3.537 \(\int \sec ^5(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=159 \[ \frac {3 a \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {b \sec ^5(c+d x) \left (4 \left (8 a^2-b^2\right )+15 a b \tan (c+d x)\right )}{70 d}+\frac {a \left (2 a^2-b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{8 d}+\frac {3 a \left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {b \sec ^5(c+d x) (a+b \tan (c+d x))^2}{7 d} \]

[Out]

3/16*a*(2*a^2-b^2)*arctanh(sin(d*x+c))/d+3/16*a*(2*a^2-b^2)*sec(d*x+c)*tan(d*x+c)/d+1/8*a*(2*a^2-b^2)*sec(d*x+
c)^3*tan(d*x+c)/d+1/7*b*sec(d*x+c)^5*(a+b*tan(d*x+c))^2/d+1/70*b*sec(d*x+c)^5*(32*a^2-4*b^2+15*a*b*tan(d*x+c))
/d

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Rubi [A]  time = 0.14, antiderivative size = 177, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3512, 743, 780, 195, 215} \[ \frac {b \sec ^5(c+d x) \left (4 \left (8 a^2-b^2\right )+15 a b \tan (c+d x)\right )}{70 d}+\frac {a \left (2 a^2-b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{8 d}+\frac {3 a \left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {3 a \left (2 a^2-b^2\right ) \sec (c+d x) \sinh ^{-1}(\tan (c+d x))}{16 d \sqrt {\sec ^2(c+d x)}}+\frac {b \sec ^5(c+d x) (a+b \tan (c+d x))^2}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]

[Out]

(3*a*(2*a^2 - b^2)*ArcSinh[Tan[c + d*x]]*Sec[c + d*x])/(16*d*Sqrt[Sec[c + d*x]^2]) + (3*a*(2*a^2 - b^2)*Sec[c
+ d*x]*Tan[c + d*x])/(16*d) + (a*(2*a^2 - b^2)*Sec[c + d*x]^3*Tan[c + d*x])/(8*d) + (b*Sec[c + d*x]^5*(a + b*T
an[c + d*x])^2)/(7*d) + (b*Sec[c + d*x]^5*(4*(8*a^2 - b^2) + 15*a*b*Tan[c + d*x]))/(70*d)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 215

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

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 3512

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(d^(2
*IntPart[m/2])*(d*Sec[e + f*x])^(2*FracPart[m/2]))/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rubi steps

\begin {align*} \int \sec ^5(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac {\sec (c+d x) \operatorname {Subst}\left (\int (a+x)^3 \left (1+\frac {x^2}{b^2}\right )^{3/2} \, dx,x,b \tan (c+d x)\right )}{b d \sqrt {\sec ^2(c+d x)}}\\ &=\frac {b \sec ^5(c+d x) (a+b \tan (c+d x))^2}{7 d}+\frac {(b \sec (c+d x)) \operatorname {Subst}\left (\int (a+x) \left (-2+\frac {7 a^2}{b^2}+\frac {9 a x}{b^2}\right ) \left (1+\frac {x^2}{b^2}\right )^{3/2} \, dx,x,b \tan (c+d x)\right )}{7 d \sqrt {\sec ^2(c+d x)}}\\ &=\frac {b \sec ^5(c+d x) (a+b \tan (c+d x))^2}{7 d}+\frac {b \sec ^5(c+d x) \left (4 \left (8 a^2-b^2\right )+15 a b \tan (c+d x)\right )}{70 d}-\frac {\left (\left (\frac {9 a}{b^2}-\frac {6 a \left (-2+\frac {7 a^2}{b^2}\right )}{b^2}\right ) b^3 \sec (c+d x)\right ) \operatorname {Subst}\left (\int \left (1+\frac {x^2}{b^2}\right )^{3/2} \, dx,x,b \tan (c+d x)\right )}{42 d \sqrt {\sec ^2(c+d x)}}\\ &=\frac {a \left (2 a^2-b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^5(c+d x) (a+b \tan (c+d x))^2}{7 d}+\frac {b \sec ^5(c+d x) \left (4 \left (8 a^2-b^2\right )+15 a b \tan (c+d x)\right )}{70 d}-\frac {\left (\left (\frac {9 a}{b^2}-\frac {6 a \left (-2+\frac {7 a^2}{b^2}\right )}{b^2}\right ) b^3 \sec (c+d x)\right ) \operatorname {Subst}\left (\int \sqrt {1+\frac {x^2}{b^2}} \, dx,x,b \tan (c+d x)\right )}{56 d \sqrt {\sec ^2(c+d x)}}\\ &=\frac {3 a \left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a \left (2 a^2-b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^5(c+d x) (a+b \tan (c+d x))^2}{7 d}+\frac {b \sec ^5(c+d x) \left (4 \left (8 a^2-b^2\right )+15 a b \tan (c+d x)\right )}{70 d}-\frac {\left (\left (\frac {9 a}{b^2}-\frac {6 a \left (-2+\frac {7 a^2}{b^2}\right )}{b^2}\right ) b^3 \sec (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{112 d \sqrt {\sec ^2(c+d x)}}\\ &=\frac {3 a \left (2 a^2-b^2\right ) \sinh ^{-1}(\tan (c+d x)) \sec (c+d x)}{16 d \sqrt {\sec ^2(c+d x)}}+\frac {3 a \left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a \left (2 a^2-b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^5(c+d x) (a+b \tan (c+d x))^2}{7 d}+\frac {b \sec ^5(c+d x) \left (4 \left (8 a^2-b^2\right )+15 a b \tan (c+d x)\right )}{70 d}\\ \end {align*}

