3.4.12 \(\int \cos ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [312]

Optimal. Leaf size=210 \[ -\frac {35 i a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} d}+\frac {35 i a^3}{128 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}-\frac {35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-35/256*I*a^(5/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)+35/128*I*a^3/d/(a+I*a*tan(d*
x+c))^(1/2)-1/6*I*a^6/d/(a+I*a*tan(d*x+c))^(1/2)/(a-I*a*tan(d*x+c))^3-7/48*I*a^5/d/(a+I*a*tan(d*x+c))^(1/2)/(a
-I*a*tan(d*x+c))^2-35/192*I*a^4/d/(a+I*a*tan(d*x+c))^(1/2)/(a-I*a*tan(d*x+c))

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Rubi [A]
time = 0.10, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3568, 44, 53, 65, 212} \begin {gather*} -\frac {35 i a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} d}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}-\frac {35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}+\frac {35 i a^3}{128 d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((-35*I)/128)*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + (((35*I)/128)*a^3)
/(d*Sqrt[a + I*a*Tan[c + d*x]]) - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]]) - (((7*I
)/48)*a^5)/(d*(a - I*a*Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]]) - (((35*I)/192)*a^4)/(d*(a - I*a*Tan[c + d*
x])*Sqrt[a + I*a*Tan[c + d*x]])

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 3568

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 \cos ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac {\left (i a^7\right ) \text {Subst}\left (\int \frac {1}{(a-x)^4 (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}}-\frac {\left (7 i a^6\right ) \text {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{12 d}\\ &=-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}-\frac {\left (35 i a^5\right ) \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{96 d}\\ &=-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}-\frac {35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}-\frac {\left (35 i a^4\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac {35 i a^3}{128 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}-\frac {35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}-\frac {\left (35 i a^3\right ) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac {35 i a^3}{128 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}-\frac {35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}-\frac {\left (35 i a^3\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{128 d}\\ &=-\frac {35 i a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} d}+\frac {35 i a^3}{128 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}-\frac {35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.01, size = 142, normalized size = 0.68 \begin {gather*} -\frac {i a^2 e^{-2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \left (\sqrt {1+e^{2 i (c+d x)}} \left (-48+87 e^{2 i (c+d x)}+38 e^{4 i (c+d x)}+8 e^{6 i (c+d x)}\right )+105 e^{i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{768 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-1/768*I)*a^2*Sqrt[1 + E^((2*I)*(c + d*x))]*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-48 + 87*E^((2*I)*(c + d*x)) + 3
8*E^((4*I)*(c + d*x)) + 8*E^((6*I)*(c + d*x))) + 105*E^(I*(c + d*x))*ArcSinh[E^(I*(c + d*x))])*Sqrt[a + I*a*Ta
n[c + d*x]])/(d*E^((2*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. 1087 vs. \(2 (170 ) = 340\).
time = 1.07, size = 1088, normalized size = 5.18

method	result	size
default	\(\text {Expression too large to display}\)	\(1088\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/24576/d*(-105*I*cos(d*x+c)^5*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)))^(11/2)*sin(d*x+c)-525*I*cos(d*x+c)^4*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)))^(11/2)*sin(d*x+c)-1050*
I*cos(d*x+c)^3*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)))^(11/2)*sin(d*x+c)-1050*I*cos(d*x+c)^2*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)))^(11/2)*sin(d*x+c)-525*I*cos(d*x+c)*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)))^(11/2)*sin(d*x+c)-3072*cos(d*x+c)^9*sin(d*x+c)-5120*I*cos(d*x+c)^10+896*I*cos(d*x+c)^8+2240*I*cos(d*x+c)^
7-6720*I*cos(d*x+c)^6-105*2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(11/2)*sin(d*x+c)-16384*sin(d*x+c)*cos(d*x+c)^11+512*I*cos(d*x+c)^9-105*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)))^(11/2)*sin(d*x+
c)+3584*sin(d*x+c)*cos(d*x+c)^8+8192*cos(d*x+c)^10*sin(d*x+c)-105*cos(d*x+c)^5*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(11/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-525*cos(d*x+c)^4*sin(d*x+c)
*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-1050*c
os(d*x+c)^3*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*2^(1/2))-1050*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctan(1/2*(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-525*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*
arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-4480*sin(d*x+c)*cos(d*x+c)^7+6720*sin(d*x+c)*cos(d*x+
c)^6+16384*I*cos(d*x+c)^12-8192*I*cos(d*x+c)^11)*(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)^5*a^2

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Maxima [A]
time = 0.50, size = 194, normalized size = 0.92 \begin {gather*} \frac {i \, {\left (105 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} - 560 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} + 924 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} - 384 \, a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 8 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3}}\right )}}{1536 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/1536*I*(105*sqrt(2)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*
tan(d*x + c) + a))) + 4*(105*(I*a*tan(d*x + c) + a)^3*a^4 - 560*(I*a*tan(d*x + c) + a)^2*a^5 + 924*(I*a*tan(d*
x + c) + a)*a^6 - 384*a^7)/((I*a*tan(d*x + c) + a)^(7/2) - 6*(I*a*tan(d*x + c) + a)^(5/2)*a + 12*(I*a*tan(d*x
+ c) + a)^(3/2)*a^2 - 8*sqrt(I*a*tan(d*x + c) + a)*a^3))/(a*d)

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Fricas [A]
time = 0.37, size = 309, normalized size = 1.47 \begin {gather*} -\frac {{\left (105 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 105 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - \sqrt {2} {\left (-8 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 46 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 125 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 39 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 48 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{768 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/768*(105*sqrt(1/2)*sqrt(-a^5/d^2)*d*e^(I*d*x + I*c)*log(4*(a^3*e^(I*d*x + I*c) - sqrt(2)*sqrt(1/2)*sqrt(-a^
5/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 105*sqrt(1/2
)*sqrt(-a^5/d^2)*d*e^(I*d*x + I*c)*log(4*(a^3*e^(I*d*x + I*c) - sqrt(2)*sqrt(1/2)*sqrt(-a^5/d^2)*(-I*d*e^(2*I*
d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - sqrt(2)*(-8*I*a^2*e^(8*I*d*x +
8*I*c) - 46*I*a^2*e^(6*I*d*x + 6*I*c) - 125*I*a^2*e^(4*I*d*x + 4*I*c) - 39*I*a^2*e^(2*I*d*x + 2*I*c) + 48*I*a^
2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - 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)**6*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [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)^6*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\cos \left (c+d\,x\right )}^6\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(5/2), x)

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