3.125 \(\int \frac{\sin ^6(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=288 \[ \frac{5 (a+b) \left (a^2+14 a b+21 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{16 a^{11/2} f}-\frac{b \left (113 a^2+420 a b+315 b^2\right ) \tan (e+f x)}{48 a^5 f \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{7 b (a+b) (7 a+15 b) \tan (e+f x)}{48 a^4 f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}+\frac{3 (a+b) \sin (e+f x) \cos ^3(e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac{(a+b) (11 a+21 b) \sin (e+f x) \cos (e+f x)}{16 a^3 f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}+\frac{\sin ^3(e+f x) \cos ^3(e+f x)}{6 a f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

[Out]

(5*(a + b)*(a^2 + 14*a*b + 21*b^2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*a^(11/2)
*f) - ((a + b)*(11*a + 21*b)*Cos[e + f*x]*Sin[e + f*x])/(16*a^3*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) + (3*(a +
b)*Cos[e + f*x]^3*Sin[e + f*x])/(8*a^2*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) + (Cos[e + f*x]^3*Sin[e + f*x]^3)/(
6*a*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) - (7*b*(a + b)*(7*a + 15*b)*Tan[e + f*x])/(48*a^4*f*(a + b + b*Tan[e +
f*x]^2)^(3/2)) - (b*(113*a^2 + 420*a*b + 315*b^2)*Tan[e + f*x])/(48*a^5*f*Sqrt[a + b + b*Tan[e + f*x]^2])

________________________________________________________________________________________

Rubi [A]  time = 0.441857, antiderivative size = 288, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {4132, 470, 578, 527, 12, 377, 203} \[ \frac{5 (a+b) \left (a^2+14 a b+21 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{16 a^{11/2} f}-\frac{b \left (113 a^2+420 a b+315 b^2\right ) \tan (e+f x)}{48 a^5 f \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{7 b (a+b) (7 a+15 b) \tan (e+f x)}{48 a^4 f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}+\frac{3 (a+b) \sin (e+f x) \cos ^3(e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac{(a+b) (11 a+21 b) \sin (e+f x) \cos (e+f x)}{16 a^3 f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}+\frac{\sin ^3(e+f x) \cos ^3(e+f x)}{6 a f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

(5*(a + b)*(a^2 + 14*a*b + 21*b^2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*a^(11/2)
*f) - ((a + b)*(11*a + 21*b)*Cos[e + f*x]*Sin[e + f*x])/(16*a^3*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) + (3*(a +
b)*Cos[e + f*x]^3*Sin[e + f*x])/(8*a^2*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) + (Cos[e + f*x]^3*Sin[e + f*x]^3)/(
6*a*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) - (7*b*(a + b)*(7*a + 15*b)*Tan[e + f*x])/(48*a^4*f*(a + b + b*Tan[e +
f*x]^2)^(3/2)) - (b*(113*a^2 + 420*a*b + 315*b^2)*Tan[e + f*x])/(48*a^5*f*Sqrt[a + b + b*Tan[e + f*x]^2])

