3.153 \(\int (a+a \sec (c+d x))^{3/2} \tan ^6(c+d x) \, dx\)

Optimal. Leaf size=258 \[ \frac{2 a^8 \tan ^{13}(c+d x)}{13 d (a \sec (c+d x)+a)^{13/2}}+\frac{14 a^7 \tan ^{11}(c+d x)}{11 d (a \sec (c+d x)+a)^{11/2}}+\frac{34 a^6 \tan ^9(c+d x)}{9 d (a \sec (c+d x)+a)^{9/2}}+\frac{30 a^5 \tan ^7(c+d x)}{7 d (a \sec (c+d x)+a)^{7/2}}+\frac{2 a^4 \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}-\frac{2 a^3 \tan ^3(c+d x)}{3 d (a \sec (c+d x)+a)^{3/2}}-\frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a^2 \tan (c+d x)}{d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(-2*a^(3/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*Tan[c + d*x])/(d*Sqrt[a + a*Se
c[c + d*x]]) - (2*a^3*Tan[c + d*x]^3)/(3*d*(a + a*Sec[c + d*x])^(3/2)) + (2*a^4*Tan[c + d*x]^5)/(5*d*(a + a*Se
c[c + d*x])^(5/2)) + (30*a^5*Tan[c + d*x]^7)/(7*d*(a + a*Sec[c + d*x])^(7/2)) + (34*a^6*Tan[c + d*x]^9)/(9*d*(
a + a*Sec[c + d*x])^(9/2)) + (14*a^7*Tan[c + d*x]^11)/(11*d*(a + a*Sec[c + d*x])^(11/2)) + (2*a^8*Tan[c + d*x]
^13)/(13*d*(a + a*Sec[c + d*x])^(13/2))

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Rubi [A]  time = 0.121799, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3887, 461, 203} \[ \frac{2 a^8 \tan ^{13}(c+d x)}{13 d (a \sec (c+d x)+a)^{13/2}}+\frac{14 a^7 \tan ^{11}(c+d x)}{11 d (a \sec (c+d x)+a)^{11/2}}+\frac{34 a^6 \tan ^9(c+d x)}{9 d (a \sec (c+d x)+a)^{9/2}}+\frac{30 a^5 \tan ^7(c+d x)}{7 d (a \sec (c+d x)+a)^{7/2}}+\frac{2 a^4 \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}-\frac{2 a^3 \tan ^3(c+d x)}{3 d (a \sec (c+d x)+a)^{3/2}}-\frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a^2 \tan (c+d x)}{d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x]^6,x]

[Out]

(-2*a^(3/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*Tan[c + d*x])/(d*Sqrt[a + a*Se
c[c + d*x]]) - (2*a^3*Tan[c + d*x]^3)/(3*d*(a + a*Sec[c + d*x])^(3/2)) + (2*a^4*Tan[c + d*x]^5)/(5*d*(a + a*Se
c[c + d*x])^(5/2)) + (30*a^5*Tan[c + d*x]^7)/(7*d*(a + a*Sec[c + d*x])^(7/2)) + (34*a^6*Tan[c + d*x]^9)/(9*d*(
a + a*Sec[c + d*x])^(9/2)) + (14*a^7*Tan[c + d*x]^11)/(11*d*(a + a*Sec[c + d*x])^(11/2)) + (2*a^8*Tan[c + d*x]
^13)/(13*d*(a + a*Sec[c + d*x])^(13/2))

Rule 3887

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]

Rule 461

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 203

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

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^{3/2} \tan ^6(c+d x) \, dx &=-\frac{\left (2 a^5\right ) \operatorname{Subst}\left (\int \frac{x^6 \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 ) \operatorname{Subst}\left (\int \left (\frac{1}{a^3}-\frac{x^2}{a^2}+\frac{x^4}{a}+15 x^6+17 a x^8+7 a^2 x^{10}+a^3 x^{12}-\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 a^2 \tan (c+d x)}{d \sqrt{a+a \sec (c+d x)}}-\frac{2 a^3 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac{2 a^4 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac{30 a^5 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\frac{34 a^6 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}}+\frac{14 a^7 \tan ^{11}(c+d x)}{11 d (a+a \sec (c+d x))^{11/2}}+\frac{2 a^8 \tan ^{13}(c+d x)}{13 d (a+a \sec (c+d x))^{13/2}}+\frac{\left (2 a^2\right ) \operatorname{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^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}+\frac{2 a^2 \tan (c+d x)}{d \sqrt{a+a \sec (c+d x)}}-\frac{2 a^3 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac{2 a^4 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac{30 a^5 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\frac{34 a^6 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}}+\frac{14 a^7 \tan ^{11}(c+d x)}{11 d (a+a \sec (c+d x))^{11/2}}+\frac{2 a^8 \tan ^{13}(c+d x)}{13 d (a+a \sec (c+d x))^{13/2}}\\ \end{align*}

