3.295 \(\int \sec ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=59 \[ \frac {2 i (a+i a \tan (c+d x))^{9/2}}{9 a^3 d}-\frac {4 i (a+i a \tan (c+d x))^{7/2}}{7 a^2 d} \]

[Out]

-4/7*I*(a+I*a*tan(d*x+c))^(7/2)/a^2/d+2/9*I*(a+I*a*tan(d*x+c))^(9/2)/a^3/d

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Rubi [A]  time = 0.07, antiderivative size = 59, 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} \[ \frac {2 i (a+i a \tan (c+d x))^{9/2}}{9 a^3 d}-\frac {4 i (a+i a \tan (c+d x))^{7/2}}{7 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-4*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^2*d) + (((2*I)/9)*(a + I*a*Tan[c + d*x])^(9/2))/(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 \sec ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac {i \operatorname {Subst}\left (\int (a-x) (a+x)^{5/2} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (2 a (a+x)^{5/2}-(a+x)^{7/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac {4 i (a+i a \tan (c+d x))^{7/2}}{7 a^2 d}+\frac {2 i (a+i a \tan (c+d x))^{9/2}}{9 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 81, normalized size = 1.37 \[ -\frac {2 a (7 \tan (c+d x)+11 i) \sec ^3(c+d x) (\cos (d x)-i \sin (d x)) \sqrt {a+i a \tan (c+d x)} (\cos (3 c+4 d x)+i \sin (3 c+4 d x))}{63 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*Sec[c + d*x]^3*(Cos[d*x] - I*Sin[d*x])*(Cos[3*c + 4*d*x] + I*Sin[3*c + 4*d*x])*(11*I + 7*Tan[c + d*x])*S
qrt[a + I*a*Tan[c + d*x]])/(63*d)

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fricas [B]  time = 0.63, size = 98, normalized size = 1.66 \[ \frac {\sqrt {2} {\left (-64 i \, a e^{\left (9 i \, d x + 9 i \, c\right )} - 288 i \, a e^{\left (7 i \, d x + 7 i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{63 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/63*sqrt(2)*(-64*I*a*e^(9*I*d*x + 9*I*c) - 288*I*a*e^(7*I*d*x + 7*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*
e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)*sec(d*x + c)^4, x)

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maple [B]  time = 1.16, size = 98, normalized size = 1.66 \[ -\frac {2 \left (16 i \left (\cos ^{4}\left (d x +c \right )\right )-16 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+2 i \left (\cos ^{2}\left (d x +c \right )\right )-10 \cos \left (d x +c \right ) \sin \left (d x +c \right )-7 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{63 d \cos \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

-2/63/d*(16*I*cos(d*x+c)^4-16*cos(d*x+c)^3*sin(d*x+c)+2*I*cos(d*x+c)^2-10*cos(d*x+c)*sin(d*x+c)-7*I)*(a*(I*sin
(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^4*a

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maxima [A]  time = 0.34, size = 40, normalized size = 0.68 \[ \frac {2 i \, {\left (7 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 18 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a\right )}}{63 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

2/63*I*(7*(I*a*tan(d*x + c) + a)^(9/2) - 18*(I*a*tan(d*x + c) + a)^(7/2)*a)/(a^3*d)

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mupad [B]  time = 6.10, size = 296, normalized size = 5.02 \[ -\frac {a\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{63\,d}-\frac {a\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{63\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}+\frac {a\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,160{}\mathrm {i}}{21\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {a\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,608{}\mathrm {i}}{63\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {a\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{9\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(3/2)/cos(c + d*x)^4,x)

[Out]

(a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*160i)/(21*d*(exp(c*2i + d*x*2i) +
1)^2) - (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*32i)/(63*d*(exp(c*2i + d*x
*2i) + 1)) - (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(63*d) - (a*(a -
(a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*608i)/(63*d*(exp(c*2i + d*x*2i) + 1)^3) +
(a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*32i)/(9*d*(exp(c*2i + d*x*2i) + 1
)^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \sec ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)*sec(c + d*x)**4, x)

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