3.365 \(\int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=55 \[ \frac {2 i \sqrt {a+i a \tan (c+d x)}}{a^3 d}+\frac {4 i}{a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

4*I/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+2*I*(a+I*a*tan(d*x+c))^(1/2)/a^3/d

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Rubi [A]  time = 0.07, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3487, 43} \[ \frac {2 i \sqrt {a+i a \tan (c+d x)}}{a^3 d}+\frac {4 i}{a^2 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(4*I)/(a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + ((2*I)*Sqrt[a + I*a*Tan[c + d*x]])/(a^3*d)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3487

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

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 36, normalized size = 0.65 \[ \frac {-2 \tan (c+d x)+6 i}{a^2 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(6*I - 2*Tan[c + d*x])/(a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [A]  time = 0.53, size = 49, normalized size = 0.89 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/(I*a*tan(d*x + c) + a)^(5/2), x)

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maple [A]  time = 1.16, size = 65, normalized size = 1.18 \[ \frac {2 \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (2 \cos \left (d x +c \right ) \sin \left (d x +c \right )+i+2 i \left (\cos ^{2}\left (d x +c \right )\right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

2/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(2*cos(d*x+c)*sin(d*x+c)+I+2*I*cos(d*x+c)^2)/a^3

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maxima [A]  time = 0.45, size = 44, normalized size = 0.80 \[ \frac {2 i \, {\left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a}}{a^{2}} + \frac {2}{\sqrt {i \, a \tan \left (d x + c\right ) + a} a}\right )}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.27, size = 72, normalized size = 1.31 \[ \frac {2\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )+2{}\mathrm {i}\right )\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

(2*(cos(2*c + 2*d*x)*1i + sin(2*c + 2*d*x) + 2i)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c +
2*d*x) + 1))^(1/2))/(a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**4/(I*a*(tan(c + d*x) - I))**(5/2), x)

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