3.339 \(\int \sec ^7(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=195 \[ \frac{a^2 (5 a+6 b) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{16 f (a+b)^{3/2}}+\frac{\tan (e+f x) \sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 f (a+b)}+\frac{(5 a+6 b) \tan (e+f x) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 f (a+b)}+\frac{a (5 a+6 b) \tan (e+f x) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{16 f (a+b)} \]

[Out]

(a^2*(5*a + 6*b)*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(16*(a + b)^(3/2)*f) + (a*(5*
a + 6*b)*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x])/(16*(a + b)*f) + ((5*a + 6*b)*Sec[e + f*x]^3*(a
+ b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x])/(24*(a + b)*f) + (Sec[e + f*x]^5*(a + b*Sin[e + f*x]^2)^(5/2)*Tan[e +
f*x])/(6*(a + b)*f)

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Rubi [A]  time = 0.176011, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3190, 382, 378, 377, 206} \[ \frac{a^2 (5 a+6 b) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{16 f (a+b)^{3/2}}+\frac{\tan (e+f x) \sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 f (a+b)}+\frac{(5 a+6 b) \tan (e+f x) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 f (a+b)}+\frac{a (5 a+6 b) \tan (e+f x) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{16 f (a+b)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^7*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(a^2*(5*a + 6*b)*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(16*(a + b)^(3/2)*f) + (a*(5*
a + 6*b)*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x])/(16*(a + b)*f) + ((5*a + 6*b)*Sec[e + f*x]^3*(a
+ b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x])/(24*(a + b)*f) + (Sec[e + f*x]^5*(a + b*Sin[e + f*x]^2)^(5/2)*Tan[e +
f*x])/(6*(a + b)*f)

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 382

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

Rule 378

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

Rule 377

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

Rule 206

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

Rubi steps

\begin{align*} \int \sec ^7(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^{3/2}}{\left (1-x^2\right )^4} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2} \tan (e+f x)}{6 (a+b) f}+\frac{(5 a+6 b) \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^{3/2}}{\left (1-x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{6 (a+b) f}\\ &=\frac{(5 a+6 b) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{24 (a+b) f}+\frac{\sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2} \tan (e+f x)}{6 (a+b) f}+\frac{(a (5 a+6 b)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{8 (a+b) f}\\ &=\frac{a (5 a+6 b) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{16 (a+b) f}+\frac{(5 a+6 b) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{24 (a+b) f}+\frac{\sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2} \tan (e+f x)}{6 (a+b) f}+\frac{\left (a^2 (5 a+6 b)\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{16 (a+b) f}\\ &=\frac{a (5 a+6 b) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{16 (a+b) f}+\frac{(5 a+6 b) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{24 (a+b) f}+\frac{\sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2} \tan (e+f x)}{6 (a+b) f}+\frac{\left (a^2 (5 a+6 b)\right ) \operatorname{Subst}\left (\int \frac{1}{1-(a+b) x^2} \, dx,x,\frac{\sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{16 (a+b) f}\\ &=\frac{a^2 (5 a+6 b) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{16 (a+b)^{3/2} f}+\frac{a (5 a+6 b) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{16 (a+b) f}+\frac{(5 a+6 b) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{24 (a+b) f}+\frac{\sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2} \tan (e+f x)}{6 (a+b) f}\\ \end{align*}

Mathematica [C]  time = 15.9361, size = 938, normalized size = 4.81 \[ \frac{a^2 \sec ^3(e+f x) \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^2 \tan (e+f x) \left (256 b \, _2F_1\left (2,5;\frac{7}{2};-\frac{(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt{\frac{\sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )}{a}} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{9/2}+256 a \, _2F_1\left (2,5;\frac{7}{2};-\frac{(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt{\frac{\sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )}{a}} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{9/2}-512 b \, _2F_1\left (2,5;\frac{7}{2};-\frac{(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt{\frac{\sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )}{a}} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}-512 a \, _2F_1\left (2,5;\frac{7}{2};-\frac{(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt{\frac{\sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )}{a}} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}-80 b \sin ^2(e+f x) \sqrt{\frac{\sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )}{a}} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}+256 b \, _2F_1\left (2,5;\frac{7}{2};-\frac{(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt{\frac{\sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )}{a}} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}-120 a \sqrt{\frac{\sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )}{a}} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}+256 a \, _2F_1\left (2,5;\frac{7}{2};-\frac{(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt{\frac{\sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )}{a}} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}+140 b \sin ^2(e+f x) \sqrt{\frac{\sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )}{a}} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}+210 a \sqrt{\frac{\sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )}{a}} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}+30 b \sin ^{-1}\left (\sqrt{-\frac{(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^2(e+f x)+45 a \sin ^{-1}\left (\sqrt{-\frac{(a+b) \tan ^2(e+f x)}{a}}\right )-30 b \sin ^2(e+f x) \sqrt{-\frac{(a+b) \sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right ) \tan ^2(e+f x)}{a^2}}-45 a \sqrt{-\frac{(a+b) \sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right ) \tan ^2(e+f x)}{a^2}}\right )}{240 f \left (b \sin ^2(e+f x)+a\right )^{3/2} \sqrt{\frac{\sec ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )}{a}} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[e + f*x]^7*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(a^2*Sec[e + f*x]^3*(1 + (b*Sin[e + f*x]^2)/a)^2*Tan[e + f*x]*(45*a*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]
] + 30*b*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^2 + 210*a*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e
+ f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2) + 140*b*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e +
f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2) - 120*a*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(
((a + b)*Tan[e + f*x]^2)/a))^(5/2) + 256*a*Hypergeometric2F1[2, 5, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sqrt[(S
ec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(5/2) - 80*b*Sin[e + f*x]^2*Sqrt[(Sec
[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(5/2) + 256*b*Hypergeometric2F1[2, 5, 7
/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*
Tan[e + f*x]^2)/a))^(5/2) - 512*a*Hypergeometric2F1[2, 5, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sqrt[(Sec[e + f*
x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(7/2) - 512*b*Hypergeometric2F1[2, 5, 7/2, -((
(a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e +
f*x]^2)/a))^(7/2) + 256*a*Hypergeometric2F1[2, 5, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sqrt[(Sec[e + f*x]^2*(a
+ b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(9/2) + 256*b*Hypergeometric2F1[2, 5, 7/2, -(((a + b)
*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2
)/a))^(9/2) - 45*a*Sqrt[-(((a + b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan[e + f*x]^2)/a^2)] - 30*b*Sin[e +
f*x]^2*Sqrt[-(((a + b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan[e + f*x]^2)/a^2)]))/(240*f*(a + b*Sin[e + f*x
]^2)^(3/2)*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2))

