\(\int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx\) [166]

Optimal result

Integrand size = 23, antiderivative size = 224 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}-\frac {2 a^3 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^4 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {6 a^5 \tan ^5(c+d x)}{d (a+a \sec (c+d x))^{5/2}}+\frac {34 a^6 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\frac {14 a^7 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}}+\frac {2 a^8 \tan ^{11}(c+d x)}{11 d (a+a \sec (c+d x))^{11/2}} \]

[Out]

2*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-2*a^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/3*a^
4*tan(d*x+c)^3/d/(a+a*sec(d*x+c))^(3/2)+6*a^5*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)+34/7*a^6*tan(d*x+c)^7/d/(a
+a*sec(d*x+c))^(7/2)+14/9*a^7*tan(d*x+c)^9/d/(a+a*sec(d*x+c))^(9/2)+2/11*a^8*tan(d*x+c)^11/d/(a+a*sec(d*x+c))^
(11/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3972, 472, 209} \[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^8 \tan ^{11}(c+d x)}{11 d (a \sec (c+d x)+a)^{11/2}}+\frac {14 a^7 \tan ^9(c+d x)}{9 d (a \sec (c+d x)+a)^{9/2}}+\frac {34 a^6 \tan ^7(c+d x)}{7 d (a \sec (c+d x)+a)^{7/2}}+\frac {6 a^5 \tan ^5(c+d x)}{d (a \sec (c+d x)+a)^{5/2}}+\frac {2 a^4 \tan ^3(c+d x)}{3 d (a \sec (c+d x)+a)^{3/2}}-\frac {2 a^3 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}} \]

[In]

Int[(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x]^4,x]

[Out]

(2*a^(5/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (2*a^3*Tan[c + d*x])/(d*Sqrt[a + a*Sec
[c + d*x]]) + (2*a^4*Tan[c + d*x]^3)/(3*d*(a + a*Sec[c + d*x])^(3/2)) + (6*a^5*Tan[c + d*x]^5)/(d*(a + a*Sec[c
+ d*x])^(5/2)) + (34*a^6*Tan[c + d*x]^7)/(7*d*(a + a*Sec[c + d*x])^(7/2)) + (14*a^7*Tan[c + d*x]^9)/(9*d*(a +
a*Sec[c + d*x])^(9/2)) + (2*a^8*Tan[c + d*x]^11)/(11*d*(a + a*Sec[c + d*x])^(11/2))

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 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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*} \text {integral}& = -\frac {\left (2 a^5\right ) \text {Subst}\left (\int \frac {x^4 \left (2+a x^2\right )^4}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d} \\ & = -\frac {\left (2 a^5\right ) \text {Subst}\left (\int \left (-\frac {1}{a^2}+\frac {x^2}{a}+15 x^4+17 a x^6+7 a^2 x^8+a^3 x^{10}+\frac {1}{a^2 \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 a^3 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^4 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {6 a^5 \tan ^5(c+d x)}{d (a+a \sec (c+d x))^{5/2}}+\frac {34 a^6 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\frac {14 a^7 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}}+\frac {2 a^8 \tan ^{11}(c+d x)}{11 d (a+a \sec (c+d x))^{11/2}}-\frac {\left (2 a^3\right ) \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 )}{d} \\ & = \frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}-\frac {2 a^3 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^4 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {6 a^5 \tan ^5(c+d x)}{d (a+a \sec (c+d x))^{5/2}}+\frac {34 a^6 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\frac {14 a^7 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}}+\frac {2 a^8 \tan ^{11}(c+d x)}{11 d (a+a \sec (c+d x))^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.81 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.67 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=\frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt {a (1+\sec (c+d x))} \left (5544 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {11}{2}}(c+d x)-1386 \sin \left (\frac {1}{2} (c+d x)\right )+1584 \sin \left (\frac {3}{2} (c+d x)\right )-1386 \sin \left (\frac {5}{2} (c+d x)\right )-143 \sin \left (\frac {7}{2} (c+d x)\right )-693 \sin \left (\frac {9}{2} (c+d x)\right )-26 \sin \left (\frac {11}{2} (c+d x)\right )\right )}{5544 d} \]

