3.560 \(\int \frac{\sec ^7(c+d x)}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=235 \[ -\frac{5 \sec (c+d x) \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{8 b^5 d}+\frac{5 a \left (a^2+b^2\right )^{3/2} \sec (c+d x) \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{b^6 d \sqrt{\sec ^2(c+d x)}}+\frac{5 \left (12 a^2 b^2+8 a^4+3 b^4\right ) \sec (c+d x) \sinh ^{-1}(\tan (c+d x))}{8 b^6 d \sqrt{\sec ^2(c+d x)}}-\frac{5 \sec ^3(c+d x) (4 a-3 b \tan (c+d x))}{12 b^3 d}-\frac{\sec ^5(c+d x)}{b d (a+b \tan (c+d x))} \]

[Out]

(5*(8*a^4 + 12*a^2*b^2 + 3*b^4)*ArcSinh[Tan[c + d*x]]*Sec[c + d*x])/(8*b^6*d*Sqrt[Sec[c + d*x]^2]) + (5*a*(a^2
+ b^2)^(3/2)*ArcTanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Sec[c + d*x])/(b^6*d*Sqrt[S
ec[c + d*x]^2]) - (5*Sec[c + d*x]^3*(4*a - 3*b*Tan[c + d*x]))/(12*b^3*d) - Sec[c + d*x]^5/(b*d*(a + b*Tan[c +
d*x])) - (5*Sec[c + d*x]*(8*a*(a^2 + b^2) - b*(4*a^2 + 3*b^2)*Tan[c + d*x]))/(8*b^5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.268137, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3512, 733, 815, 844, 215, 725, 206} \[ -\frac{5 \sec (c+d x) \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{8 b^5 d}+\frac{5 a \left (a^2+b^2\right )^{3/2} \sec (c+d x) \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{b^6 d \sqrt{\sec ^2(c+d x)}}+\frac{5 \left (12 a^2 b^2+8 a^4+3 b^4\right ) \sec (c+d x) \sinh ^{-1}(\tan (c+d x))}{8 b^6 d \sqrt{\sec ^2(c+d x)}}-\frac{5 \sec ^3(c+d x) (4 a-3 b \tan (c+d x))}{12 b^3 d}-\frac{\sec ^5(c+d x)}{b d (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7/(a + b*Tan[c + d*x])^2,x]

[Out]

(5*(8*a^4 + 12*a^2*b^2 + 3*b^4)*ArcSinh[Tan[c + d*x]]*Sec[c + d*x])/(8*b^6*d*Sqrt[Sec[c + d*x]^2]) + (5*a*(a^2
+ b^2)^(3/2)*ArcTanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Sec[c + d*x])/(b^6*d*Sqrt[S
ec[c + d*x]^2]) - (5*Sec[c + d*x]^3*(4*a - 3*b*Tan[c + d*x]))/(12*b^3*d) - Sec[c + d*x]^5/(b*d*(a + b*Tan[c +
d*x])) - (5*Sec[c + d*x]*(8*a*(a^2 + b^2) - b*(4*a^2 + 3*b^2)*Tan[c + d*x]))/(8*b^5*d)

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]

Rule 733

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

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

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 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

