3.572 \(\int \frac{\sec ^7(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=239 \[ \frac{5 \sec (c+d x) \left (4 a^2-2 a b \tan (c+d x)+b^2\right )}{2 b^5 d}-\frac{5 \sqrt{a^2+b^2} \left (4 a^2+b^2\right ) \sec (c+d x) \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{2 b^6 d \sqrt{\sec ^2(c+d x)}}-\frac{5 a \left (4 a^2+3 b^2\right ) \sec (c+d x) \sinh ^{-1}(\tan (c+d x))}{2 b^6 d \sqrt{\sec ^2(c+d x)}}+\frac{5 \sec ^3(c+d x) (4 a+b \tan (c+d x))}{6 b^3 d (a+b \tan (c+d x))}-\frac{\sec ^5(c+d x)}{2 b d (a+b \tan (c+d x))^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.243291, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3512, 733, 813, 815, 844, 215, 725, 206} \[ \frac{5 \sec (c+d x) \left (4 a^2-2 a b \tan (c+d x)+b^2\right )}{2 b^5 d}-\frac{5 \sqrt{a^2+b^2} \left (4 a^2+b^2\right ) \sec (c+d x) \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{2 b^6 d \sqrt{\sec ^2(c+d x)}}-\frac{5 a \left (4 a^2+3 b^2\right ) \sec (c+d x) \sinh ^{-1}(\tan (c+d x))}{2 b^6 d \sqrt{\sec ^2(c+d x)}}+\frac{5 \sec ^3(c+d x) (4 a+b \tan (c+d x))}{6 b^3 d (a+b \tan (c+d x))}-\frac{\sec ^5(c+d x)}{2 b d (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7/(a + b*Tan[c + d*x])^3,x]

[Out]

(-5*a*(4*a^2 + 3*b^2)*ArcSinh[Tan[c + d*x]]*Sec[c + d*x])/(2*b^6*d*Sqrt[Sec[c + d*x]^2]) - (5*Sqrt[a^2 + b^2]*
(4*a^2 + b^2)*ArcTanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Sec[c + d*x])/(2*b^6*d*Sqrt
[Sec[c + d*x]^2]) - Sec[c + d*x]^5/(2*b*d*(a + b*Tan[c + d*x])^2) + (5*Sec[c + d*x]^3*(4*a + b*Tan[c + d*x]))/
(6*b^3*d*(a + b*Tan[c + d*x])) + (5*Sec[c + d*x]*(4*a^2 + b^2 - 2*a*b*Tan[c + d*x]))/(2*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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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

Mathematica [C]  time = 2.21019, size = 688, normalized size = 2.88 \[ \frac{\sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (\frac{6 b^2 \left (a^2+b^2\right )^2 \sin (c+d x)}{a}+2 b \left (36 a^2+13 b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2+\frac{2 b \left (36 a^2+13 b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}-\frac{2 b \left (36 a^2+13 b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{6 b (a-i b) (a+i b) \left (8 a^2-b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}{a}+30 a \left (4 a^2+3 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2-30 a \left (4 a^2+3 b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2+60 \sqrt{a^2+b^2} \left (4 a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )+\frac{2 b^3 \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}-\frac{2 b^3 \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{b^2 (b-9 a) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{b^2 (9 a+b) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{12 b^6 d (a+b \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7/(a + b*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])*((6*b^2*(a^2 + b^2)^2*Sin[c + d*x])/a + (6*(a - I*b)*(a + I*
b)*b*(8*a^2 - b^2)*(a*Cos[c + d*x] + b*Sin[c + d*x]))/a + 2*b*(36*a^2 + 13*b^2)*(a*Cos[c + d*x] + b*Sin[c + d*
x])^2 + 60*Sqrt[a^2 + b^2]*(4*a^2 + b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]]*(a*Cos[c + d*x] +
b*Sin[c + d*x])^2 + 30*a*(4*a^2 + 3*b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a*Cos[c + d*x] + b*Sin[c +
d*x])^2 - 30*a*(4*a^2 + 3*b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 +
(b^2*(-9*a + b)*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*b^3*Sin[(c +
d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (2*b*(36*a^2 + 13*b^2)
*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (2*b^3*Sin[(c +
d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (b^2*(9*a + b)*(a*Cos[
c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (2*b*(36*a^2 + 13*b^2)*Sin[(c + d*x)/2
]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(12*b^6*d*(a + b*Tan[c + d*x])^
3)

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Maple [B]  time = 0.148, size = 1125, normalized size = 4.7 \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))^3,x)

[Out]

