3.3.1 \(\int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\) [201]

Optimal. Leaf size=127 \[ -\frac {2 \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {2 \tan (c+d x)}{a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {2 \tan ^3(c+d x)}{3 a d (a+a \sec (c+d x))^{3/2}}+\frac {2 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}} \]

[Out]

-2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d+2*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)-2/3*t
an(d*x+c)^3/a/d/(a+a*sec(d*x+c))^(3/2)+2/5*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)

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Rubi [A]
time = 0.06, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3972, 308, 209} \begin {gather*} -\frac {2 \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac {2 \tan (c+d x)}{a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {2 \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}-\frac {2 \tan ^3(c+d x)}{3 a d (a \sec (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d) + (2*Tan[c + d*x])/(a^2*d*Sqrt[a + a*
Sec[c + d*x]]) - (2*Tan[c + d*x]^3)/(3*a*d*(a + a*Sec[c + d*x])^(3/2)) + (2*Tan[c + d*x]^5)/(5*d*(a + a*Sec[c
+ d*x])^(5/2))

Rule 209

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

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 3972

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[-2*(a^(m/2 +
n + 1/2)/d), Subst[Int[x^m*((2 + a*x^2)^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rubi steps

\begin {align*} \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac {(2 a) \text {Subst}\left (\int \frac {x^6}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac {(2 a) \text {Subst}\left (\int \left (\frac {1}{a^3}-\frac {x^2}{a^2}+\frac {x^4}{a}-\frac {1}{a^3 \left (1+a x^2\right )}\right ) \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {2 \tan (c+d x)}{a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {2 \tan ^3(c+d x)}{3 a d (a+a \sec (c+d x))^{3/2}}+\frac {2 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac {2 \text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^2 d}\\ &=-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {2 \tan (c+d x)}{a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {2 \tan ^3(c+d x)}{3 a d (a+a \sec (c+d x))^{3/2}}+\frac {2 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(447\) vs. \(2(127)=254\).
time = 6.07, size = 447, normalized size = 3.52 \begin {gather*} \frac {\sqrt {2} \cot ^8\left (\frac {1}{2} (c+d x)\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{9/2} \sqrt {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (-1+\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^3 \left (\frac {\sqrt {2} \text {ArcSin}\left (\frac {\sqrt {2} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {1-\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}}+\frac {8 \tan ^6\left (\frac {1}{2} (c+d x)\right )}{5 \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^3 \left (-1+\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^3}+\frac {4 \tan ^4\left (\frac {1}{2} (c+d x)\right )}{3 \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2 \left (-1+\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^2}+\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (-1+\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )}\right ) \tan ^7(c+d x)}{d (a (1+\sec (c+d x)))^{5/2} \left (1-\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^{5/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[2]*Cot[(c + d*x)/2]^8*((1 + Sec[c + d*x])^(-1))^(9/2)*Sqrt[1 + Tan[(c + d*x)/2]^2]*(-1 + (2*Tan[(c + d*x
)/2]^2)/(1 + Tan[(c + d*x)/2]^2))^3*((Sqrt[2]*ArcSin[(Sqrt[2]*Tan[(c + d*x)/2])/Sqrt[1 + Tan[(c + d*x)/2]^2]]*
Tan[(c + d*x)/2])/(Sqrt[1 + Tan[(c + d*x)/2]^2]*Sqrt[1 - (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]) + (
8*Tan[(c + d*x)/2]^6)/(5*(1 + Tan[(c + d*x)/2]^2)^3*(-1 + (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))^3)
+ (4*Tan[(c + d*x)/2]^4)/(3*(1 + Tan[(c + d*x)/2]^2)^2*(-1 + (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))^
2) + (2*Tan[(c + d*x)/2]^2)/((1 + Tan[(c + d*x)/2]^2)*(-1 + (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))))
*Tan[c + d*x]^7)/(d*(a*(1 + Sec[c + d*x]))^(5/2)*(1 - (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(301\) vs. \(2(111)=222\).
time = 0.17, size = 302, normalized size = 2.38