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 \frac{\sec ^7(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=\frac{\sec (c+d x) \operatorname{Subst}\left (\int \frac{\left (1+\frac{x^2}{b^2}\right )^{5/2}}{(a+x)^2} \, dx,x,b \tan (c+d x)\right )}{b d \sqrt{\sec ^2(c+d x)}}\\ &=-\frac{\sec ^5(c+d x)}{b d (a+b \tan (c+d x))}+\frac{(5 \sec (c+d x)) \operatorname{Subst}\left (\int \frac{x \left (1+\frac{x^2}{b^2}\right )^{3/2}}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^3 d \sqrt{\sec ^2(c+d x)}}\\ &=-\frac{5 \sec ^3(c+d x) (4 a-3 b \tan (c+d x))}{12 b^3 d}-\frac{\sec ^5(c+d x)}{b d (a+b \tan (c+d x))}+\frac{(5 \sec (c+d x)) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b^2}+\frac{\left (4 a^2+3 b^2\right ) x}{b^4}\right ) \sqrt{1+\frac{x^2}{b^2}}}{a+x} \, dx,x,b \tan (c+d x)\right )}{4 b d \sqrt{\sec ^2(c+d x)}}\\ &=-\frac{5 \sec ^3(c+d x) (4 a-3 b \tan (c+d x))}{12 b^3 d}-\frac{\sec ^5(c+d x)}{b d (a+b \tan (c+d x))}-\frac{5 \sec (c+d x) \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{8 b^5 d}+\frac{(5 b \sec (c+d x)) \operatorname{Subst}\left (\int \frac{-\frac{a \left (4 a^2+5 b^2\right )}{b^6}+\frac{\left (8 a^4+12 a^2 b^2+3 b^4\right ) x}{b^8}}{(a+x) \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{8 d \sqrt{\sec ^2(c+d x)}}\\ &=-\frac{5 \sec ^3(c+d x) (4 a-3 b \tan (c+d x))}{12 b^3 d}-\frac{\sec ^5(c+d x)}{b d (a+b \tan (c+d x))}-\frac{5 \sec (c+d x) \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{8 b^5 d}-\frac{\left (5 a \left (a^2+b^2\right )^2 \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{(a+x) \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{b^7 d \sqrt{\sec ^2(c+d x)}}+\frac{\left (5 \left (8 a^4+12 a^2 b^2+3 b^4\right ) \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 b^7 d \sqrt{\sec ^2(c+d x)}}\\ &=\frac{5 \left (8 a^4+12 a^2 b^2+3 b^4\right ) \sinh ^{-1}(\tan (c+d x)) \sec (c+d x)}{8 b^6 d \sqrt{\sec ^2(c+d x)}}-\frac{5 \sec ^3(c+d x) (4 a-3 b \tan (c+d x))}{12 b^3 d}-\frac{\sec ^5(c+d x)}{b d (a+b \tan (c+d x))}-\frac{5 \sec (c+d x) \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{8 b^5 d}+\frac{\left (5 a \left (a^2+b^2\right )^2 \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a^2}{b^2}-x^2} \, dx,x,\frac{1-\frac{a \tan (c+d x)}{b}}{\sqrt{\sec ^2(c+d x)}}\right )}{b^7 d \sqrt{\sec ^2(c+d x)}}\\ &=\frac{5 \left (8 a^4+12 a^2 b^2+3 b^4\right ) \sinh ^{-1}(\tan (c+d x)) \sec (c+d x)}{8 b^6 d \sqrt{\sec ^2(c+d x)}}+\frac{5 a \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{b \left (1-\frac{a \tan (c+d x)}{b}\right )}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right ) \sec (c+d x)}{b^6 d \sqrt{\sec ^2(c+d x)}}-\frac{5 \sec ^3(c+d x) (4 a-3 b \tan (c+d x))}{12 b^3 d}-\frac{\sec ^5(c+d x)}{b d (a+b \tan (c+d x))}-\frac{5 \sec (c+d x) \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{8 b^5 d}\\ \end{align*}

Mathematica [C]  time = 6.17926, size = 1152, normalized size = 4.9 \[ \frac{10 i a (a+i b) (i a+b) \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{\sqrt{a^2+b^2} \left (a \sin \left (\frac{1}{2} (c+d x)\right )-b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\cos \left (\frac{1}{2} (c+d x)\right ) a^2+b^2 \cos \left (\frac{1}{2} (c+d x)\right )}\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{b^6 d (a+b \tan (c+d x))^2}-\frac{5 \left (8 a^4+12 b^2 a^2+3 b^4\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{8 b^6 d (a+b \tan (c+d x))^2}+\frac{5 \left (8 a^4+12 b^2 a^2+3 b^4\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{8 b^6 d (a+b \tan (c+d x))^2}-\frac{a \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{3 b^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 (a+b \tan (c+d x))^2}+\frac{\left (-12 \sin \left (\frac{1}{2} (c+d x)\right ) a^3-13 b^2 \sin \left (\frac{1}{2} (c+d x)\right ) a\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{3 b^5 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^2}+\frac{\left (12 \sin \left (\frac{1}{2} (c+d x)\right ) a^3+13 b^2 \sin \left (\frac{1}{2} (c+d x)\right ) a\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{3 b^5 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^2}-\frac{a \left (12 a^2+13 b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{3 b^5 d (a+b \tan (c+d x))^2}+\frac{\left (36 a^2-8 b a+21 b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{48 b^4 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2 (a+b \tan (c+d x))^2}+\frac{\left (-36 a^2-8 b a-21 b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{48 b^4 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^2 (a+b \tan (c+d x))^2}+\frac{a \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{3 b^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 (a+b \tan (c+d x))^2}+\frac{(a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{16 b^2 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4 (a+b \tan (c+d x))^2}-\frac{(a \cos (c+d x)+b \sin (c+d x))^2 \sec ^2(c+d x)}{16 b^2 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^4 (a+b \tan (c+d x))^2}-\frac{(a-i b)^2 (a+i b)^2 (a \cos (c+d x)+b \sin (c+d x)) \sec ^2(c+d x)}{b^5 d (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7/(a + b*Tan[c + d*x])^2,x]

