3.4.27 \(\int \cos (c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) [327]

Optimal. Leaf size=104 \[ -\frac {64 i a^3 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {16 i a^2 \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^{5/2}}{3 d} \]

[Out]

-64/3*I*a^3*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d+16/3*I*a^2*cos(d*x+c)*(a+I*a*tan(d*x+c))^(3/2)/d+2/3*I*a*cos
(d*x+c)*(a+I*a*tan(d*x+c))^(5/2)/d

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Rubi [A]
time = 0.12, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3575, 3574} \begin {gather*} -\frac {64 i a^3 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {16 i a^2 \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^{5/2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-64*I)/3)*a^3*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d + (((16*I)/3)*a^2*Cos[c + d*x]*(a + I*a*Tan[c + d*
x])^(3/2))/d + (((2*I)/3)*a*Cos[c + d*x]*(a + I*a*Tan[c + d*x])^(5/2))/d

Rule 3574

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]

Rule 3575

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

Rubi steps

\begin {align*} \int \cos (c+d x) (a+i a \tan (c+d x))^{7/2} \, dx &=\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^{5/2}}{3 d}+\frac {1}{3} (8 a) \int \cos (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\\ &=\frac {16 i a^2 \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^{5/2}}{3 d}+\frac {1}{3} \left (32 a^2\right ) \int \cos (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=-\frac {64 i a^3 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {16 i a^2 \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^{5/2}}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.50, size = 59, normalized size = 0.57 \begin {gather*} -\frac {2 i a^3 \sec (c+d x) (12+11 \cos (2 (c+d x))-5 i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-2*I)/3)*a^3*Sec[c + d*x]*(12 + 11*Cos[2*(c + d*x)] - (5*I)*Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/d

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Maple [A]
time = 0.96, size = 73, normalized size = 0.70

method	result	size
default	\(-\frac {2 \left (22 i \left (\cos ^{2}\left (d x +c \right )\right )+10 \sin \left (d x +c \right ) \cos \left (d x +c \right )+i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{3}}{3 d \cos \left (d x +c \right )}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d*(22*I*cos(d*x+c)^2+10*sin(d*x+c)*cos(d*x+c)+I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c
)*a^3

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (80) = 160\).
time = 0.58, size = 418, normalized size = 4.02 \begin {gather*} \frac {2 \, {\left (23 i \, a^{\frac {7}{2}} + \frac {20 \, a^{\frac {7}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {88 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {60 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {130 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {60 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {88 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {20 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {23 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\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}}}{-3 \, 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 {6 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {14 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {14 i \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 i \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {14 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {6 i \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {\sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(23*I*a^(7/2) + 20*a^(7/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 88*I*a^(7/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^
2 - 60*a^(7/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 130*I*a^(7/2)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 60*a^
(7/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 88*I*a^(7/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 20*a^(7/2)*sin(
d*x + c)^7/(cos(d*x + c) + 1)^7 + 23*I*a^(7/2)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*(-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)*(s
in(d*x + c)/(cos(d*x + c) + 1) - 1)^(7/2)*(-18*I*sin(d*x + c)/(cos(d*x + c) + 1) + 42*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 + 42*I*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 42*I*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 42*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 - 18*I*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 3*sin(d*x + c)^8/(cos(d*x + c) + 1)^
8 - 3))

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Fricas [A]
time = 0.37, size = 71, normalized size = 0.68 \begin {gather*} -\frac {4 \, \sqrt {2} {\left (3 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 12 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*sqrt(2)*(3*I*a^3*e^(4*I*d*x + 4*I*c) + 12*I*a^3*e^(2*I*d*x + 2*I*c) + 8*I*a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c
) + 1))/(d*e^(2*I*d*x + 2*I*c) + d)

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

[Out]

Timed out

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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(cos(d*x+c)*(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), x)

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Mupad [B]
time = 4.72, size = 102, normalized size = 0.98 \begin {gather*} -\frac {2\,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 (5\,\sin \left (c+d\,x\right )+5\,\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,35{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,11{}\mathrm {i}\right )}{3\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*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)*35i + 5*
sin(c + d*x) + cos(3*c + 3*d*x)*11i + 5*sin(3*c + 3*d*x)))/(3*d*(cos(2*c + 2*d*x) + 1))

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