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

Optimal. Leaf size=157 \[ \frac {5 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {5 i \sec (c+d x)}{64 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i \sec (c+d x)}{48 a d (a+i a \tan (c+d x))^{5/2}}+\frac {i \sec (c+d x)}{6 d (a+i a \tan (c+d x))^{7/2}} \]

[Out]

5/128*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(7/2)/d*2^(1/2)+1/6*I*sec(d*x+c)/d/
(a+I*a*tan(d*x+c))^(7/2)+5/48*I*sec(d*x+c)/a/d/(a+I*a*tan(d*x+c))^(5/2)+5/64*I*sec(d*x+c)/a^2/d/(a+I*a*tan(d*x
+c))^(3/2)

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Rubi [A]  time = 0.16, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3502, 3489, 206} \[ \frac {5 i \sec (c+d x)}{64 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {5 i \sec (c+d x)}{48 a d (a+i a \tan (c+d x))^{5/2}}+\frac {i \sec (c+d x)}{6 d (a+i a \tan (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((5*I)/64)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*a^(7/2)*d) + ((I/6)
*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((5*I)/48)*Sec[c + d*x])/(a*d*(a + I*a*Tan[c + d*x])^(5/2))
+ (((5*I)/64)*Sec[c + d*x])/(a^2*d*(a + I*a*Tan[c + d*x])^(3/2))

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 3489

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*a)/(b*f), Subst[
Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^
2, 0]

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rubi steps

\begin {align*} \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx &=\frac {i \sec (c+d x)}{6 d (a+i a \tan (c+d x))^{7/2}}+\frac {5 \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx}{12 a}\\ &=\frac {i \sec (c+d x)}{6 d (a+i a \tan (c+d x))^{7/2}}+\frac {5 i \sec (c+d x)}{48 a d (a+i a \tan (c+d x))^{5/2}}+\frac {5 \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx}{32 a^2}\\ &=\frac {i \sec (c+d x)}{6 d (a+i a \tan (c+d x))^{7/2}}+\frac {5 i \sec (c+d x)}{48 a d (a+i a \tan (c+d x))^{5/2}}+\frac {5 i \sec (c+d x)}{64 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{128 a^3}\\ &=\frac {i \sec (c+d x)}{6 d (a+i a \tan (c+d x))^{7/2}}+\frac {5 i \sec (c+d x)}{48 a d (a+i a \tan (c+d x))^{5/2}}+\frac {5 i \sec (c+d x)}{64 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{2-a x^2} \, dx,x,\frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}}\right )}{64 a^3 d}\\ &=\frac {5 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {i \sec (c+d x)}{6 d (a+i a \tan (c+d x))^{7/2}}+\frac {5 i \sec (c+d x)}{48 a d (a+i a \tan (c+d x))^{5/2}}+\frac {5 i \sec (c+d x)}{64 a^2 d (a+i a \tan (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 1.51, size = 119, normalized size = 0.76 \[ -\frac {\sec ^3(c+d x) \left (50 i \sin (2 (c+d x))+82 \cos (2 (c+d x))+\frac {30 e^{4 i (c+d x)} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+52\right )}{384 a^3 d (\tan (c+d x)-i)^3 \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/384*(Sec[c + d*x]^3*(52 + (30*E^((4*I)*(c + d*x))*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])/Sqrt[1 + E^((2*I)
*(c + d*x))] + 82*Cos[2*(c + d*x)] + (50*I)*Sin[2*(c + d*x)]))/(a^3*d*(-I + Tan[c + d*x])^3*Sqrt[a + I*a*Tan[c
+ d*x]])

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fricas [B]  time = 0.67, size = 278, normalized size = 1.77 \[ \frac {{\left (15 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (160 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + 160 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + 160 i\right )} e^{\left (-i \, d x - i \, c\right )}}{1024 \, a^{3} d}\right ) - 15 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-160 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 160 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + 160 i\right )} e^{\left (-i \, d x - i \, c\right )}}{1024 \, a^{3} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (33 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 59 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 34 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{384 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(15*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(6*I*d*x + 6*I*c)*log(1/1024*(sqrt(2)*sqrt(1/2)*(160*I*a^3*d*e
^(2*I*d*x + 2*I*c) + 160*I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) + 160*I)*e^(-I*d*x - I*c
)/(a^3*d)) - 15*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(6*I*d*x + 6*I*c)*log(1/1024*(sqrt(2)*sqrt(1/2)*(-160*I*
a^3*d*e^(2*I*d*x + 2*I*c) - 160*I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) + 160*I)*e^(-I*d*
x - I*c)/(a^3*d)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(33*I*e^(6*I*d*x + 6*I*c) + 59*I*e^(4*I*d*x + 4*
I*c) + 34*I*e^(2*I*d*x + 2*I*c) + 8*I))*e^(-6*I*d*x - 6*I*c)/(a^4*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.94, size = 373, normalized size = 2.38 \[ \frac {\left (1024 i \left (\cos ^{7}\left (d x +c \right )\right )+1024 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )-704 i \left (\cos ^{5}\left (d x +c \right )\right )+15 i \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-192 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+15 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+8 i \left (\cos ^{3}\left (d x +c \right )\right )+15 \sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+40 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-60 i \cos \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{768 d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768/d*(1024*I*cos(d*x+c)^7+1024*sin(d*x+c)*cos(d*x+c)^6-704*I*cos(d*x+c)^5+15*I*(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/
2)*cos(d*x+c)-192*sin(d*x+c)*cos(d*x+c)^4+15*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I
+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+8*I*cos(d*x+c)^3+15*2^(1/2)*sin(
d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*2^(1/2))+40*cos(d*x+c)^2*sin(d*x+c)-60*I*cos(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x
+c))^(1/2)/a^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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