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Maple [B]  time = 3.241, size = 693, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

1/96*(2*(a+b)^(7/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*(15*a^2+8*a*b-4*b^2)*sin(f*x+e)*cos(f*x+e)^4+4*(a+b)^(7/2)*(a+b
-b*cos(f*x+e)^2)^(1/2)*(5*a^2+3*a*b-2*b^2)*cos(f*x+e)^2*sin(f*x+e)+16*(a+b)^(7/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*(
a^2+2*a*b+b^2)*sin(f*x+e)+3*a^2*(5*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a
))*a^4+21*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^3*b+33*ln(2/(-1+sin(
f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2*b^2+23*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*
(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b^3+6*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/
2)+b*sin(f*x+e)+a))*b^4-5*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^4-21*
ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3*b-33*ln(2/(1+sin(f*x+e))*((a+
b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^2-23*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*
x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b^3-6*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e
)+a))*b^4)*cos(f*x+e)^6)/(a+b)^(9/2)/cos(f*x+e)^6/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sec \left (f x + e\right )^{7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*sec(f*x + e)^7, x)

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Fricas [A]  time = 25.5334, size = 1310, normalized size = 6.72 \begin{align*} \left [\frac{3 \,{\left (5 \, a^{3} + 6 \, a^{2} b\right )} \sqrt{a + b} \cos \left (f x + e\right )^{6} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \,{\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \,{\left ({\left (15 \, a^{3} + 23 \, a^{2} b + 4 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{3} + 24 \, a^{2} b + 24 \, a b^{2} + 8 \, b^{3} + 2 \,{\left (5 \, a^{3} + 8 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{192 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{6}}, -\frac{3 \,{\left (5 \, a^{3} + 6 \, a^{2} b\right )} \sqrt{-a - b} \arctan \left (\frac{{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a - b}}{2 \,{\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{6} - 2 \,{\left ({\left (15 \, a^{3} + 23 \, a^{2} b + 4 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{3} + 24 \, a^{2} b + 24 \, a b^{2} + 8 \, b^{3} + 2 \,{\left (5 \, a^{3} + 8 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{96 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/192*(3*(5*a^3 + 6*a^2*b)*sqrt(a + b)*cos(f*x + e)^6*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 8*(a^2 + 3*
a*b + 2*b^2)*cos(f*x + e)^2 - 4*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a
+ b)*sin(f*x + e) + 8*a^2 + 16*a*b + 8*b^2)/cos(f*x + e)^4) + 4*((15*a^3 + 23*a^2*b + 4*a*b^2 - 4*b^3)*cos(f*x
+ e)^4 + 8*a^3 + 24*a^2*b + 24*a*b^2 + 8*b^3 + 2*(5*a^3 + 8*a^2*b + a*b^2 - 2*b^3)*cos(f*x + e)^2)*sqrt(-b*co
s(f*x + e)^2 + a + b)*sin(f*x + e))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^6), -1/96*(3*(5*a^3 + 6*a^2*b)*sqrt(-a
- b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(((a*b +
b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*sin(f*x + e)))*cos(f*x + e)^6 - 2*((15*a^3 + 23*a^2*b + 4*a*b^2 - 4*b
^3)*cos(f*x + e)^4 + 8*a^3 + 24*a^2*b + 24*a*b^2 + 8*b^3 + 2*(5*a^3 + 8*a^2*b + a*b^2 - 2*b^3)*cos(f*x + e)^2)
*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^2 + 2*a*b + 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(sec(f*x+e)**7*(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sec \left (f x + e\right )^{7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*sec(f*x + e)^7, x)