∫ cot7 _ 2 (c + dx)(a + ia tan(c + dx))5∕2(A + B tan(c + dx)) dx [549]

3.6.49 \(\int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [549]

Optimal. Leaf size=205 \[ -\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 a^2 (38 A-35 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a^2 (8 i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d} \]

[Out]

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

(d*x+c)^(1/2)/d-2/15*a^2*(8*I*A+5*B)*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/d+2/15*a^2*(38*A-35*I*B)*cot(d*

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

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Rubi [A]
time = 0.47, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4326, 3674, 3679, 12, 3625, 211} \begin {gather*} -\frac {(4+4 i) a^{5/2} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \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^2 (5 B+8 i A) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}+\frac {2 a^2 (38 A-35 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

((-4 - 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Co

t[c + d*x]]*Sqrt[Tan[c + d*x]])/d + (2*a^2*(38*A - (35*I)*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(1

5*d) - (2*a^2*((8*I)*A + 5*B)*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d) - (2*a*A*Cot[c + d*x]^(5/2

)*(a + I*a*Tan[c + d*x])^(3/2))/(5*d)

Rule 12

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

Rule 211

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

&& PosQ[a/b]

Rule 3625

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 3674

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] && E

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

Rule 3679

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]

Rule 4326

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A

ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ

[u, x]

Rubi steps

\begin {align*} \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {1}{5} \left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^{3/2} \left (\frac {1}{2} a (8 i A+5 B)-\frac {1}{2} a (2 A-5 i B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (8 i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {1}{15} \left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (38 A-35 i B)-\frac {1}{4} a^2 (22 i A+25 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (38 A-35 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a^2 (8 i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {\left (8 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int -\frac {15 a^3 (i A+B) \sqrt {a+i a \tan (c+d x)}}{2 \sqrt {\tan (c+d x)}} \, dx}{15 a}\\ &=\frac {2 a^2 (38 A-35 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a^2 (8 i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}-\left (4 a^2 (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {2 a^2 (38 A-35 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a^2 (8 i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {\left (8 i a^4 (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {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 {(4-4 i) a^{5/2} (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 a^2 (38 A-35 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a^2 (8 i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\\ \end {align*}

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Mathematica [A]
time = 4.28, size = 306, normalized size = 1.49 \begin {gather*} \frac {\left (-4 \sqrt {2} (A-i B) e^{-3 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {\frac {i \left (1+e^{2 i (c+d x)}\right )}{-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 )-\frac {\sqrt {\cot (c+d x)} \csc ^2(c+d x) \sqrt {\sec (c+d x)} (\cos (2 c)-i \sin (2 c)) (-35 (A-i B)+(41 A-35 i B) \cos (2 (c+d x))+(11 i A+5 B) \sin (2 (c+d x)))}{15 (\cos (d x)+i \sin (d x))^2}\right ) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{d \sec ^{\frac {7}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

(((-4*Sqrt[2]*(A - I*B)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[(I

*(1 + E^((2*I)*(c + d*x))))/(-1 + E^((2*I)*(c + d*x)))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]

])/E^((3*I)*(c + d*x)) - (Sqrt[Cot[c + d*x]]*Csc[c + d*x]^2*Sqrt[Sec[c + d*x]]*(Cos[2*c] - I*Sin[2*c])*(-35*(A

- I*B) + (41*A - (35*I)*B)*Cos[2*(c + d*x)] + ((11*I)*A + 5*B)*Sin[2*(c + d*x)]))/(15*(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]

))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2213 vs. \(2 (168 ) = 336\).
time = 62.41, size = 2214, normalized size = 10.80


method	result	size

default	\(\text {Expression too large to display}\)	\(2214\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/15/d*a^2*2^(1/2)*(60*I*A*cos(d*x+c)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)*sin(d*x+c)*((-1+

cos(d*x+c))/sin(d*x+c))^(1/2)+38*A*2^(1/2)-35*I*B*2^(1/2)+30*I*B*cos(d*x+c)^2*ln((((-1+cos(d*x+c))/sin(d*x+c))

^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/(-((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos

(d*x+c)+sin(d*x+c)-1))*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+40*B*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+52*A

*cos(d*x+c)^3*2^(1/2)-49*A*cos(d*x+c)*2^(1/2)-60*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*

arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)-60*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*

x+c)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)+60*B*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)

*cos(d*x+c)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)+60*B*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c)

