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

Optimal. Leaf size=59 \[ \frac {2 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d}-\frac {4 i (a+i a \tan (c+d x))^{9/2}}{9 a^2 d} \]

[Out]

-4/9*I*(a+I*a*tan(d*x+c))^(9/2)/a^2/d+2/11*I*(a+I*a*tan(d*x+c))^(11/2)/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} \[ \frac {2 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d}-\frac {4 i (a+i a \tan (c+d x))^{9/2}}{9 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-4*I)/9)*(a + I*a*Tan[c + d*x])^(9/2))/(a^2*d) + (((2*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(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))^{5/2} \, dx &=-\frac {i \operatorname {Subst}\left (\int (a-x) (a+x)^{7/2} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (2 a (a+x)^{7/2}-(a+x)^{9/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac {4 i (a+i a \tan (c+d x))^{9/2}}{9 a^2 d}+\frac {2 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 85, normalized size = 1.44 \[ -\frac {2 a^2 (9 \tan (c+d x)+13 i) \sec ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (\cos (4 c+6 d x)+i \sin (4 c+6 d x))}{99 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Sec[c + d*x]^4*(Cos[4*c + 6*d*x] + I*Sin[4*c + 6*d*x])*(13*I + 9*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*
x]])/(99*d*(Cos[d*x] + I*Sin[d*x])^2)

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fricas [B]  time = 0.46, size = 114, normalized size = 1.93 \[ \frac {\sqrt {2} {\left (-128 i \, a^{2} e^{\left (11 i \, d x + 11 i \, c\right )} - 704 i \, a^{2} e^{\left (9 i \, d x + 9 i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{99 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/99*sqrt(2)*(-128*I*a^2*e^(11*I*d*x + 11*I*c) - 704*I*a^2*e^(9*I*d*x + 9*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) +
5*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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.30, size = 117, normalized size = 1.98 \[ \frac {2 \left (-32 i \left (\cos ^{5}\left (d x +c \right )\right )+32 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-4 i \left (\cos ^{3}\left (d x +c \right )\right )+20 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+23 i \cos \left (d x +c \right )-9 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{99 d \cos \left (d x +c \right )^{5}} \]

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]

2/99/d*(-32*I*cos(d*x+c)^5+32*sin(d*x+c)*cos(d*x+c)^4-4*I*cos(d*x+c)^3+20*cos(d*x+c)^2*sin(d*x+c)+23*I*cos(d*x
+c)-9*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^5*a^2

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maxima [A]  time = 0.53, size = 40, normalized size = 0.68 \[ \frac {2 i \, {\left (9 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 22 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a\right )}}{99 \, a^{3} d} \]

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="maxima")

[Out]

2/99*I*(9*(I*a*tan(d*x + c) + a)^(11/2) - 22*(I*a*tan(d*x + c) + a)^(9/2)*a)/(a^3*d)

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mupad [B]  time = 6.42, size = 370, normalized size = 6.27 \[ -\frac {a^2\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{99\,d}-\frac {a^2\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{99\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}+\frac {a^2\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{33\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {a^2\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,2944{}\mathrm {i}}{99\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {a^2\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,2176{}\mathrm {i}}{99\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {a^2\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{11\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/(33*d*(exp(c*2i + d*x*2i)
+ 1)^2) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(99*d*(exp(c*2i +
d*x*2i) + 1)) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*128i)/(99*d) -
(a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*2944i)/(99*d*(exp(c*2i + d*x*2i)
+ 1)^3) + (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*2176i)/(99*d*(exp(c*2
i + d*x*2i) + 1)^4) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(11*d
*(exp(c*2i + d*x*2i) + 1)^5)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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