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Mathematica [B]  time = 2.19, size = 637, normalized size = 4.01 \[ \frac {\sec ^7(c+d x) \left (4340 a^3 \sin (2 (c+d x))+2800 a^3 \sin (4 (c+d x))+420 a^3 \sin (6 (c+d x))-4410 a^3 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-1470 a^3 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-210 a^3 \cos (7 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4410 a^3 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+1470 a^3 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+210 a^3 \cos (7 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+3584 \left (3 a^2 b-b^3\right ) \cos (2 (c+d x))-3675 a \left (2 a^2-b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+10752 a^2 b+6790 a b^2 \sin (2 (c+d x))-1400 a b^2 \sin (4 (c+d x))-210 a b^2 \sin (6 (c+d x))+2205 a b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+735 a b^2 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+105 a b^2 \cos (7 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2205 a b^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-735 a b^2 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-105 a b^2 \cos (7 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+1536 b^3\right )}{35840 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^7*(10752*a^2*b + 1536*b^3 + 3584*(3*a^2*b - b^3)*Cos[2*(c + d*x)] - 4410*a^3*Cos[3*(c + d*x)]*Lo
g[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2205*a*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
- 1470*a^3*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 735*a*b^2*Cos[5*(c + d*x)]*Log[Cos[(c +
d*x)/2] - Sin[(c + d*x)/2]] - 210*a^3*Cos[7*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 105*a*b^2*C
os[7*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 3675*a*(2*a^2 - b^2)*Cos[c + d*x]*(Log[Cos[(c + d*x
)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 4410*a^3*Cos[3*(c + d*x)]*Log[Cos[(c +
d*x)/2] + Sin[(c + d*x)/2]] - 2205*a*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 1470*a^3*
Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 735*a*b^2*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]] + 210*a^3*Cos[7*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 105*a*b^2*Cos[7*(c + d
*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 4340*a^3*Sin[2*(c + d*x)] + 6790*a*b^2*Sin[2*(c + d*x)] + 2800
*a^3*Sin[4*(c + d*x)] - 1400*a*b^2*Sin[4*(c + d*x)] + 420*a^3*Sin[6*(c + d*x)] - 210*a*b^2*Sin[6*(c + d*x)]))/
(35840*d)

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fricas [A]  time = 0.84, size = 170, normalized size = 1.07 \[ \frac {105 \, {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 160 \, b^{3} + 224 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 70 \, {\left (3 \, {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{5} + 8 \, a b^{2} \cos \left (d x + c\right ) + 2 \, {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{1120 \, d \cos \left (d x + c\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/1120*(105*(2*a^3 - a*b^2)*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 105*(2*a^3 - a*b^2)*cos(d*x + c)^7*log(-sin
(d*x + c) + 1) + 160*b^3 + 224*(3*a^2*b - b^3)*cos(d*x + c)^2 + 70*(3*(2*a^3 - a*b^2)*cos(d*x + c)^5 + 8*a*b^2
*cos(d*x + c) + 2*(2*a^3 - a*b^2)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^7)

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giac [B]  time = 1.08, size = 465, normalized size = 2.92 \[ \frac {105 \, {\left (2 \, a^{3} - a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, {\left (2 \, a^{3} - a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (350 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 105 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 1680 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 840 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1540 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 3360 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 1120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 630 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1085 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 5040 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 1120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 6720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 2240 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 630 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1085 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3696 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 448 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 840 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1540 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 672 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 224 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 350 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 336 \, a^{2} b + 32 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{7}}}{560 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/560*(105*(2*a^3 - a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(2*a^3 - a*b^2)*log(abs(tan(1/2*d*x + 1/2*
c) - 1)) + 2*(350*a^3*tan(1/2*d*x + 1/2*c)^13 + 105*a*b^2*tan(1/2*d*x + 1/2*c)^13 - 1680*a^2*b*tan(1/2*d*x + 1
/2*c)^12 - 840*a^3*tan(1/2*d*x + 1/2*c)^11 + 1540*a*b^2*tan(1/2*d*x + 1/2*c)^11 + 3360*a^2*b*tan(1/2*d*x + 1/2
*c)^10 - 1120*b^3*tan(1/2*d*x + 1/2*c)^10 + 630*a^3*tan(1/2*d*x + 1/2*c)^9 + 1085*a*b^2*tan(1/2*d*x + 1/2*c)^9
- 5040*a^2*b*tan(1/2*d*x + 1/2*c)^8 - 1120*b^3*tan(1/2*d*x + 1/2*c)^8 + 6720*a^2*b*tan(1/2*d*x + 1/2*c)^6 - 2
240*b^3*tan(1/2*d*x + 1/2*c)^6 - 630*a^3*tan(1/2*d*x + 1/2*c)^5 - 1085*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 3696*a^2
*b*tan(1/2*d*x + 1/2*c)^4 - 448*b^3*tan(1/2*d*x + 1/2*c)^4 + 840*a^3*tan(1/2*d*x + 1/2*c)^3 - 1540*a*b^2*tan(1
/2*d*x + 1/2*c)^3 + 672*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 224*b^3*tan(1/2*d*x + 1/2*c)^2 - 350*a^3*tan(1/2*d*x +
1/2*c) - 105*a*b^2*tan(1/2*d*x + 1/2*c) - 336*a^2*b + 32*b^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^7)/d

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maple [B]  time = 0.54, size = 328, normalized size = 2.06 \[ \frac {a^{3} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 a^{2} b}{5 d \cos \left (d x +c \right )^{5}}+\frac {b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{6}}+\frac {3 b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{4}}+\frac {3 b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{2}}+\frac {3 a \,b^{2} \sin \left (d x +c \right )}{16 d}-\frac {3 b^{2} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{7 d \cos \left (d x +c \right )^{7}}+\frac {3 b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{35 d \cos \left (d x +c \right )^{5}}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{35 d \cos \left (d x +c \right )^{3}}-\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{35 d \cos \left (d x +c \right )}-\frac {b^{3} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{35 d}-\frac {2 b^{3} \cos \left (d x +c \right )}{35 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+b*tan(d*x+c))^3,x)

[Out]

1/4*a^3*sec(d*x+c)^3*tan(d*x+c)/d+3/8*a^3*sec(d*x+c)*tan(d*x+c)/d+3/8/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/5/d*a^
2*b/cos(d*x+c)^5+1/2/d*b^2*a*sin(d*x+c)^3/cos(d*x+c)^6+3/8/d*b^2*a*sin(d*x+c)^3/cos(d*x+c)^4+3/16/d*b^2*a*sin(
d*x+c)^3/cos(d*x+c)^2+3/16*a*b^2*sin(d*x+c)/d-3/16/d*b^2*a*ln(sec(d*x+c)+tan(d*x+c))+1/7/d*b^3*sin(d*x+c)^4/co
s(d*x+c)^7+3/35/d*b^3*sin(d*x+c)^4/cos(d*x+c)^5+1/35/d*b^3*sin(d*x+c)^4/cos(d*x+c)^3-1/35/d*b^3*sin(d*x+c)^4/c
os(d*x+c)-1/35/d*b^3*cos(d*x+c)*sin(d*x+c)^2-2/35/d*b^3*cos(d*x+c)

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maxima [A]  time = 0.39, size = 208, normalized size = 1.31 \[ \frac {35 \, a b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 70 \, a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {672 \, a^{2} b}{\cos \left (d x + c\right )^{5}} - \frac {32 \, {\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} b^{3}}{\cos \left (d x + c\right )^{7}}}{1120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/1120*(35*a*b^2*(2*(3*sin(d*x + c)^5 - 8*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4
+ 3*sin(d*x + c)^2 - 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 70*a^3*(2*(3*sin(d*x + c)^3 - 5
*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) +
672*a^2*b/cos(d*x + c)^5 - 32*(7*cos(d*x + c)^2 - 5)*b^3/cos(d*x + c)^7)/d

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mupad [B]  time = 7.34, size = 423, normalized size = 2.66 \[ \frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-b^2\right )}{8\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,a^3}{4}+\frac {3\,a\,b^2}{8}\right )+\frac {6\,a^2\,b}{5}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {11\,a\,b^2}{2}-3\,a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {11\,a\,b^2}{2}-3\,a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (\frac {5\,a^3}{4}+\frac {3\,a\,b^2}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {9\,a^3}{4}+\frac {31\,a\,b^2}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {9\,a^3}{4}+\frac {31\,a\,b^2}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (12\,a^2\,b-4\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {12\,a^2\,b}{5}-\frac {4\,b^3}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (18\,a^2\,b+4\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (24\,a^2\,b-8\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {66\,a^2\,b}{5}+\frac {8\,b^3}{5}\right )-\frac {4\,b^3}{35}+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^3/cos(c + d*x)^5,x)

[Out]

(3*a*atanh(tan(c/2 + (d*x)/2))*(2*a^2 - b^2))/(8*d) - (tan(c/2 + (d*x)/2)*((3*a*b^2)/8 + (5*a^3)/4) + (6*a^2*b
)/5 + tan(c/2 + (d*x)/2)^3*((11*a*b^2)/2 - 3*a^3) - tan(c/2 + (d*x)/2)^11*((11*a*b^2)/2 - 3*a^3) - tan(c/2 + (
d*x)/2)^13*((3*a*b^2)/8 + (5*a^3)/4) + tan(c/2 + (d*x)/2)^5*((31*a*b^2)/8 + (9*a^3)/4) - tan(c/2 + (d*x)/2)^9*
((31*a*b^2)/8 + (9*a^3)/4) - tan(c/2 + (d*x)/2)^10*(12*a^2*b - 4*b^3) - tan(c/2 + (d*x)/2)^2*((12*a^2*b)/5 - (
4*b^3)/5) + tan(c/2 + (d*x)/2)^8*(18*a^2*b + 4*b^3) - tan(c/2 + (d*x)/2)^6*(24*a^2*b - 8*b^3) + tan(c/2 + (d*x
)/2)^4*((66*a^2*b)/5 + (8*b^3)/5) - (4*b^3)/35 + 6*a^2*b*tan(c/2 + (d*x)/2)^12)/(d*(7*tan(c/2 + (d*x)/2)^2 - 2
1*tan(c/2 + (d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 - 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/2 + (d*x)/2)^10 - 7*tan(
c/2 + (d*x)/2)^12 + tan(c/2 + (d*x)/2)^14 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sec ^{5}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*sec(c + d*x)**5, x)

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