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

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

[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]])

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Rubi [A]  time = 0.107834, antiderivative size = 162, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 1.95593, 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]

-(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))])/(105*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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Maple [A]  time = 0.019, size = 116, normalized size = 0.7 \begin{align*}{\frac{2\,ia}{d} \left ( -{\frac{\sqrt{2}}{32}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{9}{2}}}}+{\frac{1}{16\,{a}^{4}}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}+{\frac{1}{24\,{a}^{3}} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{1}{20\,{a}^{2}} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}+{\frac{1}{14\,a} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{7}{2}}}} \right ) } \end{align*}

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*(-1/32/a^(9/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+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))

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

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]

Exception raised: ValueError

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Fricas [B]  time = 2.18164, size = 909, normalized size = 5.61 \begin{align*} \frac{{\left (-105 i \, \sqrt{\frac{1}{2}} a^{4} d \sqrt{\frac{1}{a^{7} d^{2}}} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a^{4} d \sqrt{\frac{1}{a^{7} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} 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 (8 i \, d x + 8 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a^{4} d \sqrt{\frac{1}{a^{7} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} 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 )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{1680 \, a^{4} d} \end{align*}

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^(8*I*d*x + 8*I*c)*log((2*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*
e^(2*I*d*x + 2*I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*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^(8*I*d*x + 8*I*c)*log(-(2*sqrt(1/2)*a^4*d*sqrt(1/(a
^7*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x +
I*c))*e^(-I*d*x - I*c)) + sqrt(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^(I*d*x + I*c))*e^(-8*I*d*x - 8*I
*c)/(a^4*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

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)