3.467 \(\int \frac {\cos ^3(c+d x)}{(a+b \tan ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=143 \[ \frac {b^2 (6 a-b) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} d (a-b)^{7/2}}-\frac {b^3 \sin (c+d x)}{2 a d (a-b)^3 \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac {\sin ^3(c+d x)}{3 d (a-b)^2}+\frac {(a-3 b) \sin (c+d x)}{d (a-b)^3} \]

[Out]

1/2*(6*a-b)*b^2*arctanh(sin(d*x+c)*(a-b)^(1/2)/a^(1/2))/a^(3/2)/(a-b)^(7/2)/d+(a-3*b)*sin(d*x+c)/(a-b)^3/d-1/3
*sin(d*x+c)^3/(a-b)^2/d-1/2*b^3*sin(d*x+c)/a/(a-b)^3/d/(a-(a-b)*sin(d*x+c)^2)

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Rubi [A]  time = 0.21, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3676, 390, 385, 208} \[ \frac {b^2 (6 a-b) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} d (a-b)^{7/2}}-\frac {b^3 \sin (c+d x)}{2 a d (a-b)^3 \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac {\sin ^3(c+d x)}{3 d (a-b)^2}+\frac {(a-3 b) \sin (c+d x)}{d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

((6*a - b)*b^2*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(2*a^(3/2)*(a - b)^(7/2)*d) + ((a - 3*b)*Sin[c + d
*x])/((a - b)^3*d) - Sin[c + d*x]^3/(3*(a - b)^2*d) - (b^3*Sin[c + d*x])/(2*a*(a - b)^3*d*(a - (a - b)*Sin[c +
d*x]^2))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^3}{\left (a-(a-b) x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a-3 b}{(a-b)^3}-\frac {x^2}{(a-b)^2}+\frac {(3 a-b) b^2-3 (a-b) b^2 x^2}{(a-b)^3 \left (a+(-a+b) x^2\right )^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {(a-3 b) \sin (c+d x)}{(a-b)^3 d}-\frac {\sin ^3(c+d x)}{3 (a-b)^2 d}+\frac {\operatorname {Subst}\left (\int \frac {(3 a-b) b^2-3 (a-b) b^2 x^2}{\left (a+(-a+b) x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{(a-b)^3 d}\\ &=\frac {(a-3 b) \sin (c+d x)}{(a-b)^3 d}-\frac {\sin ^3(c+d x)}{3 (a-b)^2 d}-\frac {b^3 \sin (c+d x)}{2 a (a-b)^3 d \left (a-(a-b) \sin ^2(c+d x)\right )}+\frac {\left ((6 a-b) b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+(-a+b) x^2} \, dx,x,\sin (c+d x)\right )}{2 a (a-b)^3 d}\\ &=\frac {(6 a-b) b^2 \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} (a-b)^{7/2} d}+\frac {(a-3 b) \sin (c+d x)}{(a-b)^3 d}-\frac {\sin ^3(c+d x)}{3 (a-b)^2 d}-\frac {b^3 \sin (c+d x)}{2 a (a-b)^3 d \left (a-(a-b) \sin ^2(c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.62, size = 147, normalized size = 1.03 \[ \frac {\frac {3 b^2 (b-6 a) \left (\log \left (\sqrt {a}-\sqrt {a-b} \sin (c+d x)\right )-\log \left (\sqrt {a-b} \sin (c+d x)+\sqrt {a}\right )\right )}{a^{3/2} (a-b)^{7/2}}+\frac {3 \sin (c+d x) \left (-\frac {4 b^3}{a ((a-b) \cos (2 (c+d x))+a+b)}+3 a-11 b\right )}{(a-b)^3}+\frac {\sin (3 (c+d x))}{(a-b)^2}}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

((3*b^2*(-6*a + b)*(Log[Sqrt[a] - Sqrt[a - b]*Sin[c + d*x]] - Log[Sqrt[a] + Sqrt[a - b]*Sin[c + d*x]]))/(a^(3/
2)*(a - b)^(7/2)) + (3*(3*a - 11*b - (4*b^3)/(a*(a + b + (a - b)*Cos[2*(c + d*x)])))*Sin[c + d*x])/(a - b)^3 +
Sin[3*(c + d*x)]/(a - b)^2)/(12*d)

