3.329 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=35 \[ -\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d} \]

[Out]

-2/5*I*a*cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(5/2)/d

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Rubi [A]  time = 0.06, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {3493} \[ -\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-2*I)/5)*a*Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(5/2))/d

Rule 3493

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

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx &=-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\\ \end {align*}

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Mathematica [B]  time = 0.57, size = 73, normalized size = 2.09 \[ \frac {2 a^3 \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (\sin (2 c+5 d x)-i \cos (2 c+5 d x))}{5 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*Cos[c + d*x]^3*((-I)*Cos[2*c + 5*d*x] + Sin[2*c + 5*d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(5*d*(Cos[d*x] +
I*Sin[d*x])^3)

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fricas [B]  time = 1.95, size = 73, normalized size = 2.09 \[ \frac {\sqrt {2} {\left (-i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*sqrt(2)*(-I*a^3*e^(6*I*d*x + 6*I*c) - 3*I*a^3*e^(4*I*d*x + 4*I*c) - 3*I*a^3*e^(2*I*d*x + 2*I*c) - I*a^3)*
sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/d

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \cos \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.52, size = 73, normalized size = 2.09 \[ -\frac {2 \left (2 i \left (\cos ^{2}\left (d x +c \right )\right )-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )-i\right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{3}}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/d*(2*I*cos(d*x+c)^2-2*cos(d*x+c)*sin(d*x+c)-I)*cos(d*x+c)^3*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2
)*a^3

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maxima [B]  time = 0.66, size = 454, normalized size = 12.97 \[ \frac {2 \, {\left (i \, a^{\frac {7}{2}} - \frac {6 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {20 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {6 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} {\left (-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac {7}{2}}}{d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}^{\frac {7}{2}} {\left (-\frac {10 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {50 i \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {25 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {100 i \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {100 i \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {25 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {50 i \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {20 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {10 i \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} + \frac {5 \, \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} - 5\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(I*a^(7/2) - 6*I*a^(7/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*I*a^(7/2)*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 - 20*I*a^(7/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*I*a^(7/2)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 6*
I*a^(7/2)*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + I*a^(7/2)*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*(-2*I*sin(d
*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(7/2)/(d*(sin(d*x + c)/(cos(d*x + c) + 1
) + 1)^(7/2)*(sin(d*x + c)/(cos(d*x + c) + 1) - 1)^(7/2)*(-10*I*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 - 50*I*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 25*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
- 100*I*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 100*I*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 25*sin(d*x + c)^8/(
cos(d*x + c) + 1)^8 - 50*I*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 20*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 10
*I*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 5*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 5))

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mupad [B]  time = 5.26, size = 112, normalized size = 3.20 \[ -\frac {a^3\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (-2\,\sin \left (c+d\,x\right )-3\,\sin \left (3\,c+3\,d\,x\right )-\sin \left (5\,c+5\,d\,x\right )+\cos \left (c+d\,x\right )\,4{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}+\cos \left (5\,c+5\,d\,x\right )\,1{}\mathrm {i}\right )}{20\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^3*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos(c + d*x)*4i - 2*sin
(c + d*x) + cos(3*c + 3*d*x)*3i + cos(5*c + 5*d*x)*1i - 3*sin(3*c + 3*d*x) - sin(5*c + 5*d*x)))/(20*d)

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

[Out]

Timed out

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