3.389 \(\int (a+b \sec ^2(e+f x))^{3/2} \tan ^5(e+f x) \, dx\)

Optimal. Leaf size=135 \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{7/2}}{7 b^2 f}-\frac{(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}+\frac{a \sqrt{a+b \sec ^2(e+f x)}}{f} \]

[Out]

-((a^(3/2)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/f) + (a*Sqrt[a + b*Sec[e + f*x]^2])/f + (a + b*Sec[e +
f*x]^2)^(3/2)/(3*f) - ((a + 2*b)*(a + b*Sec[e + f*x]^2)^(5/2))/(5*b^2*f) + (a + b*Sec[e + f*x]^2)^(7/2)/(7*b^
2*f)

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Rubi [A]  time = 0.164087, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4139, 446, 88, 50, 63, 208} \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{7/2}}{7 b^2 f}-\frac{(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}+\frac{a \sqrt{a+b \sec ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)^(3/2)*Tan[e + f*x]^5,x]

[Out]

-((a^(3/2)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/f) + (a*Sqrt[a + b*Sec[e + f*x]^2])/f + (a + b*Sec[e +
f*x]^2)^(3/2)/(3*f) - ((a + 2*b)*(a + b*Sec[e + f*x]^2)^(5/2))/(5*b^2*f) + (a + b*Sec[e + f*x]^2)^(7/2)/(7*b^
2*f)

Rule 4139

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p)/x
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 50

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

Rule 63

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 208

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

Rubi steps

\begin{align*} \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2 \left (a+b x^2\right )^{3/2}}{x} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(-1+x)^2 (a+b x)^{3/2}}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{(-a-2 b) (a+b x)^{3/2}}{b}+\frac{(a+b x)^{3/2}}{x}+\frac{(a+b x)^{5/2}}{b}\right ) \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=-\frac{(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=\frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac{a \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=\frac{a \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=\frac{a \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sec ^2(e+f x)}\right )}{b f}\\ &=-\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{f}+\frac{a \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{7/2}}{7 b^2 f}\\ \end{align*}

Mathematica [F]  time = 3.20422, size = 0, normalized size = 0. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*Sec[e + f*x]^2)^(3/2)*Tan[e + f*x]^5,x]

[Out]

Integrate[(a + b*Sec[e + f*x]^2)^(3/2)*Tan[e + f*x]^5, x]

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Maple [B]  time = 0.5, size = 2606, normalized size = 19.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x)

[Out]

