3.168 \(\int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=269 \[ \frac {(2+2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {4 a (57 B+61 i A) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a (11 A-12 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (9 B+10 i A) \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {4 a (193 A-201 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)} \]

[Out]

(2+2*I)*a^(3/2)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d-4/315*a*(193*A-201*
I*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/9*a*A*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(9/2)-2/63*a*(1
0*I*A+9*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(7/2)+4/105*a*(11*A-12*I*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*
x+c)^(5/2)+4/315*a*(61*I*A+57*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)

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Rubi [A]  time = 0.94, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3593, 3598, 12, 3544, 205} \[ \frac {(2+2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {4 a (57 B+61 i A) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a (11 A-12 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (9 B+10 i A) \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {4 a (193 A-201 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(11/2),x]

[Out]

((2 + 2*I)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a*A*Sqrt[a + I*a*Tan[c + d*x]])/(9*d*Tan[c + d*x]^(9/2)) - (2*a*((10*I)*A + 9*B)*Sqrt[a + I*a*Tan[c + d*x]])/(
63*d*Tan[c + d*x]^(7/2)) + (4*a*(11*A - (12*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(5/2)) + (4*
a*((61*I)*A + 57*B)*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*Tan[c + d*x]^(3/2)) - (4*a*(193*A - (201*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(315*d*Sqrt[Tan[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

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

Rule 3544

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

Rule 3593

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] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3598

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*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*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] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2}{9} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (10 i A+9 B)-\frac {1}{2} a (8 A-9 i B) \tan (c+d x)\right )}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 a (10 i A+9 B) \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{2} a^2 (11 A-12 i B)-\frac {3}{2} a^2 (10 i A+9 B) \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx}{63 a}\\ &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 a (10 i A+9 B) \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a (11 A-12 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} a^3 (61 i A+57 B)+3 a^3 (11 A-12 i B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{315 a^2}\\ &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 a (10 i A+9 B) \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a (11 A-12 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a (61 i A+57 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {16 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {3}{8} a^4 (193 A-201 i B)+\frac {3}{4} a^4 (61 i A+57 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{945 a^3}\\ &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 a (10 i A+9 B) \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a (11 A-12 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a (61 i A+57 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 a (193 A-201 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}+\frac {32 \int \frac {945 a^5 (i A+B) \sqrt {a+i a \tan (c+d x)}}{16 \sqrt {\tan (c+d x)}} \, dx}{945 a^4}\\ &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 a (10 i A+9 B) \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a (11 A-12 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a (61 i A+57 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 a (193 A-201 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}+(2 a (i A+B)) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 a (10 i A+9 B) \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a (11 A-12 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a (61 i A+57 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 a (193 A-201 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}+\frac {\left (4 a^3 (A-i B)\right ) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac {(2+2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 a (10 i A+9 B) \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a (11 A-12 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a (61 i A+57 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 a (193 A-201 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 15.70, size = 242, normalized size = 0.90 \[ \frac {a \sqrt {a+i a \tan (c+d x)} \left (\frac {2520 (A-i B) e^{-i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{\sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+\frac {\csc ^4(c+d x) (12 (117 A-134 i B) \cos (2 (c+d x))+(-487 A+474 i B) \cos (4 (c+d x))+144 i A \sin (2 (c+d x))-172 i A \sin (4 (c+d x))-1197 A+138 B \sin (2 (c+d x))-159 B \sin (4 (c+d x))+1134 i B)}{\sqrt {\tan (c+d x)}}\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(11/2),x]

[Out]

(a*((2520*(A - I*B)*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])/(E
^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]) + (Csc[c + d*x]^4*(-1197*A +
(1134*I)*B + 12*(117*A - (134*I)*B)*Cos[2*(c + d*x)] + (-487*A + (474*I)*B)*Cos[4*(c + d*x)] + (144*I)*A*Sin[
2*(c + d*x)] + 138*B*Sin[2*(c + d*x)] - (172*I)*A*Sin[4*(c + d*x)] - 159*B*Sin[4*(c + d*x)]))/Sqrt[Tan[c + d*x
]])*Sqrt[a + I*a*Tan[c + d*x]])/(1260*d)

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fricas [B]  time = 0.80, size = 680, normalized size = 2.53 \[ \frac {\sqrt {2} {\left ({\left (-2636 i \, A - 2532 \, B\right )} a e^{\left (11 i \, d x + 11 i \, c\right )} + {\left (3556 i \, A + 4452 \, B\right )} a e^{\left (9 i \, d x + 9 i \, c\right )} + {\left (-3384 i \, A - 2088 \, B\right )} a e^{\left (7 i \, d x + 7 i \, c\right )} + {\left (-4536 i \, A - 3192 \, B\right )} a e^{\left (5 i \, d x + 5 i \, c\right )} + {\left (3780 i \, A + 4620 \, B\right )} a e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (-1260 i \, A - 1260 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} + 315 \, \sqrt {\frac {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (2 i \, A + 2 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (2 i \, A + 2 \, B\right )} a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} + \sqrt {\frac {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) - 315 \, \sqrt {\frac {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (2 i \, A + 2 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (2 i \, A + 2 \, B\right )} a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} - \sqrt {\frac {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right )}{630 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, 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((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

1/630*(sqrt(2)*((-2636*I*A - 2532*B)*a*e^(11*I*d*x + 11*I*c) + (3556*I*A + 4452*B)*a*e^(9*I*d*x + 9*I*c) + (-3
384*I*A - 2088*B)*a*e^(7*I*d*x + 7*I*c) + (-4536*I*A - 3192*B)*a*e^(5*I*d*x + 5*I*c) + (3780*I*A + 4620*B)*a*e
^(3*I*d*x + 3*I*c) + (-1260*I*A - 1260*B)*a*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I
*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + 315*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*(d*e^(10*I*d*x
+ 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2
*I*c) - d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*
d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)) - 315*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*(d*e^(
10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*
I*d*x + 2*I*c) - d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a)*sqrt(a/(e^(2*I*d*x +
2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) - sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*
a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)))/(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8
*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) - d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\tan \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(3/2)/tan(d*x + c)^(11/2), x)

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maple [B]  time = 0.32, size = 885, normalized size = 3.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x)

[Out]

1/630/d*(a*(1+I*tan(d*x+c)))^(1/2)*a/tan(d*x+c)^(9/2)*(-1544*A*tan(d*x+c)^4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)*(I*a)^(1/2)*(-I*a)^(1/2)+630*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2
)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^5*a+488*I*A*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)
^(1/2)*(-I*a)^(1/2)+456*B*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+1608*I*B
*tan(d*x+c)^4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+1260*B*ln(1/2*(2*I*a*tan(d*x+c)+2
*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^5*a+1260*I*A*ln(1/2
*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)
^5*a-315*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*
x+c))/(tan(d*x+c)+I))*tan(d*x+c)^5*a+264*A*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*
a)^(1/2)-200*I*A*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-630*ln(1/2*(2*I*a*t
an(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^5*a-288*
I*B*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-315*I*(I*a)^(1/2)*2^(1/2)*ln(-
(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^
5*a-180*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)-140*A*(I*a)^(1/2)*(-I*a)^(
1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)

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maxima [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((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{11/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2))/tan(c + d*x)^(11/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2))/tan(c + d*x)^(11/2), x)

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sympy [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((a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(11/2),x)

[Out]

Timed out

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