3.4.15 \(\int \cos (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [315]

Optimal. Leaf size=65 \[ -\frac {8 i a^2 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{d} \]

[Out]

-8*I*a^2*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d+2*I*a*cos(d*x+c)*(a+I*a*tan(d*x+c))^(3/2)/d

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Rubi [A]
time = 0.07, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, 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 {2 i a \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}-\frac {8 i a^2 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*I)*a^2*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d + ((2*I)*a*Cos[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2))/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))^{5/2} \, dx &=\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}+(4 a) \int \cos (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=-\frac {8 i a^2 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}\\ \end {align*}

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Mathematica [A]
time = 0.35, size = 46, normalized size = 0.71 \begin {gather*} -\frac {2 i a^2 (3 \cos (c+d x)-i \sin (c+d x)) \sqrt {a+i a \tan (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.90, size = 53, normalized size = 0.82

method	result	size
default	\(-\frac {2 \left (3 i \cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{d}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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. 331 vs. \(2 (53) = 106\).
time = 0.61, size = 331, normalized size = 5.09 \begin {gather*} \frac {2 \, {\left (-3 i \, a^{\frac {5}{2}} - \frac {2 \, a^{\frac {5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {9 i \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {9 i \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {2 \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 i \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\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 {5}{2}}}{d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}^{\frac {5}{2}} {\left (\frac {4 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 i \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 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))^(5/2),x, algorithm="maxima")

[Out]

2*(-3*I*a^(5/2) - 2*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) + 9*I*a^(5/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
+ 4*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 9*I*a^(5/2)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2*a^(5/2)*
sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*I*a^(5/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*(-2*I*sin(d*x + c)/(cos
(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(5/2)/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)
*(sin(d*x + c)/(cos(d*x + c) + 1) - 1)^(5/2)*(4*I*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 - 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*I*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + sin(d*x + c)^6
/(cos(d*x + c) + 1)^6 + 1))

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Fricas [A]
time = 0.42, size = 45, normalized size = 0.69 \begin {gather*} -\frac {2 \, \sqrt {2} {\left (i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{d} \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))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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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))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.39, size = 64, normalized size = 0.98 \begin {gather*} -\frac {2\,a^2\,\left (\sin \left (c+d\,x\right )+\cos \left (c+d\,x\right )\,3{}\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}}}{d} \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)^(5/2),x)

[Out]

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

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