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

Optimal. Leaf size=246 \[ \frac {4 \sqrt {2} a^{5/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 a^2 (46 B+45 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}-\frac {8 a^2 (46 B+45 i A) \sqrt {a+i a \tan (c+d x)}}{105 d}-\frac {8 a (59 B+60 i A) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac {2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \]

[Out]

4*a^(5/2)*(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-8/105*a^2*(45*I*A+46*B)*(a+I
*a*tan(d*x+c))^(1/2)/d+2/105*a^2*(45*I*A+46*B)*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/d-2/21*a^2*(3*A-4*I*B)*(a
+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3/d-8/315*a*(60*I*A+59*B)*(a+I*a*tan(d*x+c))^(3/2)/d+2/9*I*a*B*tan(d*x+c)^3*
(a+I*a*tan(d*x+c))^(3/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.75, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3594, 3597, 3592, 3527, 3480, 206} \[ -\frac {2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 a^2 (46 B+45 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}-\frac {8 a^2 (46 B+45 i A) \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {4 \sqrt {2} a^{5/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 a (59 B+60 i A) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac {2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[2]*a^(5/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*a^2*((45*I)*A + 46*
B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d) + (2*a^2*((45*I)*A + 46*B)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(
105*d) - (2*a^2*(3*A - (4*I)*B)*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/(21*d) - (8*a*((60*I)*A + 59*B)*(a
+ I*a*Tan[c + d*x])^(3/2))/(315*d) + (((2*I)/9)*a*B*Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^(3/2))/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 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]

Rule 3597

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

Rubi steps

\begin {align*} \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac {2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac {2}{9} \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \left (\frac {3}{2} a (3 A-2 i B)+\frac {3}{2} a (3 i A+4 B) \tan (c+d x)\right ) \, dx\\ &=-\frac {2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac {4}{63} \int \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {3}{4} a^2 (39 A-38 i B)+\frac {3}{4} a^2 (45 i A+46 B) \tan (c+d x)\right ) \, dx\\ &=\frac {2 a^2 (45 i A+46 B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}-\frac {2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac {8 \int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{2} a^3 (45 i A+46 B)+\frac {3}{2} a^3 (60 A-59 i B) \tan (c+d x)\right ) \, dx}{315 a}\\ &=\frac {2 a^2 (45 i A+46 B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}-\frac {2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}-\frac {8 a (60 i A+59 B) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac {2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac {8 \int \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{2} a^3 (60 A-59 i B)-\frac {3}{2} a^3 (45 i A+46 B) \tan (c+d x)\right ) \, dx}{315 a}\\ &=-\frac {8 a^2 (45 i A+46 B) \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {2 a^2 (45 i A+46 B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}-\frac {2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}-\frac {8 a (60 i A+59 B) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac {2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}-\left (4 a^2 (A-i B)\right ) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {8 a^2 (45 i A+46 B) \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {2 a^2 (45 i A+46 B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}-\frac {2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}-\frac {8 a (60 i A+59 B) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac {2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac {\left (8 a^3 (i A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=\frac {4 \sqrt {2} a^{5/2} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 a^2 (45 i A+46 B) \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {2 a^2 (45 i A+46 B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}-\frac {2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}-\frac {8 a (60 i A+59 B) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac {2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 6.01, size = 284, normalized size = 1.15 \[ \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \left (4 \sqrt {2} (B+i A) e^{-3 i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-\frac {i (\cos (2 c)-i \sin (2 c)) \sec ^{\frac {9}{2}}(c+d x) (12 (260 A-251 i B) \cos (2 (c+d x))+(915 A-961 i B) \cos (4 (c+d x))+390 i A \sin (2 (c+d x))+285 i A \sin (4 (c+d x))+2205 A+282 B \sin (2 (c+d x))+331 B \sin (4 (c+d x))-2331 i B)}{1260 (\cos (d x)+i \sin (d x))^2}\right )}{d \sec ^{\frac {7}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((4*Sqrt[2]*(I*A + B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E
^(I*(c + d*x))])/E^((3*I)*(c + d*x)) - ((I/1260)*Sec[c + d*x]^(9/2)*(Cos[2*c] - I*Sin[2*c])*(2205*A - (2331*I)
*B + 12*(260*A - (251*I)*B)*Cos[2*(c + d*x)] + (915*A - (961*I)*B)*Cos[4*(c + d*x)] + (390*I)*A*Sin[2*(c + d*x
)] + 282*B*Sin[2*(c + d*x)] + (285*I)*A*Sin[4*(c + d*x)] + 331*B*Sin[4*(c + d*x)]))/(Cos[d*x] + I*Sin[d*x])^2)
*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(d*Sec[c + d*x]^(7/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