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rule 470

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

Rule 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 527

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 \frac{\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^4 \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 (a+b)-6 (a+b) x^2\right )}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{6 a f}\\ &=\frac{3 (a+b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{9 (a+b)^2-6 (a+b) (4 a+9 b) x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{24 a^2 f}\\ &=-\frac{(a+b) (11 a+21 b) \cos (e+f x) \sin (e+f x)}{16 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{3 (a+b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{3 (a+b)^2 (5 a+21 b)-12 b (a+b) (11 a+21 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{48 a^3 f}\\ &=-\frac{(a+b) (11 a+21 b) \cos (e+f x) \sin (e+f x)}{16 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{3 (a+b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{7 b (a+b) (7 a+15 b) \tan (e+f x)}{48 a^4 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{3 (a+b)^2 \left (15 a^2+112 a b+105 b^2\right )-42 b (a+b)^2 (7 a+15 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{144 a^4 (a+b) f}\\ &=-\frac{(a+b) (11 a+21 b) \cos (e+f x) \sin (e+f x)}{16 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{3 (a+b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{7 b (a+b) (7 a+15 b) \tan (e+f x)}{48 a^4 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b \left (113 a^2+420 a b+315 b^2\right ) \tan (e+f x)}{48 a^5 f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{45 (a+b)^3 \left (a^2+14 a b+21 b^2\right )}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{144 a^5 (a+b)^2 f}\\ &=-\frac{(a+b) (11 a+21 b) \cos (e+f x) \sin (e+f x)}{16 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{3 (a+b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{7 b (a+b) (7 a+15 b) \tan (e+f x)}{48 a^4 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b \left (113 a^2+420 a b+315 b^2\right ) \tan (e+f x)}{48 a^5 f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\left (5 (a+b) \left (a^2+14 a b+21 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{16 a^5 f}\\ &=-\frac{(a+b) (11 a+21 b) \cos (e+f x) \sin (e+f x)}{16 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{3 (a+b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{7 b (a+b) (7 a+15 b) \tan (e+f x)}{48 a^4 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b \left (113 a^2+420 a b+315 b^2\right ) \tan (e+f x)}{48 a^5 f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\left (5 (a+b) \left (a^2+14 a b+21 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{16 a^5 f}\\ &=\frac{5 (a+b) \left (a^2+14 a b+21 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{16 a^{11/2} f}-\frac{(a+b) (11 a+21 b) \cos (e+f x) \sin (e+f x)}{16 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{3 (a+b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{7 b (a+b) (7 a+15 b) \tan (e+f x)}{48 a^4 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b \left (113 a^2+420 a b+315 b^2\right ) \tan (e+f x)}{48 a^5 f \sqrt{a+b+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [B]  time = 19.5767, size = 1705, normalized size = 5.92 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-(((a + 2*b + a*Cos[2*(e + f*x)])/(a + b))^(3/2)*(a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^5*(-60*Sqrt