)^(1/2)*cos(d*x+c)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)+60*I*A*cos(d*x+c)^2*arctan(((-1+cos(

d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+30*I*A*cos(d*x+c)^2*ln((-((

-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/

2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+60*I*B*cos(d*x+c

)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+60*I*B*

cos(d*x+c)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2

)+52*I*A*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)-11*I*A*cos(d*x+c)*2^(1/2)*sin(d*x+c)-60*I*A*arctan(((-1+cos(d*x+c))/s

in(d*x+c))^(1/2)*2^(1/2)+1)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-30*A*cos(d*x+c)^2*ln((((-1+cos(d*x+c

))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/(-((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*

sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+30*B*cos(d*x+c)^2*ln((-((-1

+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)

*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-60*I*A*arctan(((-1

+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-30*I*A*ln((-((-1+cos(d

*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/

2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-60*I*B*arctan(((-1+cos(d

*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-60*I*B*arctan(((-1+cos(d*x+c

))/sin(d*x+c))^(1/2)*2^(1/2)-1)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-30*I*B*ln((((-1+cos(d*x+c))/sin(

d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/(-((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x

+c)+cos(d*x+c)+sin(d*x+c)-1))*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-5*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)-

35*2^(1/2)*B*sin(d*x+c)-41*A*2^(1/2)*cos(d*x+c)^2+60*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((

-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)+60*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+cos

(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)-60*B*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+cos(d*x+c

))/sin(d*x+c))^(1/2)*2^(1/2)+1)-60*B*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+cos(d*x+c))/sin

(d*x+c))^(1/2)*2^(1/2)-1)+30*A*ln((((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)

-1)/(-((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))*sin(d*x+c)*((-1+cos(d*x+

c))/sin(d*x+c))^(1/2)-30*B*ln((-((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)

/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))*sin(d*x+c)*((-1+cos(d*x+c))/

sin(d*x+c))^(1/2)-40*I*B*cos(d*x+c)^3*2^(1/2)+35*I*B*cos(d*x+c)^2*2^(1/2)-38*I*A*2^(1/2)*sin(d*x+c)+40*I*B*cos

(d*x+c)*2^(1/2))*(cos(d*x+c)/sin(d*x+c))^(7/2)*((I*sin(d*x+c)+cos(d*x+c))*a/cos(d*x+c))^(1/2)*sin(d*x+c)/(I*si

n(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^3

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1545 vs. \(2 (157) = 314\).
time = 0.96, size = 1545, normalized size = 7.54 \begin {gather*} \text {Too large to display} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

2/15*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((30*((I + 1)*A - (I - 1)*B)*a^2*

cos(3*d*x + 3*c) + (-(31*I + 31)*A + (25*I - 25)*B)*a^2*cos(d*x + c) + 30*((I - 1)*A + (I + 1)*B)*a^2*sin(3*d*

x + 3*c) + (-(31*I - 31)*A - (25*I + 25)*B)*a^2*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*

c) - 1)) + (30*(-(I - 1)*A - (I + 1)*B)*a^2*cos(3*d*x + 3*c) + ((31*I - 31)*A + (25*I + 25)*B)*a^2*cos(d*x + c

) + 30*((I + 1)*A - (I - 1)*B)*a^2*sin(3*d*x + 3*c) + (-(31*I + 31)*A + (25*I - 25)*B)*a^2*sin(d*x + c))*sin(3

/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a) + 15*(2*((-(I - 1)*A - (I + 1)*B)*a^2*cos(2*d*x +

2*c)^2 + (-(I - 1)*A - (I + 1)*B)*a^2*sin(2*d*x + 2*c)^2 + 2*((I - 1)*A + (I + 1)*B)*a^2*cos(2*d*x + 2*c) + (

-(I - 1)*A - (I + 1)*B)*a^2)*arctan2(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4

)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*sin(d*x + c), 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x

+ 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*cos(d*x

+ c)) + ((-(I + 1)*A + (I - 1)*B)*a^2*cos(2*d*x + 2*c)^2 + (-(I + 1)*A + (I - 1)*B)*a^2*sin(2*d*x + 2*c)^2 +

2*((I + 1)*A - (I - 1)*B)*a^2*cos(2*d*x + 2*c) + (-(I + 1)*A + (I - 1)*B)*a^2)*log(4*cos(d*x + c)^2 + 4*sin(d*

x + c)^2 + 4*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x

+ 2*c), cos(2*d*x + 2*c) - 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1))^2) + 8*(cos(2*d*x