________________________________________________________________________________________

fricas [B]  time = 0.75, size = 535, normalized size = 2.17 \[ \frac {315 \, \sqrt {-\frac {{\left (128 \, A^{2} - 256 i \, A B - 128 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left ({\left (16 i \, A + 16 \, B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {-\frac {{\left (128 \, A^{2} - 256 i \, A B - 128 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{2}}\right ) - 315 \, \sqrt {-\frac {{\left (128 \, A^{2} - 256 i \, A B - 128 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left ({\left (16 i \, A + 16 \, B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {-\frac {{\left (128 \, A^{2} - 256 i \, A B - 128 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{2}}\right ) + \sqrt {2} {\left ({\left (-9600 i \, A - 10336 \, B\right )} a^{2} e^{\left (9 i \, d x + 9 i \, c\right )} + {\left (-28080 i \, A - 26352 \, B\right )} a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + {\left (-35280 i \, A - 37296 \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + {\left (-21840 i \, A - 21840 \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (-5040 i \, A - 5040 \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{1260 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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)^2*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/1260*(315*sqrt(-(128*A^2 - 256*I*A*B - 128*B^2)*a^5/d^2)*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) +
6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log(((16*I*A + 16*B)*a^3*e^(I*d*x + I*c) + sqrt(2)*sqrt
(-(128*A^2 - 256*I*A*B - 128*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-
I*d*x - I*c)/((4*I*A + 4*B)*a^2)) - 315*sqrt(-(128*A^2 - 256*I*A*B - 128*B^2)*a^5/d^2)*(d*e^(8*I*d*x + 8*I*c)
+ 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log(((16*I*A + 16*B)*a^3*e^
(I*d*x + I*c) - sqrt(2)*sqrt(-(128*A^2 - 256*I*A*B - 128*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(
2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((4*I*A + 4*B)*a^2)) + sqrt(2)*((-9600*I*A - 10336*B)*a^2*e^(9*I*d*x
+ 9*I*c) + (-28080*I*A - 26352*B)*a^2*e^(7*I*d*x + 7*I*c) + (-35280*I*A - 37296*B)*a^2*e^(5*I*d*x + 5*I*c) + (
-21840*I*A - 21840*B)*a^2*e^(3*I*d*x + 3*I*c) + (-5040*I*A - 5040*B)*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x +
2*I*c) + 1)))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2
*I*c) + d)

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.27, size = 206, normalized size = 0.84 \[ -\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9}+\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}} a}{7}+\frac {A \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}} a}{7}-\frac {i a^{2} B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a^{3}}{3}+\frac {A \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a^{3}}{3}-2 i B \,a^{4} \sqrt {a +i a \tan \left (d x +c \right )}+2 A \,a^{4} \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {9}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/d/a^2*(-1/9*I*B*(a+I*a*tan(d*x+c))^(9/2)+1/7*I*B*(a+I*a*tan(d*x+c))^(7/2)*a+1/7*A*(a+I*a*tan(d*x+c))^(7/2
)*a-1/5*I*a^2*B*(a+I*a*tan(d*x+c))^(5/2)-1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)*a^3+1/3*A*(a+I*a*tan(d*x+c))^(3/2)*a
^3-2*I*B*a^4*(a+I*a*tan(d*x+c))^(1/2)+2*A*a^4*(a+I*a*tan(d*x+c))^(1/2)-2*a^(9/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(
a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))

________________________________________________________________________________________

maxima [A]  time = 0.96, size = 176, normalized size = 0.72 \[ -\frac {2 i \, {\left (315 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {11}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 35 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} B a + 45 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} {\left (A + i \, B\right )} a^{2} - 63 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} B a^{3} + 105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A - i \, B\right )} a^{4} + 630 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A - i \, B\right )} a^{5}\right )}}{315 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*I*(315*sqrt(2)*(A - I*B)*a^(11/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a)
+ sqrt(I*a*tan(d*x + c) + a))) - 35*I*(I*a*tan(d*x + c) + a)^(9/2)*B*a + 45*(I*a*tan(d*x + c) + a)^(7/2)*(A +
I*B)*a^2 - 63*I*(I*a*tan(d*x + c) + a)^(5/2)*B*a^3 + 105*(I*a*tan(d*x + c) + a)^(3/2)*(A - I*B)*a^4 + 630*sqrt
(I*a*tan(d*x + c) + a)*(A - I*B)*a^5)/(a^3*d)

________________________________________________________________________________________

mupad [B]  time = 7.57, size = 258, normalized size = 1.05 \[ -\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,d}-\frac {A\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}-\frac {2\,B\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}-\frac {A\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{d}-\frac {A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}\,2{}\mathrm {i}}{7\,a\,d}-\frac {4\,B\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}+\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{7\,a\,d}-\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{9/2}}{9\,a^2\,d}+\frac {\sqrt {2}\,A\,{\left (-a\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,4{}\mathrm {i}}{d}-\frac {\sqrt {2}\,B\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,4{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*(a + a*tan(c + d*x)*1i)^(7/2))/(7*a*d) - (A*a*(a + a*tan(c + d*x)*1i)^(3/2)*2i)/(3*d) - (2*B*a*(a + a*tan
(c + d*x)*1i)^(3/2))/(3*d) - (A*a^2*(a + a*tan(c + d*x)*1i)^(1/2)*4i)/d - (A*(a + a*tan(c + d*x)*1i)^(7/2)*2i)
/(7*a*d) - (4*B*a^2*(a + a*tan(c + d*x)*1i)^(1/2))/d - (2*B*(a + a*tan(c + d*x)*1i)^(5/2))/(5*d) - (2*B*(a + a
*tan(c + d*x)*1i)^(9/2))/(9*a^2*d) + (2^(1/2)*A*(-a)^(5/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a
)^(1/2)))*4i)/d - (2^(1/2)*B*a^(5/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*4i)/d

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________