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

Optimal. Leaf size=126 \[ \frac {a \left (4 a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \sec ^3(c+d x) \left (8 \left (6 a^2-b^2\right )+21 a b \tan (c+d x)\right )}{60 d}+\frac {a \left (4 a^2-3 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) (a+b \tan (c+d x))^2}{5 d} \]

[Out]

1/8*a*(4*a^2-3*b^2)*arctanh(sin(d*x+c))/d+1/8*a*(4*a^2-3*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/5*b*sec(d*x+c)^3*(a+b*
tan(d*x+c))^2/d+1/60*b*sec(d*x+c)^3*(48*a^2-8*b^2+21*a*b*tan(d*x+c))/d

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Rubi [A]  time = 0.13, antiderivative size = 144, normalized size of antiderivative = 1.14, number of steps used = 5, 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} \[ \frac {b \sec ^3(c+d x) \left (8 \left (6 a^2-b^2\right )+21 a b \tan (c+d x)\right )}{60 d}+\frac {a \left (4 a^2-3 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a \left (4 a^2-3 b^2\right ) \sec (c+d x) \sinh ^{-1}(\tan (c+d x))}{8 d \sqrt {\sec ^2(c+d x)}}+\frac {b \sec ^3(c+d x) (a+b \tan (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*a^2 - 3*b^2)*ArcSinh[Tan[c + d*x]]*Sec[c + d*x])/(8*d*Sqrt[Sec[c + d*x]^2]) + (a*(4*a^2 - 3*b^2)*Sec[c +
d*x]*Tan[c + d*x])/(8*d) + (b*Sec[c + d*x]^3*(a + b*Tan[c + d*x])^2)/(5*d) + (b*Sec[c + d*x]^3*(8*(6*a^2 - b^
2) + 21*a*b*Tan[c + d*x]))/(60*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 ^3(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac {\sec (c+d x) \operatorname {Subst}\left (\int (a+x)^3 \sqrt {1+\frac {x^2}{b^2}} \, dx,x,b \tan (c+d x)\right )}{b d \sqrt {\sec ^2(c+d x)}}\\ &=\frac {b \sec ^3(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac {(b \sec (c+d x)) \operatorname {Subst}\left (\int (a+x) \left (-2+\frac {5 a^2}{b^2}+\frac {7 a x}{b^2}\right ) \sqrt {1+\frac {x^2}{b^2}} \, dx,x,b \tan (c+d x)\right )}{5 d \sqrt {\sec ^2(c+d x)}}\\ &=\frac {b \sec ^3(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac {b \sec ^3(c+d x) \left (8 \left (6 a^2-b^2\right )+21 a b \tan (c+d x)\right )}{60 d}-\frac {\left (a \left (3-\frac {4 a^2}{b^2}\right ) b \sec (c+d x)\right ) \operatorname {Subst}\left (\int \sqrt {1+\frac {x^2}{b^2}} \, dx,x,b \tan (c+d x)\right )}{4 d \sqrt {\sec ^2(c+d x)}}\\ &=\frac {a \left (4 a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac {b \sec ^3(c+d x) \left (8 \left (6 a^2-b^2\right )+21 a b \tan (c+d x)\right )}{60 d}-\frac {\left (a \left (3-\frac {4 a^2}{b^2}\right ) b \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 )}{8 d \sqrt {\sec ^2(c+d x)}}\\ &=\frac {a \left (4 a^2-3 b^2\right ) \sinh ^{-1}(\tan (c+d x)) \sec (c+d x)}{8 d \sqrt {\sec ^2(c+d x)}}+\frac {a \left (4 a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac {b \sec ^3(c+d x) \left (8 \left (6 a^2-b^2\right )+21 a b \tan (c+d x)\right )}{60 d}\\ \end {align*}

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Mathematica [B]  time = 1.30, size = 464, normalized size = 3.68 \[ \frac {\sec ^5(c+d x) \left (240 a^3 \sin (2 (c+d x))+120 a^3 \sin (4 (c+d x))-300 a^3 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-60 a^3 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+300 a^3 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+60 a^3 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+320 \left (3 a^2 b-b^3\right ) \cos (2 (c+d x))-150 a \left (4 a^2-3 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+960 a^2 b+540 a b^2 \sin (2 (c+d x))-90 a b^2 \sin (4 (c+d x))+225 a b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+45 a b^2 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-225 a b^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-45 a b^2 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+64 b^3\right )}{1920 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^5*(960*a^2*b + 64*b^3 + 320*(3*a^2*b - b^3)*Cos[2*(c + d*x)] - 300*a^3*Cos[3*(c + d*x)]*Log[Cos[
(c + d*x)/2] - Sin[(c + d*x)/2]] + 225*a*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 60*a^
3*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 45*a*b^2*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]] - 150*a*(4*a^2 - 3*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]]) + 300*a^3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 225*a*b^
2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 60*a^3*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]] - 45*a*b^2*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 240*a^3*Sin[2*(c + d*x
)] + 540*a*b^2*Sin[2*(c + d*x)] + 120*a^3*Sin[4*(c + d*x)] - 90*a*b^2*Sin[4*(c + d*x)]))/(1920*d)