[In]

Integrate[(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x]^4,x]

[Out]

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^5*Sqrt[a*(1 + Sec[c + d*x])]*(5544*Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]
*Cos[c + d*x]^(11/2) - 1386*Sin[(c + d*x)/2] + 1584*Sin[(3*(c + d*x))/2] - 1386*Sin[(5*(c + d*x))/2] - 143*Sin
[(7*(c + d*x))/2] - 693*Sin[(9*(c + d*x))/2] - 26*Sin[(11*(c + d*x))/2]))/(5544*d)

Maple [A] (verified)

Time = 129.41 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (693 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+693 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-52 \sin \left (d x +c \right )-719 \tan \left (d x +c \right )-366 \sec \left (d x +c \right ) \tan \left (d x +c \right )+157 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}+224 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}+63 \sec \left (d x +c \right )^{4} \tan \left (d x +c \right )\right )}{693 d \left (\cos \left (d x +c \right )+1\right )}\)	\(236\)

[In]

int((a+a*sec(d*x+c))^(5/2)*tan(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

2/693/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(693*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)
/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+693*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(
sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-52*sin(d*x+c)-719*tan(d*x+c)-366*sec(d*x+c)*tan(
d*x+c)+157*tan(d*x+c)*sec(d*x+c)^2+224*tan(d*x+c)*sec(d*x+c)^3+63*sec(d*x+c)^4*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.90 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=\left [\frac {693 \, {\left (a^{2} \cos \left (d x + c\right )^{6} + a^{2} \cos \left (d x + c\right )^{5}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left (52 \, a^{2} \cos \left (d x + c\right )^{5} + 719 \, a^{2} \cos \left (d x + c\right )^{4} + 366 \, a^{2} \cos \left (d x + c\right )^{3} - 157 \, a^{2} \cos \left (d x + c\right )^{2} - 224 \, a^{2} \cos \left (d x + c\right ) - 63 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}}, -\frac {2 \, {\left (693 \, {\left (a^{2} \cos \left (d x + c\right )^{6} + a^{2} \cos \left (d x + c\right )^{5}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (52 \, a^{2} \cos \left (d x + c\right )^{5} + 719 \, a^{2} \cos \left (d x + c\right )^{4} + 366 \, a^{2} \cos \left (d x + c\right )^{3} - 157 \, a^{2} \cos \left (d x + c\right )^{2} - 224 \, a^{2} \cos \left (d x + c\right ) - 63 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{693 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}}\right ] \]

[In]

integrate((a+a*sec(d*x+c))^(5/2)*tan(d*x+c)^4,x, algorithm="fricas")

[Out]

[1/693*(693*(a^2*cos(d*x + c)^6 + a^2*cos(d*x + c)^5)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*co
s(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*(52*a^2*
cos(d*x + c)^5 + 719*a^2*cos(d*x + c)^4 + 366*a^2*cos(d*x + c)^3 - 157*a^2*cos(d*x + c)^2 - 224*a^2*cos(d*x +
c) - 63*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5), -2/6
93*(693*(a^2*cos(d*x + c)^6 + a^2*cos(d*x + c)^5)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d
*x + c)/(sqrt(a)*sin(d*x + c))) + (52*a^2*cos(d*x + c)^5 + 719*a^2*cos(d*x + c)^4 + 366*a^2*cos(d*x + c)^3 - 1
57*a^2*cos(d*x + c)^2 - 224*a^2*cos(d*x + c) - 63*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(
d*cos(d*x + c)^6 + d*cos(d*x + c)^5)]

Sympy [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**(5/2)*tan(d*x+c)**4,x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))^(5/2)*tan(d*x+c)^4,x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{4} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^(5/2)*tan(d*x+c)^4,x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

[In]

int(tan(c + d*x)^4*(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(tan(c + d*x)^4*(a + a/cos(c + d*x))^(5/2), x)