3.4.89 \(\int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [389]

Optimal. Leaf size=121 \[ -\frac {3 i \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{a^{7/2} d}-\frac {2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^{5/2}}+\frac {6 i \sec (c+d x)}{a^2 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

-3*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*2^(1/2)/a^(7/2)/d-2*I*sec(d*x+c)^3/a/d/(
a+I*a*tan(d*x+c))^(5/2)+6*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 = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3582, 3583, 3570, 212} \begin {gather*} -\frac {3 i \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{a^{7/2} d}+\frac {6 i \sec (c+d x)}{a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac {2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

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

Rule 212

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

Rule 3570

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 3582

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

Rule 3583

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

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Mathematica [A]
time = 1.11, size = 126, normalized size = 1.04 \begin {gather*} \frac {16 e^{5 i (c+d x)} \left (-1-3 e^{2 i (c+d x)}+3 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{a^3 d \left (1+e^{2 i (c+d x)}\right )^4 (-i+\tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(16*E^((5*I)*(c + d*x))*(-1 - 3*E^((2*I)*(c + d*x)) + 3*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcT
anh[Sqrt[1 + E^((2*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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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (102 ) = 204\).
time = 0.82, size = 318, normalized size = 2.63

method	result	size
default	\(-\frac {\left (3 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )+3 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+3 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sin \left (d x +c \right )-8 i \left (\cos ^{3}\left (d x +c \right )\right )-8 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{2 d \,a^{4}}\)	\(318\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/d*(3*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))*cos(d*x+c)*2^(1/2)+3*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*c
os(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)+3*2^(1/2)*(-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))*sin(d*x+c)-8*I*cos(d*x+c)^3-8*cos(d*x+c)^2*sin(d*x+c)-4*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/c
os(d*x+c))^(1/2)/a^4

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (96) = 192\).
time = 0.40, size = 245, normalized size = 2.02 \begin {gather*} \frac {{\left (-3 i \, \sqrt {2} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {12 \, {\left ({\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + 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}}} + i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3} d}\right ) + 3 i \, \sqrt {2} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {12 \, {\left ({\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 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}}} + i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3} d}\right ) - 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/2*(-3*I*sqrt(2)*a^4*d*sqrt(1/(a^7*d^2))*e^(2*I*d*x + 2*I*c)*log(-12*((I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)
*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) + I)*e^(-I*d*x - I*c)/(a^3*d)) + 3*I*sqrt(2)*a^4*d*sqrt(1
/(a^7*d^2))*e^(2*I*d*x + 2*I*c)*log(-12*((-I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c)
+ 1))*sqrt(1/(a^7*d^2)) + I)*e^(-I*d*x - I*c)/(a^3*d)) - 2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-3*I*e^(
2*I*d*x + 2*I*c) - I))*e^(-2*I*d*x - 2*I*c)/(a^4*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Integral(sec(c + d*x)**5/(I*a*(tan(c + d*x) - I))**(7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^5/(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 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(7/2)),x)

[Out]

int(1/(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(7/2)), x)

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