3.138 \(\int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=161 \[ \frac {2 \left (3 a^2+b^2\right ) \log (\tan (c+d x))}{b^5 d}+\frac {2 \left (3 a^2+b^2\right ) \log (a \cot (c+d x)+b)}{b^5 d}-\frac {\left (3 a^2-b^2\right ) \left (a^2+b^2\right )}{a^2 b^4 d (a \cot (c+d x)+b)}-\frac {\left (a^2+b^2\right )^2}{2 a^2 b^3 d (a \cot (c+d x)+b)^2}-\frac {3 a \tan (c+d x)}{b^4 d}+\frac {\tan ^2(c+d x)}{2 b^3 d} \]

[Out]

-1/2*(a^2+b^2)^2/a^2/b^3/d/(b+a*cot(d*x+c))^2-(3*a^2-b^2)*(a^2+b^2)/a^2/b^4/d/(b+a*cot(d*x+c))+2*(3*a^2+b^2)*l
n(b+a*cot(d*x+c))/b^5/d+2*(3*a^2+b^2)*ln(tan(d*x+c))/b^5/d-3*a*tan(d*x+c)/b^4/d+1/2*tan(d*x+c)^2/b^3/d

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Rubi [A]  time = 0.17, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3088, 894} \[ -\frac {\left (a^2+b^2\right )^2}{2 a^2 b^3 d (a \cot (c+d x)+b)^2}-\frac {\left (3 a^2-b^2\right ) \left (a^2+b^2\right )}{a^2 b^4 d (a \cot (c+d x)+b)}+\frac {2 \left (3 a^2+b^2\right ) \log (\tan (c+d x))}{b^5 d}+\frac {2 \left (3 a^2+b^2\right ) \log (a \cot (c+d x)+b)}{b^5 d}-\frac {3 a \tan (c+d x)}{b^4 d}+\frac {\tan ^2(c+d x)}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2 + b^2)^2/(2*a^2*b^3*d*(b + a*Cot[c + d*x])^2) - ((3*a^2 - b^2)*(a^2 + b^2))/(a^2*b^4*d*(b + a*Cot[c + d*
x])) + (2*(3*a^2 + b^2)*Log[b + a*Cot[c + d*x]])/(b^5*d) + (2*(3*a^2 + b^2)*Log[Tan[c + d*x]])/(b^5*d) - (3*a*
Tan[c + d*x])/(b^4*d) + Tan[c + d*x]^2/(2*b^3*d)

Rule 894

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

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^3 (b+a x)^3} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {1}{b^3 x^3}-\frac {3 a}{b^4 x^2}+\frac {2 \left (3 a^2+b^2\right )}{b^5 x}-\frac {\left (a^2+b^2\right )^2}{a b^3 (b+a x)^3}+\frac {-3 a^4-2 a^2 b^2+b^4}{a b^4 (b+a x)^2}-\frac {2 a \left (3 a^2+b^2\right )}{b^5 (b+a x)}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {\left (a^2+b^2\right )^2}{2 a^2 b^3 d (b+a \cot (c+d x))^2}-\frac {\left (3 a^2-b^2\right ) \left (a^2+b^2\right )}{a^2 b^4 d (b+a \cot (c+d x))}+\frac {2 \left (3 a^2+b^2\right ) \log (b+a \cot (c+d x))}{b^5 d}+\frac {2 \left (3 a^2+b^2\right ) \log (\tan (c+d x))}{b^5 d}-\frac {3 a \tan (c+d x)}{b^4 d}+\frac {\tan ^2(c+d x)}{2 b^3 d}\\ \end {align*}