Mathematica [A]  time = 8.41438, size = 147, normalized size = 0.57 \[ \frac{a \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^6(c+d x) \sqrt{a (\sec (c+d x)+1)} \left (164736 \sin \left (\frac{1}{2} (c+d x)\right )+81081 \sin \left (\frac{3}{2} (c+d x)\right )+134849 \sin \left (\frac{5}{2} (c+d x)\right )+98176 \sin \left (\frac{9}{2} (c+d x)\right )+45045 \sin \left (\frac{11}{2} (c+d x)\right )+32429 \sin \left (\frac{13}{2} (c+d x)\right )-1441440 \sqrt{2} \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^{\frac{13}{2}}(c+d x)\right )}{1441440 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x]^6,x]

[Out]

(a*Sec[(c + d*x)/2]*Sec[c + d*x]^6*Sqrt[a*(1 + Sec[c + d*x])]*(-1441440*Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2
]]*Cos[c + d*x]^(13/2) + 164736*Sin[(c + d*x)/2] + 81081*Sin[(3*(c + d*x))/2] + 134849*Sin[(5*(c + d*x))/2] +
98176*Sin[(9*(c + d*x))/2] + 45045*Sin[(11*(c + d*x))/2] + 32429*Sin[(13*(c + d*x))/2]))/(1441440*d)

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Maple [B]  time = 0.259, size = 656, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^6,x)

[Out]

1/2882880/d*a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(45045*2^(1/2)*sin(d*x+c)*cos(d*x+c)^6*arctanh(1/2*2^(1/2)*(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)+270270*2^(1/2
)*sin(d*x+c)*cos(d*x+c)^5*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(13/2)+675675*2^(1/2)*sin(d*x+c)*cos(d*x+c)^4*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)+900900*2^(1/2)*sin(d*x+c)*cos
(d*x+c)^3*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(13/2)+675675*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)+270270*2^(1/2)*sin(d*x+c)*cos(d*x+c)*arctanh(
1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)+
45045*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(13/2)*sin(d*x+c)-4150912*cos(d*x+c)^7-807424*cos(d*x+c)^6+5563904*cos(d*x+c)^5+2781952*cos(d*x
+c)^4-2585600*cos(d*x+c)^3-1809920*cos(d*x+c)^2+564480*cos(d*x+c)+443520)/cos(d*x+c)^6/sin(d*x+c)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^6,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.98966, size = 1156, normalized size = 4.48 \begin{align*} \left [\frac{45045 \,{\left (a \cos \left (d x + c\right )^{7} + a \cos \left (d x + c\right )^{6}\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 (32429 \, a \cos \left (d x + c\right )^{6} + 38737 \, a \cos \left (d x + c\right )^{5} - 4731 \, a \cos \left (d x + c\right )^{4} - 26465 \, a \cos \left (d x + c\right )^{3} - 6265 \, a \cos \left (d x + c\right )^{2} + 7875 \, a \cos \left (d x + c\right ) + 3465 \, a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \,{\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}}, \frac{2 \,{\left (45045 \,{\left (a \cos \left (d x + c\right )^{7} + a \cos \left (d x + c\right )^{6}\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 (32429 \, a \cos \left (d x + c\right )^{6} + 38737 \, a \cos \left (d x + c\right )^{5} - 4731 \, a \cos \left (d x + c\right )^{4} - 26465 \, a \cos \left (d x + c\right )^{3} - 6265 \, a \cos \left (d x + c\right )^{2} + 7875 \, a \cos \left (d x + c\right ) + 3465 \, a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{45045 \,{\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^6,x, algorithm="fricas")

[Out]

[1/45045*(45045*(a*cos(d*x + c)^7 + a*cos(d*x + c)^6)*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*(32429*a
*cos(d*x + c)^6 + 38737*a*cos(d*x + c)^5 - 4731*a*cos(d*x + c)^4 - 26465*a*cos(d*x + c)^3 - 6265*a*cos(d*x + c
)^2 + 7875*a*cos(d*x + c) + 3465*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^7 +
d*cos(d*x + c)^6), 2/45045*(45045*(a*cos(d*x + c)^7 + a*cos(d*x + c)^6)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (32429*a*cos(d*x + c)^6 + 38737*a*cos(d*x + c)^5 - 473
1*a*cos(d*x + c)^4 - 26465*a*cos(d*x + c)^3 - 6265*a*cos(d*x + c)^2 + 7875*a*cos(d*x + c) + 3465*a)*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^7 + d*cos(d*x + c)^6)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(3/2)*tan(d*x+c)**6,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^6,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError