3.522 \(\int (a+b \sec (d+e x)) (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x))^{3/2} \, dx\)

Optimal. Leaf size=359 \[ \frac{a^4 b^3 x \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{\left (a^2 \sec (d+e x)+a b\right )^3}+\frac{a^5 \left (3 a^2+5 b^2\right ) \tan (d+e x) \sec (d+e x) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{6 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{a^4 b \left (11 a^2+8 b^2\right ) \tan (d+e x) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{3 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{b \tan (d+e x) \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{3 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{\left (9 a^2 b^2+a^4+2 b^4\right ) \tanh ^{-1}(\sin (d+e x)) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{2 e (a \sec (d+e x)+b)^3} \]

[Out]

((a^4 + 9*a^2*b^2 + 2*b^4)*ArcTanh[Sin[d + e*x]]*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2)^(3/2))/(2*e*(
b + a*Sec[d + e*x])^3) + (a^4*b^3*x*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2)^(3/2))/(a*b + a^2*Sec[d +
e*x])^3 + (a^4*b*(11*a^2 + 8*b^2)*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2)^(3/2)*Tan[d + e*x])/(3*e*(a*
b + a^2*Sec[d + e*x])^3) + (a^5*(3*a^2 + 5*b^2)*Sec[d + e*x]*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2)^(
3/2)*Tan[d + e*x])/(6*e*(a*b + a^2*Sec[d + e*x])^3) + (b*(a^2*b + a^3*Sec[d + e*x])^2*(b^2 + 2*a*b*Sec[d + e*x
] + a^2*Sec[d + e*x]^2)^(3/2)*Tan[d + e*x])/(3*e*(a*b + a^2*Sec[d + e*x])^3)

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Rubi [A]  time = 0.287014, antiderivative size = 359, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4174, 3918, 4048, 3770, 3767, 8} \[ \frac{a^4 b^3 x \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{\left (a^2 \sec (d+e x)+a b\right )^3}+\frac{a^5 \left (3 a^2+5 b^2\right ) \tan (d+e x) \sec (d+e x) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{6 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{a^4 b \left (11 a^2+8 b^2\right ) \tan (d+e x) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{3 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{b \tan (d+e x) \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{3 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{\left (9 a^2 b^2+a^4+2 b^4\right ) \tanh ^{-1}(\sin (d+e x)) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{2 e (a \sec (d+e x)+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^4 + 9*a^2*b^2 + 2*b^4)*ArcTanh[Sin[d + e*x]]*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2)^(3/2))/(2*e*(
b + a*Sec[d + e*x])^3) + (a^4*b^3*x*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2)^(3/2))/(a*b + a^2*Sec[d +
e*x])^3 + (a^4*b*(11*a^2 + 8*b^2)*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2)^(3/2)*Tan[d + e*x])/(3*e*(a*
b + a^2*Sec[d + e*x])^3) + (a^5*(3*a^2 + 5*b^2)*Sec[d + e*x]*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2)^(
3/2)*Tan[d + e*x])/(6*e*(a*b + a^2*Sec[d + e*x])^3) + (b*(a^2*b + a^3*Sec[d + e*x])^2*(b^2 + 2*a*b*Sec[d + e*x
] + a^2*Sec[d + e*x]^2)^(3/2)*Tan[d + e*x])/(3*e*(a*b + a^2*Sec[d + e*x])^3)

Rule 4174

Int[((A_) + (B_.)*sec[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*sec[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[(a + b*Sec[d + e*x] + c*Sec[d + e*x]^2)^n/(b + 2*c*Sec[d + e*x])^(2*n), Int[(A +
B*Sec[d + e*x])*(b + 2*c*Sec[d + e*x])^(2*n), x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0
] &&  !IntegerQ[n]

Rule 3918

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2*c
*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (a+b \sec (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \, dx &=\frac{\left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \int \left (2 a b+2 a^2 \sec (d+e x)\right )^3 (a+b \sec (d+e x)) \, dx}{\left (2 a b+2 a^2 \sec (d+e x)\right )^3}\\ &=\frac{b \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{\left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \int \left (2 a b+2 a^2 \sec (d+e x)\right ) \left (12 a^3 b^2+4 a^2 b \left (8 a^2+3 b^2\right ) \sec (d+e x)+4 a^3 \left (3 a^2+5 b^2\right ) \sec ^2(d+e x)\right ) \, dx}{3 \left (2 a b+2 a^2 \sec (d+e x)\right )^3}\\ &=\frac{a^5 \left (3 a^2+5 b^2\right ) \sec (d+e x) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{6 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{b \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{\left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \int \left (48 a^4 b^3+24 a^3 \left (a^4+9 a^2 b^2+2 b^4\right ) \sec (d+e x)+16 a^4 b \left (11 a^2+8 b^2\right ) \sec ^2(d+e x)\right ) \, dx}{6 \left (2 a b+2 a^2 \sec (d+e x)\right )^3}\\ &=\frac{a^4 b^3 x \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}{\left (a b+a^2 \sec (d+e x)\right )^3}+\frac{a^5 \left (3 a^2+5 b^2\right ) \sec (d+e x) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{6 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{b \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{\left (8 a^4 b \left (11 a^2+8 b^2\right ) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}\right ) \int \sec ^2(d+e x) \, dx}{3 \left (2 a b+2 a^2 \sec (d+e x)\right )^3}+\frac{\left (4 a^3 \left (a^4+9 a^2 b^2+2 b^4\right ) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}\right ) \int \sec (d+e x) \, dx}{\left (2 a b+2 a^2 \sec (d+e x)\right )^3}\\ &=\frac{\left (a^4+9 a^2 b^2+2 b^4\right ) \tanh ^{-1}(\sin (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}{2 e (b+a \sec (d+e x))^3}+\frac{a^4 b^3 x \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}{\left (a b+a^2 \sec (d+e x)\right )^3}+\frac{a^5 \left (3 a^2+5 b^2\right ) \sec (d+e x) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{6 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{b \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}-\frac{\left (8 a^4 b \left (11 a^2+8 b^2\right ) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (d+e x))}{3 e \left (2 a b+2 a^2 \sec (d+e x)\right )^3}\\ &=\frac{\left (a^4+9 a^2 b^2+2 b^4\right ) \tanh ^{-1}(\sin (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}{2 e (b+a \sec (d+e x))^3}+\frac{a^4 b^3 x \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}{\left (a b+a^2 \sec (d+e x)\right )^3}+\frac{a^4 b \left (11 a^2+8 b^2\right ) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{a^5 \left (3 a^2+5 b^2\right ) \sec (d+e x) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{6 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{b \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.837774, size = 128, normalized size = 0.36 \[ \frac{\cos (d+e x) \sqrt{(a \sec (d+e x)+b)^2} \left (3 \left (9 a^2 b^2+a^4+2 b^4\right ) \tanh ^{-1}(\sin (d+e x))+3 a \tan (d+e x) \left (a \left (a^2+3 b^2\right ) \sec (d+e x)+8 a^2 b+6 b^3\right )+2 a^3 b \tan ^3(d+e x)+6 a b^3 e x\right )}{6 e (a+b \cos (d+e x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[d + e*x]*Sqrt[(b + a*Sec[d + e*x])^2]*(6*a*b^3*e*x + 3*(a^4 + 9*a^2*b^2 + 2*b^4)*ArcTanh[Sin[d + e*x]] +
3*a*(8*a^2*b + 6*b^3 + a*(a^2 + 3*b^2)*Sec[d + e*x])*Tan[d + e*x] + 2*a^3*b*Tan[d + e*x]^3))/(6*e*(a + b*Cos[d
+ e*x]))

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Maple [A]  time = 0.264, size = 387, normalized size = 1.1 \begin{align*}{\frac{1}{6\,e \left ( b\cos \left ( ex+d \right ) +a \right ) ^{3}} \left ( 3\,\ln \left ({\frac{\sin \left ( ex+d \right ) +1-\cos \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{a}^{4}+27\,\ln \left ({\frac{\sin \left ( ex+d \right ) +1-\cos \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{a}^{2}{b}^{2}+6\,\ln \left ({\frac{\sin \left ( ex+d \right ) +1-\cos \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{b}^{4}-3\,\ln \left ( -{\frac{\cos \left ( ex+d \right ) -1+\sin \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{a}^{4}-27\,\ln \left ( -{\frac{\cos \left ( ex+d \right ) -1+\sin \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{a}^{2}{b}^{2}-6\,\ln \left ( -{\frac{\cos \left ( ex+d \right ) -1+\sin \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{b}^{4}+6\, \left ( \cos \left ( ex+d \right ) \right ) ^{3} \left ( ex+d \right ) a{b}^{3}+22\,\sin \left ( ex+d \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{2}{a}^{3}b+18\,\sin \left ( ex+d \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{2}a{b}^{3}+3\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ){a}^{4}+9\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ){a}^{2}{b}^{2}+2\,{a}^{3}b\sin \left ( ex+d \right ) \right ) \left ({\frac{ \left ( b\cos \left ( ex+d \right ) +a \right ) ^{2}}{ \left ( \cos \left ( ex+d \right ) \right ) ^{2}}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(e*x+d))*(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2)^(3/2),x)

[Out]

1/6/e*(3*ln((sin(e*x+d)+1-cos(e*x+d))/sin(e*x+d))*cos(e*x+d)^3*a^4+27*ln((sin(e*x+d)+1-cos(e*x+d))/sin(e*x+d))
*cos(e*x+d)^3*a^2*b^2+6*ln((sin(e*x+d)+1-cos(e*x+d))/sin(e*x+d))*cos(e*x+d)^3*b^4-3*ln(-(cos(e*x+d)-1+sin(e*x+
d))/sin(e*x+d))*cos(e*x+d)^3*a^4-27*ln(-(cos(e*x+d)-1+sin(e*x+d))/sin(e*x+d))*cos(e*x+d)^3*a^2*b^2-6*ln(-(cos(
e*x+d)-1+sin(e*x+d))/sin(e*x+d))*cos(e*x+d)^3*b^4+6*cos(e*x+d)^3*(e*x+d)*a*b^3+22*sin(e*x+d)*cos(e*x+d)^2*a^3*
b+18*sin(e*x+d)*cos(e*x+d)^2*a*b^3+3*sin(e*x+d)*cos(e*x+d)*a^4+9*sin(e*x+d)*cos(e*x+d)*a^2*b^2+2*a^3*b*sin(e*x
+d))*((b*cos(e*x+d)+a)^2/cos(e*x+d)^2)^(3/2)/(b*cos(e*x+d)+a)^3

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Maxima [A]  time = 1.66739, size = 594, normalized size = 1.65 \begin{align*} \frac{3 \,{\left (4 \, b^{3} \arctan \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) +{\left (a^{3} + 6 \, a b^{2}\right )} \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right ) -{\left (a^{3} + 6 \, a b^{2}\right )} \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - 1\right ) - \frac{2 \,{\left (\frac{{\left (a^{3} + 6 \, a^{2} b\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac{{\left (a^{3} - 6 \, a^{2} b\right )} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}\right )}}{\frac{2 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac{\sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} - 1}\right )} a +{\left (3 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right ) - 3 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - 1\right ) - \frac{2 \,{\left (\frac{3 \,{\left (2 \, a^{3} + 3 \, a^{2} b + 6 \, a b^{2}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - \frac{4 \,{\left (a^{3} + 9 \, a b^{2}\right )} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}} + \frac{3 \,{\left (2 \, a^{3} - 3 \, a^{2} b + 6 \, a b^{2}\right )} \sin \left (e x + d\right )^{5}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{5}}\right )}}{\frac{3 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac{3 \, \sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} + \frac{\sin \left (e x + d\right )^{6}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{6}} - 1}\right )} b}{6 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(4*b^3*arctan(sin(e*x + d)/(cos(e*x + d) + 1)) + (a^3 + 6*a*b^2)*log(sin(e*x + d)/(cos(e*x + d) + 1) +
1) - (a^3 + 6*a*b^2)*log(sin(e*x + d)/(cos(e*x + d) + 1) - 1) - 2*((a^3 + 6*a^2*b)*sin(e*x + d)/(cos(e*x + d)
+ 1) + (a^3 - 6*a^2*b)*sin(e*x + d)^3/(cos(e*x + d) + 1)^3)/(2*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 - sin(e*x +
d)^4/(cos(e*x + d) + 1)^4 - 1))*a + (3*(3*a^2*b + 2*b^3)*log(sin(e*x + d)/(cos(e*x + d) + 1) + 1) - 3*(3*a^2*
b + 2*b^3)*log(sin(e*x + d)/(cos(e*x + d) + 1) - 1) - 2*(3*(2*a^3 + 3*a^2*b + 6*a*b^2)*sin(e*x + d)/(cos(e*x +
d) + 1) - 4*(a^3 + 9*a*b^2)*sin(e*x + d)^3/(cos(e*x + d) + 1)^3 + 3*(2*a^3 - 3*a^2*b + 6*a*b^2)*sin(e*x + d)^
5/(cos(e*x + d) + 1)^5)/(3*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 - 3*sin(e*x + d)^4/(cos(e*x + d) + 1)^4 + sin(e
*x + d)^6/(cos(e*x + d) + 1)^6 - 1))*b)/e