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Mathematica [A]  time = 3.00, size = 140, normalized size = 0.87 \[ \frac {-2 a \left (-\frac {a^2+b^2}{a+b \tan (c+d x)}-2 a \log (a+b \tan (c+d x))+b \tan (c+d x)\right )+2 \left (a^2+b^2\right ) \left (\frac {3 a^2+4 a b \tan (c+d x)-b^2}{2 (a+b \tan (c+d x))^2}+\log (a+b \tan (c+d x))\right )+\frac {b^4 \sec ^4(c+d x)}{2 (a+b \tan (c+d x))^2}}{b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.69, size = 354, normalized size = 2.20 \[ \frac {24 \, a^{2} b^{2} \cos \left (d x + c\right )^{4} + b^{4} - 2 \, {\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left ({\left (3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 2 \, {\left ({\left (3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) - 4 \, {\left (a b^{3} \cos \left (d x + c\right ) + 3 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (2 \, a b^{6} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + b^{7} d \cos \left (d x + c\right )^{2} + {\left (a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(24*a^2*b^2*cos(d*x + c)^4 + b^4 - 2*(9*a^2*b^2 + b^4)*cos(d*x + c)^2 + 2*((3*a^4 - 2*a^2*b^2 - b^4)*cos(d
*x + c)^4 + 2*(3*a^3*b + a*b^3)*cos(d*x + c)^3*sin(d*x + c) + (3*a^2*b^2 + b^4)*cos(d*x + c)^2)*log(2*a*b*cos(
d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 2*((3*a^4 - 2*a^2*b^2 - b^4)*cos(d*x + c)^4 + 2*(3
*a^3*b + a*b^3)*cos(d*x + c)^3*sin(d*x + c) + (3*a^2*b^2 + b^4)*cos(d*x + c)^2)*log(cos(d*x + c)^2) - 4*(a*b^3
*cos(d*x + c) + 3*(a^3*b - a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/(2*a*b^6*d*cos(d*x + c)^3*sin(d*x + c) + b^7*d
*cos(d*x + c)^2 + (a^2*b^5 - b^7)*d*cos(d*x + c)^4)

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giac [A]  time = 0.60, size = 140, normalized size = 0.87 \[ \frac {\frac {4 \, {\left (3 \, a^{2} + b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{5}} + \frac {b^{3} \tan \left (d x + c\right )^{2} - 6 \, a b^{2} \tan \left (d x + c\right )}{b^{6}} - \frac {18 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 6 \, b^{4} \tan \left (d x + c\right )^{2} + 28 \, a^{3} b \tan \left (d x + c\right ) + 4 \, a b^{3} \tan \left (d x + c\right ) + 11 \, a^{4} + b^{4}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} b^{5}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*(3*a^2 + b^2)*log(abs(b*tan(d*x + c) + a))/b^5 + (b^3*tan(d*x + c)^2 - 6*a*b^2*tan(d*x + c))/b^6 - (18*
a^2*b^2*tan(d*x + c)^2 + 6*b^4*tan(d*x + c)^2 + 28*a^3*b*tan(d*x + c) + 4*a*b^3*tan(d*x + c) + 11*a^4 + b^4)/(
(b*tan(d*x + c) + a)^2*b^5))/d

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maple [A]  time = 0.40, size = 184, normalized size = 1.14 \[ \frac {\tan ^{2}\left (d x +c \right )}{2 b^{3} d}-\frac {3 a \tan \left (d x +c \right )}{b^{4} d}+\frac {4 a^{3}}{d \,b^{5} \left (a +b \tan \left (d x +c \right )\right )}+\frac {4 a}{d \,b^{3} \left (a +b \tan \left (d x +c \right )\right )}+\frac {6 \ln \left (a +b \tan \left (d x +c \right )\right ) a^{2}}{d \,b^{5}}+\frac {2 \ln \left (a +b \tan \left (d x +c \right )\right )}{d \,b^{3}}-\frac {a^{4}}{2 d \,b^{5} \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {a^{2}}{d \,b^{3} \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {1}{2 b d \left (a +b \tan \left (d x +c \right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*tan(d*x+c)^2/b^3/d-3*a*tan(d*x+c)/b^4/d+4/d*a^3/b^5/(a+b*tan(d*x+c))+4/d*a/b^3/(a+b*tan(d*x+c))+6/d/b^5*ln
(a+b*tan(d*x+c))*a^2+2/d/b^3*ln(a+b*tan(d*x+c))-1/2/d/b^5/(a+b*tan(d*x+c))^2*a^4-1/d/b^3/(a+b*tan(d*x+c))^2*a^
2-1/2/b/d/(a+b*tan(d*x+c))^2