method	result	size
default	\(\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (15 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {2}+30 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {2}+15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )-184 \left (\cos ^{3}\left (d x +c \right )\right )+272 \left (\cos ^{2}\left (d x +c \right )\right )-112 \cos \left (d x +c \right )+24\right )}{60 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} a^{3}}\)	\(302\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(15*cos(d*x+c)^2*sin(d*x+c)*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)))^(5/2)*2^(1/2)+30*cos(d*x+c)*sin(d*x+c)
*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))
)^(5/2)*2^(1/2)+15*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)))^(5/2)*sin(d*x+c)-184*cos(d*x+c)^3+272*cos(d*x+c)^2-112*cos(d*x+c)+24)/sin(d*x+c)/c
os(d*x+c)^2/a^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(20*(9*sin(4*d*x + 4*c) + 16*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))
- 4*(45*cos(4*d*x + 4*c) + 80*cos(2*d*x + 2*c) + 23)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))
- 15*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d
*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d
*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c) + 1)) + 1) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*arctan2((cos(2*d*x + 2*c
)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c)
, cos(2*d*x + 2*c) + 1)) - 1) + 2*(a^3*d*cos(2*d*x + 2*c)^2 + a^3*d*sin(2*d*x + 2*c)^2 + 2*a^3*d*cos(2*d*x + 2
*c) + a^3*d)*integrate(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(((cos(2*d*x
+ 2*c)^4 + sin(2*d*x + 2*c)^4 + 2*cos(2*d*x + 2*c)^3 + (2*cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d
*x + 2*c)^2 + (cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2
*c))*cos(14*d*x + 14*c) + 6*(cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 +
cos(2*d*x + 2*c))*cos(12*d*x + 12*c) + 15*(cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2
*d*x + 2*c)^2 + cos(2*d*x + 2*c))*cos(10*d*x + 10*c) + 20*(cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2
*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*cos(8*d*x + 8*c) + 15*(cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*
sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*cos(6*d*x + 6*c) + 6*(cos(2*d*x + 2*c)^3 + cos(2
*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*cos(4*d*x + 4*c) + cos(2*d*x + 2*c)^
2 + (sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*sin(14*d*x + 14*c) +
6*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*sin(12*d*x + 12*c) +
15*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*sin(10*d*x + 10*c) +
20*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 15
*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 6*(s
in(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*sin(4*d*x + 4*c))*cos(7/2*
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - ((sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)*sin(2*d*x + 2*c))*cos(14*d*x + 14*c) + 6*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)*sin(2*d*x + 2*c))*cos(12*d*x + 12*c) + 15*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)*sin(2*d*x + 2*c))*cos(10*d*x + 10*c) + 20*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)*sin(2*d*x + 2*c))*cos(8*d*x + 8*c) + 15*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) +
1)*sin(2*d*x + 2*c))*cos(6*d*x + 6*c) + 6*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*
sin(2*d*x + 2*c))*cos(4*d*x + 4*c) - (cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c)^2 + cos(2*d*x + 2*c))*sin(14*d*x + 14*c) - 6*(cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 +
2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*sin(12*d*x + 12*c) - 15*(cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2
*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*sin(10*d*x + 10*c) - 20*(cos(2*d*x + 2*c)^3 + cos(2*d
*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*sin(8*d*x + 8*c) - 15*(cos(2*d*x + 2*c
)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*sin(6*d*x + 6*c) - 6*(cos
(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*sin(4*d*x + 4
*c))*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
+ 1)) - (((sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*cos(14*d*x + 1
4*c) + 6*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*cos(12*d*x + 12
*c) + 15*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*cos(10*d*x + 10
*c) + 20*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*cos(8*d*x + 8*c
) + 15*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*cos(6*d*x + 6*c)
+ 6*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*cos(4*d*x + 4*c) - (
cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*sin(14*d*x
+ 14*c) - 6*(cos(2*d*x + 2*c)^3 + cos(2*d*x + ...

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Fricas [A]
time = 2.67, size = 323, normalized size = 2.54 \begin {gather*} \left [-\frac {15 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left (23 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}}, \frac {2 \, {\left (15 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (23 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/15*(15*(cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c
) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 2*(23*cos(d*x + c)^
2 - 11*cos(d*x + c) + 3)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + a^3*d*c
os(d*x + c)^2), 2/15*(15*(cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (23*cos(d*x + c)^2 - 11*cos(d*x + c) + 3)*sqrt((a*cos(d*x + c) + a)
/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{6}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**6/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral(tan(c + d*x)**6/(a*(sec(c + d*x) + 1))**(5/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (111) = 222\).
time = 4.18, size = 257, normalized size = 2.02 \begin {gather*} \frac {15 \, \sqrt {-a} {\left (\frac {\log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} + \frac {2 \, {\left ({\left (\frac {37 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {40 \, \sqrt {2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {15 \, \sqrt {2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/15*(15*sqrt(-a)*(log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt
(2) + 3)))/(a^3*sgn(cos(d*x + c))) - log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 +
a))^2 + a*(2*sqrt(2) - 3)))/(a^3*sgn(cos(d*x + c)))) + 2*((37*sqrt(2)*tan(1/2*d*x + 1/2*c)^2/sgn(cos(d*x + c)
) - 40*sqrt(2)/sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 + 15*sqrt(2)/sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)/
((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^6}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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