3.210 \(\int \frac {\tan ^{\frac {7}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=218 \[ \frac {11 \sqrt [4]{-1} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a d}+\frac {7 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a d}+\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

11/4*(-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))/d/a^(1/2)+(1/2-1/2*I)*arc
tanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d/a^(1/2)+7/4*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c)
)^(1/2)/a/d-3/2*I*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(3/2)/a/d-tan(d*x+c)^(5/2)/d/(a+I*a*tan(d*x+c))^(1/2)

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Rubi [A]  time = 0.69, 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 = {3558, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac {\tan ^{\frac {5}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a d}+\frac {11 \sqrt [4]{-1} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 \sqrt {a} d}+\frac {7 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a d}+\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*(-1)^(1/4)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(4*Sqrt[a]*d) + ((1
/2 - I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) - Tan[c + d*x]
^(5/2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (7*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(4*a*d) - (((3*I)/2)
*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d)

Rule 63

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 203

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

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 217

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 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 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((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 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]

Rule 3599

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 3601

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)}{\sqrt {a+i a \tan (c+d x)}} \, dx &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {5 a}{2}+3 i a \tan (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a d}-\frac {\int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (-\frac {9 i a^2}{2}-\frac {7}{2} a^2 \tan (c+d x)\right ) \, dx}{2 a^3}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a d}-\frac {3 i \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a d}-\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {7 a^3}{4}-\frac {11}{4} i a^3 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{2 a^4}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a d}-\frac {3 i \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a d}-\frac {11 \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^2}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a d}-\frac {3 i \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a d}-\frac {11 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac {(i a) \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 {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a d}-\frac {3 i \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a d}-\frac {11 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 d}\\ &=\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a d}-\frac {3 i \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a d}-\frac {11 \operatorname {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 )}{4 d}\\ &=\frac {11 \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 )}{4 \sqrt {a} d}+\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a d}-\frac {3 i \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a d}\\ \end {align*}

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Mathematica [A]  time = 3.27, size = 224, normalized size = 1.03 \[ \frac {\sqrt {\tan (c+d x)} \sec ^2(c+d x) (i \sin (2 (c+d x))+5 \cos (2 (c+d x))+9)-\frac {2 e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \left (4 \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )-11 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{\left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^2}}{8 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^(3/2)*(4*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))
]] - 11*Sqrt[2]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/((((-I)*(-1 + E^((2*I)*(c
+ d*x))))/(1 + E^((2*I)*(c + d*x))))^(3/2)*(1 + E^((2*I)*(c + d*x)))^2) + Sec[c + d*x]^2*(9 + 5*Cos[2*(c + d*x
)] + I*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]])/(8*d*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.73, size = 658, normalized size = 3.02 \[ \frac {\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 (3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 9 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2\right )} + {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {2 i}{a d^{2}}} \log \left (\frac {1}{4} \, a d \sqrt {-\frac {2 i}{a 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 ) - {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {2 i}{a d^{2}}} \log \left (-\frac {1}{4} \, a d \sqrt {-\frac {2 i}{a 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 ) - {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {121 i}{16 \, a d^{2}}} \log \left (\frac {208 \, {\left (11 \, \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 )} + 2 \, {\left (3 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {-\frac {121 i}{16 \, a d^{2}}}\right )}}{6655 \, {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}\right ) + {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {121 i}{16 \, a d^{2}}} \log \left (\frac {208 \, {\left (11 \, \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 )} - 2 \, {\left (3 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {-\frac {121 i}{16 \, a d^{2}}}\right )}}{6655 \, {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}\right )}{4 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

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))^(1/2),x, algorithm="fricas")

[Out]

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))*(3
*e^(4*I*d*x + 4*I*c) + 9*e^(2*I*d*x + 2*I*c) + 2) + (a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(-2*I/
(a*d^2))*log(1/4*a*d*sqrt(-2*I/(a*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)) - (a*d*e^(3*I*d*x + 3*I*c) +
a*d*e^(I*d*x + I*c))*sqrt(-2*I/(a*d^2))*log(-1/4*a*d*sqrt(-2*I/(a*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)) - (a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(-121/16*I/(a*d^2))*log(208/6655*(11*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)) + 2*(3*a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt(-121/16*I/(a*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) +
(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(-121/16*I/(a*d^2))*log(208/6655*(11*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)) - 2*(3*a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt(-121/16*I/(a*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)))/(a*d*e
^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))

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

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))^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.23, size = 667, normalized size = 3.06 \[ -\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (2 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a +4 i \left (\tan ^{3}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-2 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a -22 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )+16 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+4 \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, \tan \left (d x +c \right ) a +11 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a +2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )-11 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +14 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{8 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}} \]

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))^(1/2),x)

[Out]

-1/8/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*(2*I*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))*(I*a)^(1/2)*tan(d*x+c)^2*a+4*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)^3-2*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-22*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*ta
n(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)*a+16*I*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)+4*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)*tan(d*x+c)*a+11*(-I*a)^(1/2)*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))*tan(d*x+c)^2*a+2*(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)^2-11*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+14*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I
*a)^(1/2)*(-I*a)^(1/2))/a/(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)+I)^2

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

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))^(1/2),x, algorithm="maxima")

[Out]

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

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

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)^(1/2),x)

[Out]

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

[Out]

Timed out

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