1/420/f/(a+b)^(9/2)/b^2*4^(1/2)*(-1+cos(f*x+e))^3*(-210*cos(f*x+e)^7*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*
x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2
)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^6*b^2-945*cos(f*x+e)^7*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)
*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^5*b^3-1575*cos(f*x+e)^7*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*co
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)
^(1/2)+b)/sin(f*x+e)^2)*a^4*b^4-1155*cos(f*x+e)^7*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+
e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)
+b)/sin(f*x+e)^2)*a^3*b^5-315*cos(f*x+e)^7*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(
1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin
(f*x+e)^2)*a^2*b^6+210*cos(f*x+e)^7*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f
*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)
^2)*a^6*b^2+945*cos(f*x+e)^7*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^
2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^5
*b^3-70*cos(f*x+e)^5*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^4-70*cos(f*x+e)^4*(a+b)^(7/2)*(
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^4+84*cos(f*x+e)^3*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^
2)^(1/2)*b^4+84*cos(f*x+e)^2*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^4-30*cos(f*x+e)*(a+b)^(
7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^4-30*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2
)*a*b^3+1575*cos(f*x+e)^7*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^
(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^4*b^
4+210*cos(f*x+e)^7*(a+b)^(7/2)*a^(3/2)*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a
*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*b^3+96*cos(f*x+e)^7*(a+b)^(7/2)*((b+a*cos(f
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b-196*cos(f*x+e)^7*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2
)*a^2*b^2-280*cos(f*x+e)^7*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^3+96*cos(f*x+e)^6*(a+b)
^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b-196*cos(f*x+e)^6*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+c
os(f*x+e))^2)^(1/2)*a^2*b^2-280*cos(f*x+e)^6*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^3-6*c
os(f*x+e)^5*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b+162*cos(f*x+e)^5*(a+b)^(7/2)*((b+a*c
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^2+98*cos(f*x+e)^5*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)
^(1/2)*a*b^3-6*cos(f*x+e)^4*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b+162*cos(f*x+e)^4*(a+
b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^2+98*cos(f*x+e)^4*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(
1+cos(f*x+e))^2)^(1/2)*a*b^3-48*cos(f*x+e)^3*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^2+1
155*cos(f*x+e)^7*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+
b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^3*b^5+315*cos
(f*x+e)^7*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2
)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^2*b^6+12*cos(f*x+e)^
7*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^4+12*cos(f*x+e)^6*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/
(1+cos(f*x+e))^2)^(1/2)*a^4-30*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^4+36*cos(f*x+e)^3*(a+
b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^3-48*cos(f*x+e)^2*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+
cos(f*x+e))^2)^(1/2)*a^2*b^2+36*cos(f*x+e)^2*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^3-30*
cos(f*x+e)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^3+210*cos(f*x+e)^7*(a+b)^(7/2)*a^(5/2)*
ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^
2)/(1+cos(f*x+e))^2)^(1/2))*b^2)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)/cos(f*x+e)^4/sin(f*x+e)^6/((b+a*cos(f
*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*tan(f*x + e)^5, x)

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Fricas [B]  time = 24.0816, size = 1289, normalized size = 9.55 \begin{align*} \left [\frac{105 \, a^{\frac{3}{2}} b^{2} \cos \left (f x + e\right )^{6} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \,{\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt{a} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) - 8 \,{\left (2 \,{\left (3 \, a^{3} + 21 \, a^{2} b - 70 \, a b^{2}\right )} \cos \left (f x + e\right )^{6} -{\left (3 \, a^{2} b - 84 \, a b^{2} + 35 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 15 \, b^{3} - 6 \,{\left (4 \, a b^{2} - 7 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{840 \, b^{2} f \cos \left (f x + e\right )^{6}}, \frac{105 \, \sqrt{-a} a b^{2} \arctan \left (\frac{{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \,{\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{6} - 4 \,{\left (2 \,{\left (3 \, a^{3} + 21 \, a^{2} b - 70 \, a b^{2}\right )} \cos \left (f x + e\right )^{6} -{\left (3 \, a^{2} b - 84 \, a b^{2} + 35 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 15 \, b^{3} - 6 \,{\left (4 \, a b^{2} - 7 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{420 \, b^{2} f \cos \left (f x + e\right )^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x, algorithm="fricas")

[Out]

[1/840*(105*a^(3/2)*b^2*cos(f*x + e)^6*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos
(f*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 - 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*co
s(f*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) - 8*(2*(3*a^3 + 21*a^2
*b - 70*a*b^2)*cos(f*x + e)^6 - (3*a^2*b - 84*a*b^2 + 35*b^3)*cos(f*x + e)^4 - 15*b^3 - 6*(4*a*b^2 - 7*b^3)*co
s(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b^2*f*cos(f*x + e)^6), 1/420*(105*sqrt(-a)*a*b^2*a
rctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e
)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2))*cos(f*x + e)^6 - 4*(2*(3*a^3 + 21*a^2*b - 70*a*b
^2)*cos(f*x + e)^6 - (3*a^2*b - 84*a*b^2 + 35*b^3)*cos(f*x + e)^4 - 15*b^3 - 6*(4*a*b^2 - 7*b^3)*cos(f*x + e)^
2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b^2*f*cos(f*x + e)^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+b*sec(f*x+e)**2)**(3/2)*tan(f*x+e)**5,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*tan(f*x + e)^5, x)