3.4.5 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [305]

Optimal. Leaf size=192 \[ \frac {7 i a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16 \sqrt {2} d}+\frac {7 i a^2 \cos (c+d x)}{24 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16 d}-\frac {7 i a \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{30 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d} \]

[Out]

7/32*I*a^(3/2)*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d*2^(1/2)+7/24*I*a^2*cos(d*x+c
)/d/(a+I*a*tan(d*x+c))^(1/2)-7/16*I*a*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-7/30*I*a*cos(d*x+c)^3*(a+I*a*tan(d
*x+c))^(1/2)/d-1/5*I*cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2)/d

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Rubi [A]
time = 0.21, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3578, 3583, 3571, 3570, 212} \begin {gather*} \frac {7 i a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16 \sqrt {2} d}+\frac {7 i a^2 \cos (c+d x)}{24 d \sqrt {a+i a \tan (c+d x)}}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}-\frac {7 i a \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{30 d}-\frac {7 i a \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((7*I)/16)*a^(3/2)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) + (((7*I
)/24)*a^2*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((7*I)/16)*a*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]]
)/d - (((7*I)/30)*a*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - ((I/5)*Cos[c + d*x]^5*(a + I*a*Tan[c + d*x]
)^(3/2))/d

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3570

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2*(a/(b*f)), Subst[
Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^
2, 0]

Rule 3571

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/(a*f*m)), x] + Dist[a/(2*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(a + b*Tan
[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && EqQ[m/2 + n, 0] && GtQ[n, 0]

Rule 3578

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

Rule 3583

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

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {1}{10} (7 a) \int \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {7 i a \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{30 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {1}{12} \left (7 a^2\right ) \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {7 i a^2 \cos (c+d x)}{24 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{30 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {1}{16} (7 a) \int \cos (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {7 i a^2 \cos (c+d x)}{24 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16 d}-\frac {7 i a \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{30 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {1}{32} \left (7 a^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {7 i a^2 \cos (c+d x)}{24 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16 d}-\frac {7 i a \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{30 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {\left (7 i a^2\right ) \text {Subst}\left (\int \frac {1}{2-a x^2} \, dx,x,\frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}}\right )}{16 d}\\ &=\frac {7 i a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16 \sqrt {2} d}+\frac {7 i a^2 \cos (c+d x)}{24 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16 d}-\frac {7 i a \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{30 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\\ \end {align*}

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Mathematica [A]
time = 1.63, size = 160, normalized size = 0.83 \begin {gather*} -\frac {i a e^{-3 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-15+101 e^{2 i (c+d x)}+148 e^{4 i (c+d x)}+38 e^{6 i (c+d x)}+6 e^{8 i (c+d x)}-105 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{240 \sqrt {2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((-1/240*I)*a*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-15 + 101*E^((2*I)*(c + d*x)) + 148*E^(
(4*I)*(c + d*x)) + 38*E^((6*I)*(c + d*x)) + 6*E^((8*I)*(c + d*x)) - 105*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*
(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/(Sqrt[2]*d*E^((3*I)*(c + d*x)))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 913 vs. \(2 (155 ) = 310\).
time = 0.82, size = 914, normalized size = 4.76

method	result	size
default	\(\text {Expression too large to display}\)	\(914\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/7680/d*(420*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x
+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)-1536*I*cos(d*x+c)^9-105*cos(d*x+c)^4*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(9/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*2^(1/2)+448*I*cos(d*x+
c)^7-420*cos(d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(
1/2))*sin(d*x+c)*2^(1/2)+420*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)*2^(1/2)-630*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x
+c)))^(9/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*2^(1/2)-3360*I*cos(d*x+c)^5-42
0*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin
(d*x+c)*2^(1/2)-105*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*2^(1/2))*sin(d*x+c)+105*I*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^
(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*sin(d*x+c)+630*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arctanh(1/2*
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)-3072*cos(d
*x+c)^9*sin(d*x+c)+256*I*cos(d*x+c)^8+1536*sin(d*x+c)*cos(d*x+c)^8+1120*I*cos(d*x+c)^6-1792*sin(d*x+c)*cos(d*x
+c)^7+3072*I*cos(d*x+c)^10+2240*sin(d*x+c)*cos(d*x+c)^6+105*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arctanh(1/2
*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)-3360*sin(
d*x+c)*cos(d*x+c)^5)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^4*a

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Maxima [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)^5*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 0.40, size = 274, normalized size = 1.43 \begin {gather*} \frac {{\left (105 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {7 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + i \, a^{2}\right )} e^{\left (-i \, d x - i \, c\right )}}{8 \, d}\right ) - 105 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {7 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - i \, a^{2}\right )} e^{\left (-i \, d x - i \, c\right )}}{8 \, d}\right ) + \sqrt {2} {\left (-6 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 38 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 148 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 101 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{480 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/480*(105*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(2*I*d*x + 2*I*c)*log(7/8*(sqrt(2)*sqrt(1/2)*(d*e^(2*I*d*x + 2*I*c) +
d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) + I*a^2)*e^(-I*d*x - I*c)/d) - 105*sqrt(1/2)*sqrt(-a^3/d^2
)*d*e^(2*I*d*x + 2*I*c)*log(-7/8*(sqrt(2)*sqrt(1/2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*
d*x + 2*I*c) + 1)) - I*a^2)*e^(-I*d*x - I*c)/d) + sqrt(2)*(-6*I*a*e^(8*I*d*x + 8*I*c) - 38*I*a*e^(6*I*d*x + 6*
I*c) - 148*I*a*e^(4*I*d*x + 4*I*c) - 101*I*a*e^(2*I*d*x + 2*I*c) + 15*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*
e^(-2*I*d*x - 2*I*c)/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)**5*(a+I*a*tan(d*x+c))**(3/2),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)^5*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)*cos(d*x + c)^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(3/2), x)

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