3.3.26 \(\int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [226]

Optimal. Leaf size=218 \[ \frac {2 \sqrt [4]{-1} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

2*(-1)^(1/4)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(5/2)/d+(1/8-1/8*I)*arctan
h((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(5/2)/d+7/4*tan(d*x+c)^(1/2)/a^2/d/(a+I*a*tan(d*x
+c))^(1/2)-1/5*tan(d*x+c)^(5/2)/d/(a+I*a*tan(d*x+c))^(5/2)+1/2*I*tan(d*x+c)^(3/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.46, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3639, 3676, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {2 \sqrt [4]{-1} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-1)^(1/4)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + ((1/8
- I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) - Tan[c + d*x]^(5
/2)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I/2)*Tan[c + d*x]^(3/2))/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (7*Sq
rt[Tan[c + d*x]])/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

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

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(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 - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)
) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], 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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m
, 2*n])

Rule 3676

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]

Rule 3680

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

Rule 3682

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (-\frac {5 a}{2}+5 i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-\frac {45 i a^2}{4}-15 a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {105 a^3}{8}-15 i a^3 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{15 a^6}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{a^4}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {i \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 )}{4 a d}\\ &=\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d}\\ &=\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^2 d}\\ &=\frac {2 \sqrt [4]{-1} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 2.81, size = 241, normalized size = 1.11 \begin {gather*} \frac {i e^{-6 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (1-8 e^{2 i (c+d x)}+48 e^{4 i (c+d x)}-41 e^{6 i (c+d x)}-5 e^{5 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 )+40 \sqrt {2} e^{5 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{20 \sqrt {2} a^3 d \sqrt {\tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/20)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 - 8*E^((2*I)*(c + d*x)) + 48*E^((4*I)*(c +
d*x)) - 41*E^((6*I)*(c + d*x)) - 5*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/
Sqrt[-1 + E^((2*I)*(c + d*x))]] + 40*Sqrt[2]*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[(Sqrt[
2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(Sqrt[2]*a^3*d*E^((6*I)*(c + d*x))*Sqrt[Tan[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. 962 vs. \(2 (170 ) = 340\).
time = 0.20, size = 963, normalized size = 4.42 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(-320*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)-516*(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)^2+460*I*(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)-5*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))*a+30*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))*a*tan(d*x+c)^2-80*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)*a*tan(d*x+c)^4-20*(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))*a*tan(d*x+c)^3-5
*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))*a*tan(d*x+c)^4-196*I*(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)
^3+320*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)*a*tan(d*x+c)^3+480*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)*a*tan(d*x+c)^2+20*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))*(I*a)^(1/2)*2^(1/2)*a*tan(d*x+c)-80*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))*a*(-I*a)^(1/2)+140*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*
a)^(1/2)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^4/(I*a)^(1/2)/(-I*a)^(1/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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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. 621 vs. \(2 (160) = 320\).
time = 0.67, size = 621, normalized size = 2.85 \begin {gather*} \frac {{\left (10 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \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}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 10 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \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}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 10 \, a^{3} d \sqrt {-\frac {4 i}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {52 \, {\left (4 \, \sqrt {2} \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}} {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} + {\left (3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {-\frac {4 i}{a^{5} d^{2}}}\right )}}{605 \, {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}\right ) + 10 \, a^{3} d \sqrt {-\frac {4 i}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {52 \, {\left (4 \, \sqrt {2} \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}} {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} - {\left (3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {-\frac {4 i}{a^{5} d^{2}}}\right )}}{605 \, {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}\right ) + \sqrt {2} \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}} {\left (41 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 34 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 6 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(10*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(I*d*x + I*c) + 1
/4*sqrt(2)*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^(
2*I*d*x + 2*I*c) + 1)) - 10*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-a^3*d*sqrt(-1/8*I/(a^5*d^2))
*e^(I*d*x + I*c) + 1/4*sqrt(2)*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^(2*I*d*x + 2*I*c) + 1)) - 10*a^3*d*sqrt(-4*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(52/605*(4*s
qrt(2)*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^(3*I*
d*x + 3*I*c) + e^(I*d*x + I*c)) + (3*a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt(-4*I/(a^5*d^2)))/(e^(2*I*d*x + 2*
I*c) + 1)) + 10*a^3*d*sqrt(-4*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(52/605*(4*sqrt(2)*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^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c))
- (3*a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt(-4*I/(a^5*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) + sqrt(2)*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))*(41*e^(6*I*d*x + 6*I*c) +
34*e^(4*I*d*x + 4*I*c) - 6*e^(2*I*d*x + 2*I*c) + 1))*e^(-5*I*d*x - 5*I*c)/(a^3*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3062 deep

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Giac [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(tan(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

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

[Out]

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

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