 1527*B)*e^(6*I*d*x + 6*I*c) + (168*I*A - 348*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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.24, size = 181, normalized size = 0.71 \[ \frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+3 i B a \sqrt {a +i a \tan \left (d x +c \right )}+A \sqrt {a +i a \tan \left (d x +c \right )}\, a -\frac {a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16}+\frac {a^{2} \left (31 i B +17 A \right )}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{3} \left (9 i B +7 A \right )}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{4} \left (i B +A \right )}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.97, size = 185, normalized size = 0.73 \[ \frac {i \, {\left (15 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 160 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} B a + 480 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + 3 i \, B\right )} a^{2} + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (17 \, A + 31 i \, B\right )} a^{3} - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (7 \, A + 9 i \, B\right )} a^{4} + 12 \, {\left (A + i \, B\right )} a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\right )}}{240 \, a^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 6.96, size = 279, normalized size = 1.09 \[ \frac {\frac {A\,1{}\mathrm {i}}{5\,d}-\frac {A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{6\,a\,d}+\frac {A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,17{}\mathrm {i}}{4\,a^2\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a^3\,d}-\frac {6\,B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^3\,d}+\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^4\,d}-\frac {\frac {B\,a^2}{5}+\frac {31\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4}-\frac {3\,B\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2}}{a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d}+\frac {\sqrt {2}\,B\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{5/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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