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fricas [A]  time = 0.70, size = 147, normalized size = 1.17 \[ \frac {15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 48 \, b^{3} + 80 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (6 \, a b^{2} \cos \left (d x + c\right ) + {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(4*a^3 - 3*a*b^2)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*a^3 - 3*a*b^2)*cos(d*x + c)^5*log(-si
n(d*x + c) + 1) + 48*b^3 + 80*(3*a^2*b - b^3)*cos(d*x + c)^2 + 30*(6*a*b^2*cos(d*x + c) + (4*a^3 - 3*a*b^2)*co
s(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [B]  time = 2.61, size = 333, normalized size = 2.64 \[ \frac {15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (60 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 360 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 270 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 240 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 480 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 80 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 270 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 240 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 80 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 60 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 45 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, a^{2} b + 16 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) + 2*(60*a^3*tan(1/2*d*x + 1/2*c)^9 + 45*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 360*a^2*b*tan(1/2*d*x + 1/2*
c)^8 - 120*a^3*tan(1/2*d*x + 1/2*c)^7 + 270*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*a^2*b*tan(1/2*d*x + 1/2*c)^6 -
240*b^3*tan(1/2*d*x + 1/2*c)^6 - 480*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 80*b^3*tan(1/2*d*x + 1/2*c)^4 + 120*a^3*ta
n(1/2*d*x + 1/2*c)^3 - 270*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 240*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 80*b^3*tan(1/2*d*
x + 1/2*c)^2 - 60*a^3*tan(1/2*d*x + 1/2*c) - 45*a*b^2*tan(1/2*d*x + 1/2*c) - 120*a^2*b + 16*b^3)/(tan(1/2*d*x
+ 1/2*c)^2 - 1)^5)/d

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maple [B]  time = 0.53, size = 256, normalized size = 2.03 \[ \frac {a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a^{2} b}{d \cos \left (d x +c \right )^{3}}+\frac {3 b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {3 a \,b^{2} \sin \left (d x +c \right )}{8 d}-\frac {3 b^{2} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}-\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )}-\frac {b^{3} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{15 d}-\frac {2 b^{3} \cos \left (d x +c \right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^3*sec(d*x+c)*tan(d*x+c)/d+1/2/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^2*b/cos(d*x+c)^3+3/4/d*b^2*a*sin(d*x
+c)^3/cos(d*x+c)^4+3/8/d*b^2*a*sin(d*x+c)^3/cos(d*x+c)^2+3/8*a*b^2*sin(d*x+c)/d-3/8/d*b^2*a*ln(sec(d*x+c)+tan(
d*x+c))+1/5/d*b^3*sin(d*x+c)^4/cos(d*x+c)^5+1/15/d*b^3*sin(d*x+c)^4/cos(d*x+c)^3-1/15/d*b^3*sin(d*x+c)^4/cos(d
*x+c)-1/15/d*b^3*cos(d*x+c)*sin(d*x+c)^2-2/15/d*b^3*cos(d*x+c)

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maxima [A]  time = 0.38, size = 157, normalized size = 1.25 \[ \frac {45 \, a b^{2} {\left (\frac {2 \, {\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {240 \, a^{2} b}{\cos \left (d x + c\right )^{3}} - \frac {16 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} b^{3}}{\cos \left (d x + c\right )^{5}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(45*a*b^2*(2*(sin(d*x + c)^3 + sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) - 60*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(
d*x + c) - 1)) + 240*a^2*b/cos(d*x + c)^3 - 16*(5*cos(d*x + c)^2 - 3)*b^3/cos(d*x + c)^5)/d

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mupad [B]  time = 7.32, size = 293, normalized size = 2.33 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (a^3+\frac {3\,a\,b^2}{4}\right )-2\,a^2\,b-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {9\,a\,b^2}{2}-2\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {9\,a\,b^2}{2}-2\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (4\,a^2\,b-\frac {4\,b^3}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,a^2\,b+\frac {4\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (12\,a^2\,b-4\,b^3\right )+\frac {4\,b^3}{15}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^3+\frac {3\,a\,b^2}{4}\right )-6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a\,b^2}{4}-a^3\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^9*((3*a*b^2)/4 + a^3) - 2*a^2*b - tan(c/2 + (d*x)/2)^3*((9*a*b^2)/2 - 2*a^3) + tan(c/2 + (
d*x)/2)^7*((9*a*b^2)/2 - 2*a^3) + tan(c/2 + (d*x)/2)^2*(4*a^2*b - (4*b^3)/3) - tan(c/2 + (d*x)/2)^4*(8*a^2*b +
(4*b^3)/3) + tan(c/2 + (d*x)/2)^6*(12*a^2*b - 4*b^3) + (4*b^3)/15 - tan(c/2 + (d*x)/2)*((3*a*b^2)/4 + a^3) -
6*a^2*b*tan(c/2 + (d*x)/2)^8)/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 -
5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1)) - (atanh(tan(c/2 + (d*x)/2))*((3*a*b^2)/4 - a^3))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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