[Out]

-(((a - I*b)^2*(a + I*b)^2*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(b^5*d*(a + b*Tan[c + d*x])^2)) -
(a*(12*a^2 + 13*b^2)*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(3*b^5*d*(a + b*Tan[c + d*x])^2) + (
(10*I)*a*(a + I*b)*(I*a + b)*Sqrt[a^2 + b^2]*ArcTanh[(Sqrt[a^2 + b^2]*(-(b*Cos[(c + d*x)/2]) + a*Sin[(c + d*x)
/2]))/(a^2*Cos[(c + d*x)/2] + b^2*Cos[(c + d*x)/2])]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(b^6*
d*(a + b*Tan[c + d*x])^2) - (5*(8*a^4 + 12*a^2*b^2 + 3*b^4)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sec[c + d
*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(8*b^6*d*(a + b*Tan[c + d*x])^2) + (5*(8*a^4 + 12*a^2*b^2 + 3*b^4)*
Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(8*b^6*d*(a + b*T
an[c + d*x])^2) + (Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(16*b^2*d*(Cos[(c + d*x)/2] - Sin[(c +
d*x)/2])^4*(a + b*Tan[c + d*x])^2) + ((36*a^2 - 8*a*b + 21*b^2)*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x
])^2)/(48*b^4*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a + b*Tan[c + d*x])^2) - (a*Sec[c + d*x]^2*Sin[(c + d
*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(3*b^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(a + b*Tan[c + d*
x])^2) - (Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(16*b^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^
4*(a + b*Tan[c + d*x])^2) + (a*Sec[c + d*x]^2*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(3*b^3*d*(
Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(a + b*Tan[c + d*x])^2) + ((-36*a^2 - 8*a*b - 21*b^2)*Sec[c + d*x]^2*(a
*Cos[c + d*x] + b*Sin[c + d*x])^2)/(48*b^4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a + b*Tan[c + d*x])^2) +
(Sec[c + d*x]^2*(-12*a^3*Sin[(c + d*x)/2] - 13*a*b^2*Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(
3*b^5*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^2) + (Sec[c + d*x]^2*(12*a^3*Sin[(c + d*x)/
2] + 13*a*b^2*Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(3*b^5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2])*(a + b*Tan[c + d*x])^2)

________________________________________________________________________________________

Maple [B]  time = 0.129, size = 989, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7/(a+b*tan(d*x+c))^2,x)

[Out]

