3.134 \(\int \frac {1}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=162 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} a^{7/2} d}+\frac {i}{8 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {i}{12 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {i}{10 a d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}} \]

[Out]

-1/16*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(7/2)/d*2^(1/2)+1/8*I/a^3/d/(a+I*a*tan(d*x+c))
^(1/2)+1/7*I/d/(a+I*a*tan(d*x+c))^(7/2)+1/10*I/a/d/(a+I*a*tan(d*x+c))^(5/2)+1/12*I/a^2/d/(a+I*a*tan(d*x+c))^(3
/2)

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Rubi [A]  time = 0.11, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3479, 3480, 206} \[ \frac {i}{8 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {i}{12 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} a^{7/2} d}+\frac {i}{10 a d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(-7/2),x]

[Out]

((-I/8)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(7/2)*d) + (I/7)/(d*(a + I*a*Tan[c +
d*x])^(7/2)) + (I/10)/(a*d*(a + I*a*Tan[c + d*x])^(5/2)) + (I/12)/(a^2*d*(a + I*a*Tan[c + d*x])^(3/2)) + (I/8
)/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 206

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

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (c+d x))^{7/2}} \, dx &=\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}+\frac {\int \frac {1}{(a+i a \tan (c+d x))^{5/2}} \, dx}{2 a}\\ &=\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}+\frac {i}{10 a d (a+i a \tan (c+d x))^{5/2}}+\frac {\int \frac {1}{(a+i a \tan (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}+\frac {i}{10 a d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{12 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {1}{\sqrt {a+i a \tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}+\frac {i}{10 a d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{12 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {i}{8 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \sqrt {a+i a \tan (c+d x)} \, dx}{16 a^4}\\ &=\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}+\frac {i}{10 a d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{12 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {i}{8 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {i \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{8 a^3 d}\\ &=-\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} a^{7/2} d}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}+\frac {i}{10 a d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{12 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {i}{8 a^3 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 2.06, size = 150, normalized size = 0.93 \[ -\frac {81 e^{2 i (c+d x)}+188 e^{4 i (c+d x)}+298 e^{6 i (c+d x)}+176 e^{8 i (c+d x)}-105 e^{7 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+15}{105 a^3 d \left (1+e^{2 i (c+d x)}\right )^4 (\tan (c+d x)-i)^3 \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/105*(15 + 81*E^((2*I)*(c + d*x)) + 188*E^((4*I)*(c + d*x)) + 298*E^((6*I)*(c + d*x)) + 176*E^((8*I)*(c + d*
x)) - 105*E^((7*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])/(a^3*d*(1 + E^((2*I)*(c
+ d*x)))^4*(-I + Tan[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.45, size = 294, normalized size = 1.81 \[ \frac {{\left (-105 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 105 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (176 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 298 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 188 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 81 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i\right )}\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{1680 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1680*(-105*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(7*I*d*x + 7*I*c)*log(4*(sqrt(2)*sqrt(1/2)*(a^4*d*e^(2*I*d*
x + 2*I*c) + a^4*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c))
+ 105*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(7*I*d*x + 7*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a^4*d*e^(2*I*d*x + 2
*I*c) + a^4*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sq
rt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(176*I*e^(8*I*d*x + 8*I*c) + 298*I*e^(6*I*d*x + 6*I*c) + 188*I*e^(4*I*
d*x + 4*I*c) + 81*I*e^(2*I*d*x + 2*I*c) + 15*I))*e^(-7*I*d*x - 7*I*c)/(a^4*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 116, normalized size = 0.72 \[ \frac {2 i a \left (\frac {1}{16 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{24 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{20 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {1}{14 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{32 a^{\frac {9}{2}}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/d*a*(1/16/a^4/(a+I*a*tan(d*x+c))^(1/2)+1/24/a^3/(a+I*a*tan(d*x+c))^(3/2)+1/20/a^2/(a+I*a*tan(d*x+c))^(5/2)
+1/14/a/(a+I*a*tan(d*x+c))^(7/2)-1/32/a^(9/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))

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maxima [A]  time = 0.50, size = 137, normalized size = 0.85 \[ \frac {i \, {\left (\frac {105 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} + \frac {4 \, {\left (105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} + 70 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a + 84 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} + 120 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2}}\right )}}{3360 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3360*I*(105*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x
+ c) + a)))/a^(5/2) + 4*(105*(I*a*tan(d*x + c) + a)^3 + 70*(I*a*tan(d*x + c) + a)^2*a + 84*(I*a*tan(d*x + c) +
a)*a^2 + 120*a^3)/((I*a*tan(d*x + c) + a)^(7/2)*a^2))/(a*d)

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mupad [B]  time = 3.98, size = 129, normalized size = 0.80 \[ \frac {\frac {1{}\mathrm {i}}{7\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,1{}\mathrm {i}}{12\,a^2\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{8\,a^3\,d}+\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{10\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{16\,{\left (-a\right )}^{7/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1i/(7*d) + ((a + a*tan(c + d*x)*1i)^2*1i)/(12*a^2*d) + ((a + a*tan(c + d*x)*1i)^3*1i)/(8*a^3*d) + ((a + a*tan
(c + d*x)*1i)*1i)/(10*a*d))/(a + a*tan(c + d*x)*1i)^(7/2) - (2^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/
2))/(2*(-a)^(1/2)))*1i)/(16*(-a)^(7/2)*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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