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maxima [B]  time = 0.37, size = 652, normalized size = 4.05 \[ -\frac {2 \, {\left (\frac {\frac {{\left (6 \, a^{5} + 2 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {{\left (18 \, a^{4} b + 6 \, a^{2} b^{3} - b^{5}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {{\left (18 \, a^{5} - 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {2 \, {\left (18 \, a^{4} b + 8 \, a^{2} b^{3} - b^{5}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (18 \, a^{5} - 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {{\left (18 \, a^{4} b + 6 \, a^{2} b^{3} - b^{5}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {{\left (6 \, a^{5} + 2 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4} b^{4} + \frac {4 \, a^{3} b^{5} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {12 \, a^{3} b^{5} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {12 \, a^{3} b^{5} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {4 \, a^{3} b^{5} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {a^{4} b^{4} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {4 \, {\left (a^{4} b^{4} - a^{2} b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, {\left (3 \, a^{4} b^{4} - 4 \, a^{2} b^{6}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, {\left (a^{4} b^{4} - a^{2} b^{6}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {{\left (3 \, a^{2} + b^{2}\right )} \log \left (-a - \frac {2 \, b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{b^{5}} + \frac {{\left (3 \, a^{2} + b^{2}\right )} \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b^{5}} + \frac {{\left (3 \, a^{2} + b^{2}\right )} \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b^{5}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(((6*a^5 + 2*a^3*b^2 - a*b^4)*sin(d*x + c)/(cos(d*x + c) + 1) + (18*a^4*b + 6*a^2*b^3 - b^5)*sin(d*x + c)^2
/(cos(d*x + c) + 1)^2 - (18*a^5 - 2*a^3*b^2 - 3*a*b^4)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 2*(18*a^4*b + 8*a
^2*b^3 - b^5)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + (18*a^5 - 2*a^3*b^2 - 3*a*b^4)*sin(d*x + c)^5/(cos(d*x + c
) + 1)^5 + (18*a^4*b + 6*a^2*b^3 - b^5)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - (6*a^5 + 2*a^3*b^2 - a*b^4)*sin(
d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^4*b^4 + 4*a^3*b^5*sin(d*x + c)/(cos(d*x + c) + 1) - 12*a^3*b^5*sin(d*x + c
)^3/(cos(d*x + c) + 1)^3 + 12*a^3*b^5*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 4*a^3*b^5*sin(d*x + c)^7/(cos(d*x
+ c) + 1)^7 + a^4*b^4*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 4*(a^4*b^4 - a^2*b^6)*sin(d*x + c)^2/(cos(d*x + c)
+ 1)^2 + 2*(3*a^4*b^4 - 4*a^2*b^6)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*(a^4*b^4 - a^2*b^6)*sin(d*x + c)^6
/(cos(d*x + c) + 1)^6) - (3*a^2 + b^2)*log(-a - 2*b*sin(d*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*
x + c) + 1)^2)/b^5 + (3*a^2 + b^2)*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/b^5 + (3*a^2 + b^2)*log(sin(d*x +
c)/(cos(d*x + c) + 1) - 1)/b^5)/d

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mupad [B]  time = 4.74, size = 1204, normalized size = 7.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a*cos(c + d*x) + b*sin(c + d*x))^3),x)

[Out]