11/8/d/b^2/(tan(1/2*d*x+1/2*c)-1)^2-15/8/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)+9/8/d/b^2/(tan(1/2*d*x+1/2*c)-1)-1/4/d
/b^2/(tan(1/2*d*x+1/2*c)+1)^4+1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)^3-11/8/d/b^2/(tan(1/2*d*x+1/2*c)+1)^2+15/8/d/b^
2*ln(tan(1/2*d*x+1/2*c)+1)+9/8/d/b^2/(tan(1/2*d*x+1/2*c)+1)+2/d/b/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)
*b-a)+1/4/d/b^2/(tan(1/2*d*x+1/2*c)-1)^4+1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)^3-2/3/d/b^3/(tan(1/2*d*x+1/2*c)+1)^3
*a+2/3/d/b^3/(tan(1/2*d*x+1/2*c)-1)^3*a+3/2/d/b^4/(tan(1/2*d*x+1/2*c)-1)^2*a^2+1/d/b^3/(tan(1/2*d*x+1/2*c)-1)^
2*a-5/d/b^6*ln(tan(1/2*d*x+1/2*c)-1)*a^4-15/2/d/b^4*ln(tan(1/2*d*x+1/2*c)-1)*a^2+4/d/b^5/(tan(1/2*d*x+1/2*c)-1
)*a^3+3/2/d/b^4/(tan(1/2*d*x+1/2*c)-1)*a^2-20/d/b^4*a^3/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*
b)/(a^2+b^2)^(1/2))-10/d/b^2*a/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))+2/d/b
^4/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)*a^3*tan(1/2*d*x+1/2*c)+4/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*
tan(1/2*d*x+1/2*c)*b-a)*a*tan(1/2*d*x+1/2*c)-10/d/b^6*a^5/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-
2*b)/(a^2+b^2)^(1/2))+5/d/b^6*ln(tan(1/2*d*x+1/2*c)+1)*a^4+15/2/d/b^4*ln(tan(1/2*d*x+1/2*c)+1)*a^2-4/d/b^5/(ta
n(1/2*d*x+1/2*c)+1)*a^3+3/2/d/b^4/(tan(1/2*d*x+1/2*c)+1)*a^2-5/d/b^3/(tan(1/2*d*x+1/2*c)+1)*a+5/d/b^3/(tan(1/2
*d*x+1/2*c)-1)*a+2/d/b^5/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)*a^4+4/d/b^3/(tan(1/2*d*x+1/2*c)^2*a
-2*tan(1/2*d*x+1/2*c)*b-a)*a^2-3/2/d/b^4/(tan(1/2*d*x+1/2*c)+1)^2*a^2+1/d/b^3/(tan(1/2*d*x+1/2*c)+1)^2*a+2/d/(
tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)/a*tan(1/2*d*x+1/2*c)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.88144, size = 1118, normalized size = 4.76 \begin{align*} \frac{12 \, b^{5} - 30 \,{\left (8 \, a^{4} b + 12 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} + 10 \,{\left (4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} + 120 \,{\left ({\left (a^{4} + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{5} +{\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 15 \,{\left ({\left (8 \, a^{5} + 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} +{\left (8 \, a^{4} b + 12 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (8 \, a^{5} + 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} +{\left (8 \, a^{4} b + 12 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \,{\left (2 \, a b^{4} \cos \left (d x + c\right ) + 3 \,{\left (4 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{48 \,{\left (a b^{6} d \cos \left (d x + c\right )^{5} + b^{7} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/48*(12*b^5 - 30*(8*a^4*b + 12*a^2*b^3 + 3*b^5)*cos(d*x + c)^4 + 10*(4*a^2*b^3 + 3*b^5)*cos(d*x + c)^2 + 120*
((a^4 + a^2*b^2)*cos(d*x + c)^5 + (a^3*b + a*b^3)*cos(d*x + c)^4*sin(d*x + c))*sqrt(a^2 + b^2)*log((2*a*b*cos(
d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d
*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)) + 15*((8*a^5 + 12*a^3*b^2 + 3*
a*b^4)*cos(d*x + c)^5 + (8*a^4*b + 12*a^2*b^3 + 3*b^5)*cos(d*x + c)^4*sin(d*x + c))*log(sin(d*x + c) + 1) - 15
*((8*a^5 + 12*a^3*b^2 + 3*a*b^4)*cos(d*x + c)^5 + (8*a^4*b + 12*a^2*b^3 + 3*b^5)*cos(d*x + c)^4*sin(d*x + c))*
log(-sin(d*x + c) + 1) - 10*(2*a*b^4*cos(d*x + c) + 3*(4*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^3)*sin(d*x + c))/(a*b
^6*d*cos(d*x + c)^5 + b^7*d*cos(d*x + c)^4*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{7}{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7/(a+b*tan(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)**7/(a + b*tan(c + d*x))**2, x)

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Giac [B]  time = 1.6479, size = 716, normalized size = 3.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(15*(8*a^4 + 12*a^2*b^2 + 3*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^6 - 15*(8*a^4 + 12*a^2*b^2 + 3*b^4)
*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^6 + 120*(a^5 + 2*a^3*b^2 + a*b^4)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b
- 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^6) + 48*(a^4
*b*tan(1/2*d*x + 1/2*c) + 2*a^2*b^3*tan(1/2*d*x + 1/2*c) + b^5*tan(1/2*d*x + 1/2*c) + a^5 + 2*a^3*b^2 + a*b^4)
/((a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)*a*b^5) + 2*(36*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 27*b
^3*tan(1/2*d*x + 1/2*c)^7 + 96*a^3*tan(1/2*d*x + 1/2*c)^6 + 144*a*b^2*tan(1/2*d*x + 1/2*c)^6 - 36*a^2*b*tan(1/
2*d*x + 1/2*c)^5 - 3*b^3*tan(1/2*d*x + 1/2*c)^5 - 288*a^3*tan(1/2*d*x + 1/2*c)^4 - 336*a*b^2*tan(1/2*d*x + 1/2
*c)^4 - 36*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 3*b^3*tan(1/2*d*x + 1/2*c)^3 + 288*a^3*tan(1/2*d*x + 1/2*c)^2 + 304*
a*b^2*tan(1/2*d*x + 1/2*c)^2 + 36*a^2*b*tan(1/2*d*x + 1/2*c) + 27*b^3*tan(1/2*d*x + 1/2*c) - 96*a^3 - 112*a*b^
2)/((tan(1/2*d*x + 1/2*c)^2 - 1)^4*b^5))/d