[a + b]*(3*a^3 + 17*a^2*b + 28*a*b^2 + 14*b^3)*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*(a + 2*b + a*Cos[2*(
e + f*x)])^2 + Sqrt[a]*Sin[e + f*x]*Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]*(3*(239*a^5 + 1839*a^4*b + 5200*a
^3*b^2 + 6960*a^2*b^3 + 4480*a*b^4 + 1120*b^5) - 2*a*(459*a^4 + 3180*a^3*b + 7200*a^2*b^2 + 6720*a*b^3 + 2240*
b^4)*Sin[e + f*x]^2 + 672*a^2*b*(a + b)^2*Sin[e + f*x]^4 + 192*a^3*(a + b)^2*Sin[e + f*x]^6)))/(3072*Sqrt[2]*a
^(9/2)*f*(a + 2*b + a*Cos[2*(e + f*x)])^(7/2)*(a + b*Sec[e + f*x]^2)^(5/2)) + (((a + 2*b + a*Cos[2*(e + f*x)])
/(a + b))^(3/2)*(a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^5*(420*Sqrt[a + b]*(a^4 + 9*a^3*b + 26*a^2*b
^2 + 30*a*b^3 + 12*b^4)*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 - Sqrt[a]*
Sin[e + f*x]*Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]*(3*(561*a^6 + 6161*a^5*b + 25200*a^4*b^2 + 50960*a^3*b^3
+ 54880*a^2*b^4 + 30240*a*b^5 + 6720*b^6) - 2*a*(1151*a^5 + 11230*a^4*b + 39200*a^3*b^2 + 62720*a^2*b^3 + 470
40*a*b^4 + 13440*b^5)*Sin[e + f*x]^2 + 672*a^2*(a + b)^2*(a^2 + 3*a*b + 6*b^2)*Sin[e + f*x]^4 - 576*a^3*(a - 2
*b)*(a + b)^2*Sin[e + f*x]^6 + 512*a^4*(a + b)^2*Sin[e + f*x]^8)))/(3072*Sqrt[2]*a^(11/2)*f*(a + 2*b + a*Cos[2
*(e + f*x)])^(7/2)*(a + b*Sec[e + f*x]^2)^(5/2)) - (5*(a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Csc[e + f*x]*Sec[e
+ f*x]^5*(Sin[e + f*x]^2/(a + b) + ((a + 2*b + a*Cos[2*(e + f*x)])*Sin[e + f*x]^2)/(a + b)^2 - (12*Sin[e + f*x
]^4)/(a + b) + (16*(a + b - a*Sin[e + f*x]^2)*(1 - (a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b)*Sin[e + f*x]^2)/
(a + 2*b + a*Cos[2*(e + f*x)]) + (a^2*(a + b)*Sin[e + f*x]^4)/(a + b - a*Sin[e + f*x]^2)^2 + (3*Sqrt[a]*Sqrt[a
+ b]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*Sin[e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]))/a^3)
)/(12288*Sqrt[2]*f*(a + b*Sec[e + f*x]^2)^(5/2)*(a + b - a*Sin[e + f*x]^2)^(3/2)) + (5*(a + 2*b + a*Cos[2*e +
2*f*x])^(5/2)*Csc[e + f*x]*Sec[e + f*x]^5*(Sin[e + f*x]^2/(a + b) + ((a + 2*b + a*Cos[2*(e + f*x)])*Sin[e + f*
x]^2)/(a + b)^2 - (24*Sin[e + f*x]^4)/(a + b) + (96*Sin[e + f*x]^6)/a + (80*(a + b - a*Sin[e + f*x]^2)*(1 - (a
*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b)*Sin[e + f*x]^2)/(a + 2*b + a*Cos[2*(e + f*x)]) + (a^2*(a + b)*Sin[e +
f*x]^4)/(a + b - a*Sin[e + f*x]^2)^2 + (3*Sqrt[a]*Sqrt[a + b]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*Sin[
e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]))/a^3 - (160*(a + b - a*Sin[e + f*x]^2)*(1 - (a*Sin[e + f*x
]^2)/(a + b))*((-6*a*(a + b)^2*Sin[e + f*x]^2)/(a + 2*b + a*Cos[2*(e + f*x)]) + (3*Sqrt[a]*(a + b)^(3/2)*ArcSi
n[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*Sin[e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)] + (a^2*Sin[e + f*
x]^4)/(-1 + (a*Sin[e + f*x]^2)/(a + b))^2))/a^4))/(12288*Sqrt[2]*f*(a + b*Sec[e + f*x]^2)^(5/2)*(a + b - a*Sin
[e + f*x]^2)^(3/2)) + (5*(2*a + 3*b + a*Cos[2*(e + f*x)])*(a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^4*
Tan[e + f*x])/(3072*(a + b)^2*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2)*(a + b*Sec[e + f*x]^2)^(5/2)) - (5*(b + (
3*a + 2*b)*Cos[2*(e + f*x)])*(a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^4*Tan[e + f*x])/(3072*(a + b)^2
*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2)*(a + b*Sec[e + f*x]^2)^(5/2))

