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

Optimal. Leaf size=154 \[ -\frac {2 i \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {26 \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]

[Out]

(-1-I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*a^(1/2)/d+26/15*(a+I*a*tan(d*x+c))^(1/
2)/d/tan(d*x+c)^(1/2)-2/5*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)-2/15*I*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*
x+c)^(3/2)

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Rubi [A]  time = 0.41, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3561, 3598, 12, 3544, 205} \[ -\frac {2 i \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {26 \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1 - I)*Sqrt[a]*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*Sqrt[a + I*
a*Tan[c + d*x]])/(5*d*Tan[c + d*x]^(5/2)) - (((2*I)/15)*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/2)) + (
26*Sqrt[a + I*a*Tan[c + d*x]])/(15*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 3561

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

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 {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \int \frac {\left (\frac {i a}{2}-2 a \tan (c+d x)\right ) \sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{5 a}\\ &=-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {13 a^2}{4}-\frac {1}{2} i a^2 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^2}\\ &=-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {26 \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}+\frac {8 \int -\frac {15 i a^3 \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{15 a^3}\\ &=-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {26 \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-i \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {26 \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {\left (2 a^2\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 {(1+i) \sqrt {a} \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 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {26 \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [F]  time = 3.41, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.55, size = 423, normalized size = 2.75 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (68 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 12 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 20 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 60 i \, e^{\left (i \, d x + i \, c\right )}\right )} + 15 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{30 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/30*(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))*(
68*I*e^(7*I*d*x + 7*I*c) - 12*I*e^(5*I*d*x + 5*I*c) - 20*I*e^(3*I*d*x + 3*I*c) + 60*I*e^(I*d*x + I*c)) + 15*(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)*sqrt(2*I*a/d^2)*log((sqrt(2)*sqr
t(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) + I*d*sqrt(2*I*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 15*(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)*sqrt(2*I*a/d^2)*log((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) - I*d*sqrt(2*I*a/d^2)*e^(I*d*x
+ I*c))*e^(-I*d*x - I*c)))/(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 \[ \int \frac {\sqrt {i \, a \tan \left (d x + c\right ) + a}}{\tan \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.23, size = 357, normalized size = 2.32 \[ -\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (15 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{3}\left (d x +c \right )\right ) a -15 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{4}\left (d x +c \right )\right ) a +52 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (\tan ^{3}\left (d x +c \right )\right )-16 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}-56 i \left (\tan ^{2}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}+12 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{30 d \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30/d*(a*(1+I*tan(d*x+c)))^(1/2)/tan(d*x+c)^(5/2)*(15*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))*tan(d*x+c)^3*a-15*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)^4*a+52*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3-16*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)
-56*I*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)+12*I*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/
2)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-I*a)^(1/2)/(-tan(d*x+c)+I)

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maxima [B]  time = 0.75, size = 1144, normalized size = 7.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

-1/900*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((-(900*I - 900)*cos(3*d*x + 3*
c) + (1170*I - 1170)*cos(d*x + c) + (900*I + 900)*sin(3*d*x + 3*c) - (1170*I + 1170)*sin(d*x + c))*cos(3/2*arc
tan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + ((900*I + 900)*cos(3*d*x + 3*c) - (1170*I + 1170)*cos(d*x + c
) + (900*I - 900)*sin(3*d*x + 3*c) - (1170*I - 1170)*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*
x + 2*c) + 1)))*sqrt(a) + (((900*I - 900)*cos(2*d*x + 2*c)^2 + (900*I - 900)*sin(2*d*x + 2*c)^2 - (1800*I - 18
00)*cos(2*d*x + 2*c) + 900*I - 900)*arctan2((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)) - cos(d*x + c), (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)) - sin(d*
x + c)) + (-(450*I + 450)*cos(2*d*x + 2*c)^2 - (450*I + 450)*sin(2*d*x + 2*c)^2 + (900*I + 900)*cos(2*d*x + 2*
c) - 450*I - 450)*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 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) - 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)*(c
os(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))*sin(d*x + c) + cos(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)*s
qrt(a) + (((900*I - 900)*cos(5*d*x + 5*c) + (150*I - 150)*cos(3*d*x + 3*c) + (390*I - 390)*cos(d*x + c) - (900
*I + 900)*sin(5*d*x + 5*c) - (150*I + 150)*sin(3*d*x + 3*c) - (390*I + 390)*sin(d*x + c))*cos(5/2*arctan2(sin(
2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (((240*I - 240)*cos(d*x + c) - (240*I + 240)*sin(d*x + c))*cos(2*d*x +
2*c)^2 + ((240*I - 240)*cos(d*x + c) - (240*I + 240)*sin(d*x + c))*sin(2*d*x + 2*c)^2 + (-(480*I - 480)*cos(d
*x + c) + (480*I + 480)*sin(d*x + c))*cos(2*d*x + 2*c) + (240*I - 240)*cos(d*x + c) - (240*I + 240)*sin(d*x +
c))*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(900*I + 900)*cos(5*d*x + 5*c) - (150*I + 15
0)*cos(3*d*x + 3*c) - (390*I + 390)*cos(d*x + c) - (900*I - 900)*sin(5*d*x + 5*c) - (150*I - 150)*sin(3*d*x +
3*c) - (390*I - 390)*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + ((-(240*I + 240
)*cos(d*x + c) - (240*I - 240)*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (-(240*I + 240)*cos(d*x + c) - (240*I - 240)
*sin(d*x + c))*sin(2*d*x + 2*c)^2 + ((480*I + 480)*cos(d*x + c) + (480*I - 480)*sin(d*x + c))*cos(2*d*x + 2*c)
- (240*I + 240)*cos(d*x + c) - (240*I - 240)*sin(d*x + 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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(1/2)/tan(d*x+c)**(7/2),x)

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))/tan(c + d*x)**(7/2), x)

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