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Fricas [A]  time = 2.20469, size = 394, normalized size = 1.1 \begin{align*} \frac{12 \, a b^{3} e x \cos \left (e x + d\right )^{3} + 3 \,{\left (a^{4} + 9 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (e x + d\right )^{3} \log \left (\sin \left (e x + d\right ) + 1\right ) - 3 \,{\left (a^{4} + 9 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (e x + d\right )^{3} \log \left (-\sin \left (e x + d\right ) + 1\right ) + 2 \,{\left (2 \, a^{3} b + 2 \,{\left (11 \, a^{3} b + 9 \, a b^{3}\right )} \cos \left (e x + d\right )^{2} + 3 \,{\left (a^{4} + 3 \, a^{2} b^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{12 \, e \cos \left (e x + d\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*a*b^3*e*x*cos(e*x + d)^3 + 3*(a^4 + 9*a^2*b^2 + 2*b^4)*cos(e*x + d)^3*log(sin(e*x + d) + 1) - 3*(a^4
+ 9*a^2*b^2 + 2*b^4)*cos(e*x + d)^3*log(-sin(e*x + d) + 1) + 2*(2*a^3*b + 2*(11*a^3*b + 9*a*b^3)*cos(e*x + d)^
2 + 3*(a^4 + 3*a^2*b^2)*cos(e*x + d))*sin(e*x + d))/(e*cos(e*x + d)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(e*x+d))*(b**2+2*a*b*sec(e*x+d)+a**2*sec(e*x+d)**2)**(3/2),x)

[Out]

Integral((a + b*sec(d + e*x))*((a*sec(d + e*x) + b)**2)**(3/2), x)

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Giac [A]  time = 1.66667, size = 880, normalized size = 2.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(x*e + d)*a*b^3*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d)) + 3*(a^4*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d))
+ 9*a^2*b^2*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d)) + 2*b^4*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d)))*log(abs(tan
(1/2*x*e + 1/2*d) + 1)) - 3*(a^4*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d)) + 9*a^2*b^2*sgn(b*cos(x*e + d)^2 + a*c
os(x*e + d)) + 2*b^4*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d)))*log(abs(tan(1/2*x*e + 1/2*d) - 1)) + 2*(3*a^4*sgn
(b*cos(x*e + d)^2 + a*cos(x*e + d))*tan(1/2*x*e + 1/2*d)^5 - 24*a^3*b*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d))*t
an(1/2*x*e + 1/2*d)^5 + 9*a^2*b^2*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d))*tan(1/2*x*e + 1/2*d)^5 - 18*a*b^3*sgn
(b*cos(x*e + d)^2 + a*cos(x*e + d))*tan(1/2*x*e + 1/2*d)^5 + 40*a^3*b*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d))*t
an(1/2*x*e + 1/2*d)^3 + 36*a*b^3*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d))*tan(1/2*x*e + 1/2*d)^3 - 3*a^4*sgn(b*c
os(x*e + d)^2 + a*cos(x*e + d))*tan(1/2*x*e + 1/2*d) - 24*a^3*b*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d))*tan(1/2
*x*e + 1/2*d) - 9*a^2*b^2*sgn(b*cos(x*e + d)^2 + a*cos(x*e + d))*tan(1/2*x*e + 1/2*d) - 18*a*b^3*sgn(b*cos(x*e
+ d)^2 + a*cos(x*e + d))*tan(1/2*x*e + 1/2*d))/(tan(1/2*x*e + 1/2*d)^2 - 1)^3)*e^(-1)