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Maple [C]  time = 1.371, size = 4477, normalized size = 15.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x)

[Out]

-1/48/f/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/a^5*sin(f*x+e)*(b+a*cos(f*x+e)^2)*(-630*cos(f*x+e)^2*sin(f*x+e
)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(
a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(
f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b
^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*b^3+315*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos
(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2
)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1
/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^4
-630*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)
))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*Ellip
ticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-
(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b^4-113*((2*I*a^(1/2)*b^(1/2)+
a-b)/(a+b))^(1/2)*a^2*b^2-420*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^3+33*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1
/2)+a-b)/(a+b))^(1/2)*a^4+8*cos(f*x+e)^9*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^4+315*cos(f*x+e)*((2*I*a^(1
/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4-33*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^4-8*cos(f*x+e)^8*(
(2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^4+26*cos(f*x+e)^6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^4-26*cos(
f*x+e)^7*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^4-315*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4+420*cos(f
*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^3-162*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)
*a^3*b-574*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b^2-420*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)
+a-b)/(a+b))^(1/2)*a*b^3+113*cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b^2+420*cos(f*x+e)*((2*I*a
^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^3-30*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*
b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(
f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),
-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^
(1/2))*a^3*b-450*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(
1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))
^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a
-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^2*b^2-1050*sin(f
*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(
-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+
cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/
2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*b^3+15*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*
cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(
1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b
^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*
a^3*b+225*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f
*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*
EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2
)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^2*b^2+525*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)
-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(
1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/s
in(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b^3-30*cos(f*x+e)^2*sin(
f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*
(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1
+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1
/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^4+15*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*
(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*co
s(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2
*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b
)^2)^(1/2))*a^4+225*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a
*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)
/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(
3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^3*b+525*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(1/(
a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*
x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a
^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)
^(1/2))*a^2*b^2+315*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a
*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)
/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(
3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b^3-450*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(1/(
a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*
x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*
a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a
+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^3*b-1050*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos
(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2
)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(
1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/
((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^2*b^2-18*cos(f*x+e)^7*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3*b
+18*cos(f*x+e)^6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3*b+96*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b
))^(1/2)*a^3*b+63*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b^2-96*cos(f*x+e)^4*((2*I*a^(1/2)*b
^(1/2)+a-b)/(a+b))^(1/2)*a^3*b-63*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b^2+162*cos(f*x+e)^
3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3*b+574*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b
^2)/(-1+cos(f*x+e))/cos(f*x+e)^5/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(5/2), x)

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Fricas [A]  time = 72.5447, size = 2375, normalized size = 8.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(a^3*b^2 + 15*a^2*b^3 + 35*a*b^4 + 21*b^5 + (a^5 + 15*a^4*b + 35*a^3*b^2 + 21*a^2*b^3)*cos(f*x + e
)^4 + 2*(a^4*b + 15*a^3*b^2 + 35*a^2*b^3 + 21*a*b^4)*cos(f*x + e)^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256
*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2
- 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3
- a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos
(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(8*a^5*cos(f*x + e)^9 - 2*(1
3*a^5 + 9*a^4*b)*cos(f*x + e)^7 + 3*(11*a^5 + 32*a^4*b + 21*a^3*b^2)*cos(f*x + e)^5 + 2*(81*a^4*b + 287*a^3*b^
2 + 210*a^2*b^3)*cos(f*x + e)^3 + (113*a^3*b^2 + 420*a^2*b^3 + 315*a*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2
+ b)/cos(f*x + e)^2)*sin(f*x + e))/(a^8*f*cos(f*x + e)^4 + 2*a^7*b*f*cos(f*x + e)^2 + a^6*b^2*f), -1/192*(15*
(a^3*b^2 + 15*a^2*b^3 + 35*a*b^4 + 21*b^5 + (a^5 + 15*a^4*b + 35*a^3*b^2 + 21*a^2*b^3)*cos(f*x + e)^4 + 2*(a^4
*b + 15*a^3*b^2 + 35*a^2*b^3 + 21*a*b^4)*cos(f*x + e)^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a
*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2
*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) + 4*(8*a^5*cos(f*x + e)^9
- 2*(13*a^5 + 9*a^4*b)*cos(f*x + e)^7 + 3*(11*a^5 + 32*a^4*b + 21*a^3*b^2)*cos(f*x + e)^5 + 2*(81*a^4*b + 287
*a^3*b^2 + 210*a^2*b^3)*cos(f*x + e)^3 + (113*a^3*b^2 + 420*a^2*b^3 + 315*a*b^4)*cos(f*x + e))*sqrt((a*cos(f*x
+ e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^8*f*cos(f*x + e)^4 + 2*a^7*b*f*cos(f*x + e)^2 + a^6*b^2*f)]

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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(sin(f*x+e)**6/(a+b*sec(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(5/2), x)