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

Optimal. Leaf size=127 \[ \frac {4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (5 B+7 i A)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (A-i B)}{d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[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)

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Rubi [A]  time = 0.26, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3593, 3591, 3529, 3533, 205} \[ \frac {4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (5 B+7 i A)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (A-i B)}{d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[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 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 3529

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 3533

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 3591

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 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]

Rubi steps

\begin {align*} \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 \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 ) \operatorname {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) \tan ^{-1}\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*}

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Mathematica [B]  time = 5.56, size = 272, normalized size = 2.14 \[ \frac {\cos ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left (-\frac {4 i e^{-2 i c} (A-i B) \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac {(\cos (2 c)-i \sin (2 c)) \csc ^2(c+d x) (5 (B+2 i A) \sin (2 (c+d x))+(33 A-30 i B) \cos (2 (c+d x))-27 A+30 i B)}{15 \sqrt {\tan (c+d x)}}\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^3*(((-4*I)*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[S
qrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])/(E^((2*I)*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 +
E^((2*I)*(c + d*x)))]) - (Csc[c + d*x]^2*(Cos[2*c] - I*Sin[2*c])*(-27*A + (30*I)*B + (33*A - (30*I)*B)*Cos[2*(
c + d*x)] + 5*((2*I)*A + B)*Sin[2*(c + d*x)]))/(15*Sqrt[Tan[c + d*x]]))*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c
+ d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [B]  time = 0.56, size = 498, normalized size = 3.92 \[ -\frac {15 \, \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 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 {{\left (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 15 \, \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 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 {{\left (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) - {\left ({\left (344 i \, A + 280 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-88 i \, A - 200 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-248 i \, A - 280 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (184 i \, A + 200 \, 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}}}{60 \, {\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 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

-1/60*(15*sqrt((16*I*A^2 + 32*A*B - 16*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((4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + sqrt((16*I*A^2 + 32*A*B - 16*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)/((2*I*A + 2*B)*a^2)) - 15*sqrt((16*I*A^2 + 32*A*B - 16*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((4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 + 32*A*
B - 16*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)/((2*I*A + 2*B)*a^2)) - ((344*I*A + 280*B)*a^2*e^(6*I*d*x + 6*I*c) + (-88*I*A - 200*B)*
a^2*e^(4*I*d*x + 4*I*c) + (-248*I*A - 280*B)*a^2*e^(2*I*d*x + 2*I*c) + (184*I*A + 200*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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\tan \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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maple [B]  time = 0.11, size = 537, normalized size = 4.23 \[ -\frac {2 a^{2} A}{5 d \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {i a^{2} A \sqrt {2}\, \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 d}+\frac {4 a^{2} A}{d \sqrt {\tan \left (d x +c \right )}}-\frac {4 i a^{2} A}{3 d \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 a^{2} B}{3 d \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {i a^{2} A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {i a^{2} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {i a^{2} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{2} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{2} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{2} B \sqrt {2}\, \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 d}-\frac {i a^{2} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {i a^{2} B \sqrt {2}\, \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 d}-\frac {4 i a^{2} B}{d \sqrt {\tan \left (d x +c \right )}}+\frac {a^{2} A \sqrt {2}\, \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 d}+\frac {a^{2} A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {a^{2} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.90, size = 195, normalized size = 1.54 \[ -\frac {15 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2} - \frac {4 \, {\left (30 \, {\left (A - i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} - {\left (10 i \, A + 5 \, B\right )} a^{2} \tan \left (d x + c\right ) - 3 \, A a^{2}\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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mupad [B]  time = 7.99, size = 258, normalized size = 2.03 \[ -\frac {\frac {2\,A\,a^2}{5\,d}-\frac {4\,A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d}+\frac {A\,a^2\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{3\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}-\frac {\frac {2\,B\,a^2}{3\,d}+\frac {B\,a^2\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}+\frac {\sqrt {2}\,A\,a^2\,\ln \left (4\,A\,a^2\,d+\sqrt {2}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d}-\frac {\sqrt {4{}\mathrm {i}}\,A\,a^2\,\ln \left (4\,A\,a^2\,d+2\,\sqrt {4{}\mathrm {i}}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^2\,\ln \left (-B\,a^2\,d\,4{}\mathrm {i}+\sqrt {2}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (1-\mathrm {i}\right )}{d}-\frac {\sqrt {-4{}\mathrm {i}}\,B\,a^2\,\ln \left (-B\,a^2\,d\,4{}\mathrm {i}+2\,\sqrt {-4{}\mathrm {i}}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]

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)^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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \left (- \frac {A}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\right )\, dx + \int \frac {A}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {B}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\right )\, dx + \int \frac {B}{\sqrt {\tan {\left (c + d x \right )}}}\, dx + \int \left (- \frac {2 i A}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\right )\, dx + \int \left (- \frac {2 i B}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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