- ((2*tan(c/2 + (d*x)/2)*(6*a^4 - b^4 + 2*a^2*b^2))/(a*b^4) - (2*tan(c/2 + (d*x)/2)^7*(6*a^4 - b^4 + 2*a^2*b^2
))/(a*b^4) + (2*tan(c/2 + (d*x)/2)^3*(3*b^4 - 18*a^4 + 2*a^2*b^2))/(a*b^4) + (2*tan(c/2 + (d*x)/2)^2*(18*a^4 -
b^4 + 6*a^2*b^2))/(a^2*b^3) - (2*tan(c/2 + (d*x)/2)^5*(3*b^4 - 18*a^4 + 2*a^2*b^2))/(a*b^4) - (4*tan(c/2 + (d
*x)/2)^4*(18*a^4 - b^4 + 8*a^2*b^2))/(a^2*b^3) + (2*tan(c/2 + (d*x)/2)^6*(18*a^4 - b^4 + 6*a^2*b^2))/(a^2*b^3)
)/(d*(tan(c/2 + (d*x)/2)^4*(6*a^2 - 8*b^2) - tan(c/2 + (d*x)/2)^6*(4*a^2 - 4*b^2) - tan(c/2 + (d*x)/2)^2*(4*a^
2 - 4*b^2) + a^2*tan(c/2 + (d*x)/2)^8 + a^2 - 12*a*b*tan(c/2 + (d*x)/2)^3 + 12*a*b*tan(c/2 + (d*x)/2)^5 - 4*a*
b*tan(c/2 + (d*x)/2)^7 + 4*a*b*tan(c/2 + (d*x)/2))) - (atan((((3*a^2 + b^2)*((2*(3*a^2 + b^2)*((4*(a*b^10 + 4*
a^3*b^8))/b^8 - (4*tan(c/2 + (d*x)/2)^2*(3*a*b^10 + 4*a^3*b^8))/b^8 + 16*a^2*b*tan(c/2 + (d*x)/2)))/b^5 - (4*(
4*a*b^7 + 12*a^3*b^5))/b^8 + (4*tan(c/2 + (d*x)/2)^2*(4*a*b^7 + 12*a^3*b^5))/b^8 + (16*tan(c/2 + (d*x)/2)*(6*a
^4 + 2*a^2*b^2))/b^4)*2i)/b^5 - ((3*a^2 + b^2)*((4*(4*a*b^7 + 12*a^3*b^5))/b^8 + (2*(3*a^2 + b^2)*((4*(a*b^10
+ 4*a^3*b^8))/b^8 - (4*tan(c/2 + (d*x)/2)^2*(3*a*b^10 + 4*a^3*b^8))/b^8 + 16*a^2*b*tan(c/2 + (d*x)/2)))/b^5 -
(4*tan(c/2 + (d*x)/2)^2*(4*a*b^7 + 12*a^3*b^5))/b^8 - (16*tan(c/2 + (d*x)/2)*(6*a^4 + 2*a^2*b^2))/b^4)*2i)/b^5
)/((8*(4*a*b^4 + 36*a^5 + 24*a^3*b^2))/b^8 + (2*(3*a^2 + b^2)*((2*(3*a^2 + b^2)*((4*(a*b^10 + 4*a^3*b^8))/b^8
- (4*tan(c/2 + (d*x)/2)^2*(3*a*b^10 + 4*a^3*b^8))/b^8 + 16*a^2*b*tan(c/2 + (d*x)/2)))/b^5 - (4*(4*a*b^7 + 12*a
^3*b^5))/b^8 + (4*tan(c/2 + (d*x)/2)^2*(4*a*b^7 + 12*a^3*b^5))/b^8 + (16*tan(c/2 + (d*x)/2)*(6*a^4 + 2*a^2*b^2
))/b^4))/b^5 + (2*(3*a^2 + b^2)*((4*(4*a*b^7 + 12*a^3*b^5))/b^8 + (2*(3*a^2 + b^2)*((4*(a*b^10 + 4*a^3*b^8))/b
^8 - (4*tan(c/2 + (d*x)/2)^2*(3*a*b^10 + 4*a^3*b^8))/b^8 + 16*a^2*b*tan(c/2 + (d*x)/2)))/b^5 - (4*tan(c/2 + (d
*x)/2)^2*(4*a*b^7 + 12*a^3*b^5))/b^8 - (16*tan(c/2 + (d*x)/2)*(6*a^4 + 2*a^2*b^2))/b^4))/b^5 + (8*tan(c/2 + (d
*x)/2)^2*(4*a*b^4 + 36*a^5 + 24*a^3*b^2))/b^8))*(3*a^2 + b^2)*4i)/(b^5*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**3/(a*cos(c + d*x) + b*sin(c + d*x))**3, x)

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