+ 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c),

cos(2*d*x + 2*c) - 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))))*(cos(2*d*x

+ 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + ((30*((I + 1)*A - (I - 1)*B)*a^2*cos(5

*d*x + 5*c) + 5*(-(5*I + 5)*A + (11*I - 11)*B)*a^2*cos(3*d*x + 3*c) + ((7*I + 7)*A - (25*I - 25)*B)*a^2*cos(d*

x + c) + 30*((I - 1)*A + (I + 1)*B)*a^2*sin(5*d*x + 5*c) + 5*(-(5*I - 5)*A - (11*I + 11)*B)*a^2*sin(3*d*x + 3*

c) + ((7*I - 7)*A + (25*I + 25)*B)*a^2*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1))

+ 4*((-(2*I + 2)*A + (5*I - 5)*B)*a^2*cos(d*x + c) + (-(2*I - 2)*A - (5*I + 5)*B)*a^2*sin(d*x + c) + ((-(2*I +

2)*A + (5*I - 5)*B)*a^2*cos(d*x + c) + (-(2*I - 2)*A - (5*I + 5)*B)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((

-(2*I + 2)*A + (5*I - 5)*B)*a^2*cos(d*x + c) + (-(2*I - 2)*A - (5*I + 5)*B)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)

^2 + 2*(((2*I + 2)*A - (5*I - 5)*B)*a^2*cos(d*x + c) + ((2*I - 2)*A + (5*I + 5)*B)*a^2*sin(d*x + c))*cos(2*d*x

+ 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (30*(-(I - 1)*A - (I + 1)*B)*a^2*cos(5*d*x

+ 5*c) + 5*((5*I - 5)*A + (11*I + 11)*B)*a^2*cos(3*d*x + 3*c) + (-(7*I - 7)*A - (25*I + 25)*B)*a^2*cos(d*x +

c) + 30*((I + 1)*A - (I - 1)*B)*a^2*sin(5*d*x + 5*c) + 5*(-(5*I + 5)*A + (11*I - 11)*B)*a^2*sin(3*d*x + 3*c) +

((7*I + 7)*A - (25*I - 25)*B)*a^2*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 4*

(((2*I - 2)*A + (5*I + 5)*B)*a^2*cos(d*x + c) + (-(2*I + 2)*A + (5*I - 5)*B)*a^2*sin(d*x + c) + (((2*I - 2)*A

+ (5*I + 5)*B)*a^2*cos(d*x + c) + (-(2*I + 2)*A + (5*I - 5)*B)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (((2*I -

2)*A + (5*I + 5)*B)*a^2*cos(d*x + c) + (-(2*I + 2)*A + (5*I - 5)*B)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^2 + 2*

((-(2*I - 2)*A - (5*I + 5)*B)*a^2*cos(d*x + c) + ((2*I + 2)*A - (5*I - 5)*B)*a^2*sin(d*x + c))*cos(2*d*x + 2*c

))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a))/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^

2 - 2*cos(2*d*x + 2*c) + 1)^(5/4)*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 515 vs. \(2 (157) = 314\).
time = 1.21, size = 515, normalized size = 2.51 \begin {gather*} -\frac {2 \, {\left (15 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 15 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 2 \, \sqrt {2} {\left (2 \, {\left (13 \, A - 10 i \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 35 \, {\left (A - i \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, {\left (A - i \, 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}} \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 )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2/15*(15*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d

)*log(4*((A - I*B)*a^3*e^(I*d*x + I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*(I*d*e^(2*I*d*x + 2*I*c) - I*

d)*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)))*e^(-I*d*x -

I*c)/((-I*A - B)*a^2)) - 15*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*

I*d*x + 2*I*c) + d)*log(4*((A - I*B)*a^3*e^(I*d*x + I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*(-I*d*e^(2*

I*d*x + 2*I*c) + I*d)*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)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 2*sqrt(2)*(2*(13*A - 10*I*B)*a^2*e^(5*I*d*x + 5*I*c) - 35*(A - I*B

)*a^2*e^(3*I*d*x + 3*I*c) + 15*(A - I*B)*a^2*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)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**(7/2)*(a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\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 )}^{5/2} \,d x \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2), x)

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