-2/d*b/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2/a^2*tan(1/2*d*x+1/2*c)^2+25/d/b^4/(tan(1/2*d*x+1/2*
c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a^3*tan(1/2*d*x+1/2*c)-7/d/b^4/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)
*b-a)^2*a^3*tan(1/2*d*x+1/2*c)^3-5/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a*tan(1/2*d*x+1/2
*c)^3-1/d/b/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2-1/3/d/b^3/(tan(1/2*d*x+1/2*c)-1)^3-1/2/d/b^3/(
tan(1/2*d*x+1/2*c)-1)^2-5/2/d/b^3/(tan(1/2*d*x+1/2*c)-1)+1/3/d/b^3/(tan(1/2*d*x+1/2*c)+1)^3-1/2/d/b^3/(tan(1/2
*d*x+1/2*c)+1)^2+5/2/d/b^3/(tan(1/2*d*x+1/2*c)+1)+25/d/b^4/(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))*a^2+23/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a*tan(1/2*d*x+1/2*c)-8
/d/b^5/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a^4*tan(1/2*d*x+1/2*c)^2+9/d/b^3/(tan(1/2*d*x+1/2*c
)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a^2*tan(1/2*d*x+1/2*c)^2+8/d/b^5/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c
)*b-a)^2*a^4+7/d/b^3/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a^2+5/d/b^2/(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))-3/2/d/b^4/(tan(1/2*d*x+1/2*c)-1)^2*a-6/d/b^5/(tan(1/2*d*x+1/2
*c)-1)*a^2-3/2/d/b^4/(tan(1/2*d*x+1/2*c)-1)*a+10/d*a^3/b^6*ln(tan(1/2*d*x+1/2*c)-1)+15/2/d*a/b^4*ln(tan(1/2*d*
x+1/2*c)-1)+2/d/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2/a*tan(1/2*d*x+1/2*c)^3-2/d/(tan(1/2*d*x+1/
2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2/a*tan(1/2*d*x+1/2*c)-10/d*a^3/b^6*ln(tan(1/2*d*x+1/2*c)+1)-15/2/d*a/b^4*l
n(tan(1/2*d*x+1/2*c)+1)+3/2/d/b^4/(tan(1/2*d*x+1/2*c)+1)^2*a+6/d/b^5/(tan(1/2*d*x+1/2*c)+1)*a^2-3/2/d/b^4/(tan
(1/2*d*x+1/2*c)+1)*a+15/d/b/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*tan(1/2*d*x+1/2*c)^2+20/d/b^6/
(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))*a^4

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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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.12065, size = 1299, normalized size = 5.44 \begin{align*} \frac{4 \, b^{5} + 30 \,{\left (4 \, a^{4} b + a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} + 20 \,{\left (2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left ({\left (4 \, a^{4} - 3 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (4 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) +{\left (4 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{3}\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 (4 \, a^{5} - a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (4 \, a^{4} b + 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) +{\left (4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \,{\left ({\left (4 \, a^{5} - a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (4 \, a^{4} b + 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) +{\left (4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \,{\left (a b^{4} \cos \left (d x + c\right ) - 6 \,{\left (3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \,{\left (2 \, a b^{7} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + b^{8} d \cos \left (d x + c\right )^{3} +{\left (a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{5}\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))^3,x, algorithm="fricas")

[Out]

1/12*(4*b^5 + 30*(4*a^4*b + a^2*b^3 - b^5)*cos(d*x + c)^4 + 20*(2*a^2*b^3 + b^5)*cos(d*x + c)^2 + 15*((4*a^4 -
3*a^2*b^2 - b^4)*cos(d*x + c)^5 + 2*(4*a^3*b + a*b^3)*cos(d*x + c)^4*sin(d*x + c) + (4*a^2*b^2 + b^4)*cos(d*x
+ c)^3)*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*((4*a^5 - a^3*b^2 - 3*a*b^4)*cos(d*x + c)^5 + 2*(4*a^4*b + 3*a^2*b^3)*cos(d*x + c)^4*sin(d*x +
c) + (4*a^3*b^2 + 3*a*b^4)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) + 15*((4*a^5 - a^3*b^2 - 3*a*b^4)*cos(d*x +
c)^5 + 2*(4*a^4*b + 3*a^2*b^3)*cos(d*x + c)^4*sin(d*x + c) + (4*a^3*b^2 + 3*a*b^4)*cos(d*x + c)^3)*log(-sin(d*
x + c) + 1) - 10*(a*b^4*cos(d*x + c) - 6*(3*a^3*b^2 + 2*a*b^4)*cos(d*x + c)^3)*sin(d*x + c))/(2*a*b^7*d*cos(d*
x + c)^4*sin(d*x + c) + b^8*d*cos(d*x + c)^3 + (a^2*b^6 - b^8)*d*cos(d*x + c)^5)

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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(sec(d*x+c)**7/(a+b*tan(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.63557, size = 689, normalized size = 2.88 \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))^3,x, algorithm="giac")

[Out]

-1/6*(15*(4*a^3 + 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^6 - 15*(4*a^3 + 3*a*b^2)*log(abs(tan(1/2*d*x +
1/2*c) - 1))/b^6 + 15*(4*a^4 + 5*a^2*b^2 + b^4)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/a
bs(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) + 2*(9*a*b*tan(1/2*d*x + 1/2*c)^
5 + 36*a^2*tan(1/2*d*x + 1/2*c)^4 + 18*b^2*tan(1/2*d*x + 1/2*c)^4 - 72*a^2*tan(1/2*d*x + 1/2*c)^2 - 24*b^2*tan
(1/2*d*x + 1/2*c)^2 - 9*a*b*tan(1/2*d*x + 1/2*c) + 36*a^2 + 14*b^2)/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*b^5) + 6*(
7*a^5*b*tan(1/2*d*x + 1/2*c)^3 + 5*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 8*a^6*tan
(1/2*d*x + 1/2*c)^2 - 9*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 + 2*b^6*tan(1/2*d*x
+ 1/2*c)^2 - 25*a^5*b*tan(1/2*d*x + 1/2*c) - 23*a^3*b^3*tan(1/2*d*x + 1/2*c) + 2*a*b^5*tan(1/2*d*x + 1/2*c) -
8*a^6 - 7*a^4*b^2 + a^2*b^4)/((a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^2*a^2*b^5))/d