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fricas [B]  time = 0.70, size = 600, normalized size = 4.20 \[ \left [\frac {3 \, {\left (6 \, a b^{3} - b^{4} + {\left (6 \, a^{2} b^{2} - 7 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a^{2} - a b} \log \left (-\frac {{\left (a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - a b} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) + 2 \, {\left (4 \, a^{4} b - 20 \, a^{3} b^{2} + 13 \, a^{2} b^{3} + 3 \, a b^{4} + 2 \, {\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, a^{5} - 11 \, a^{4} b + 16 \, a^{3} b^{2} - 7 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left ({\left (a^{7} - 5 \, a^{6} b + 10 \, a^{5} b^{2} - 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} b - 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} - 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} d\right )}}, -\frac {3 \, {\left (6 \, a b^{3} - b^{4} + {\left (6 \, a^{2} b^{2} - 7 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a^{2} + a b} \arctan \left (\frac {\sqrt {-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) - {\left (4 \, a^{4} b - 20 \, a^{3} b^{2} + 13 \, a^{2} b^{3} + 3 \, a b^{4} + 2 \, {\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, a^{5} - 11 \, a^{4} b + 16 \, a^{3} b^{2} - 7 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{7} - 5 \, a^{6} b + 10 \, a^{5} b^{2} - 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} b - 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} - 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/12*(3*(6*a*b^3 - b^4 + (6*a^2*b^2 - 7*a*b^3 + b^4)*cos(d*x + c)^2)*sqrt(a^2 - a*b)*log(-((a - b)*cos(d*x +
c)^2 - 2*sqrt(a^2 - a*b)*sin(d*x + c) - 2*a + b)/((a - b)*cos(d*x + c)^2 + b)) + 2*(4*a^4*b - 20*a^3*b^2 + 13*
a^2*b^3 + 3*a*b^4 + 2*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*cos(d*x + c)^4 + 2*(2*a^5 - 11*a^4*b + 16*a^3*b^2
- 7*a^2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/((a^7 - 5*a^6*b + 10*a^5*b^2 - 10*a^4*b^3 + 5*a^3*b^4 - a^2*b^5)*d*
cos(d*x + c)^2 + (a^6*b - 4*a^5*b^2 + 6*a^4*b^3 - 4*a^3*b^4 + a^2*b^5)*d), -1/6*(3*(6*a*b^3 - b^4 + (6*a^2*b^2
- 7*a*b^3 + b^4)*cos(d*x + c)^2)*sqrt(-a^2 + a*b)*arctan(sqrt(-a^2 + a*b)*sin(d*x + c)/a) - (4*a^4*b - 20*a^3
*b^2 + 13*a^2*b^3 + 3*a*b^4 + 2*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*cos(d*x + c)^4 + 2*(2*a^5 - 11*a^4*b + 1
6*a^3*b^2 - 7*a^2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/((a^7 - 5*a^6*b + 10*a^5*b^2 - 10*a^4*b^3 + 5*a^3*b^4 - a
^2*b^5)*d*cos(d*x + c)^2 + (a^6*b - 4*a^5*b^2 + 6*a^4*b^3 - 4*a^3*b^4 + a^2*b^5)*d)]

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giac [B]  time = 2.55, size = 329, normalized size = 2.30 \[ \frac {\frac {3 \, b^{3} \sin \left (d x + c\right )}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} {\left (a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right )^{2} - a\right )}} + \frac {3 \, {\left (6 \, a b^{2} - b^{3}\right )} \arctan \left (-\frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt {-a^{2} + a b}}\right )}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \sqrt {-a^{2} + a b}} - \frac {2 \, {\left (a^{4} \sin \left (d x + c\right )^{3} - 4 \, a^{3} b \sin \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \sin \left (d x + c\right )^{3} - 4 \, a b^{3} \sin \left (d x + c\right )^{3} + b^{4} \sin \left (d x + c\right )^{3} - 3 \, a^{4} \sin \left (d x + c\right ) + 18 \, a^{3} b \sin \left (d x + c\right ) - 36 \, a^{2} b^{2} \sin \left (d x + c\right ) + 30 \, a b^{3} \sin \left (d x + c\right ) - 9 \, b^{4} \sin \left (d x + c\right )\right )}}{a^{6} - 6 \, a^{5} b + 15 \, a^{4} b^{2} - 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/6*(3*b^3*sin(d*x + c)/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*(a*sin(d*x + c)^2 - b*sin(d*x + c)^2 - a)) + 3*(6
*a*b^2 - b^3)*arctan(-(a*sin(d*x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)
*sqrt(-a^2 + a*b)) - 2*(a^4*sin(d*x + c)^3 - 4*a^3*b*sin(d*x + c)^3 + 6*a^2*b^2*sin(d*x + c)^3 - 4*a*b^3*sin(d
*x + c)^3 + b^4*sin(d*x + c)^3 - 3*a^4*sin(d*x + c) + 18*a^3*b*sin(d*x + c) - 36*a^2*b^2*sin(d*x + c) + 30*a*b
^3*sin(d*x + c) - 9*b^4*sin(d*x + c))/(a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6))/
d

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maple [A]  time = 0.81, size = 164, normalized size = 1.15 \[ \frac {-\frac {\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {b \left (\sin ^{3}\left (d x +c \right )\right )}{3}-a \sin \left (d x +c \right )+3 b \sin \left (d x +c \right )}{\left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )}-\frac {b^{2} \left (-\frac {b \sin \left (d x +c \right )}{2 a \left (a \left (\sin ^{2}\left (d x +c \right )\right )-b \left (\sin ^{2}\left (d x +c \right )\right )-a \right )}-\frac {\left (6 a -b \right ) \arctanh \left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{2 a \sqrt {a \left (a -b \right )}}\right )}{\left (a -b \right )^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x)

[Out]

1/d*(-1/(a^2-2*a*b+b^2)/(a-b)*(1/3*a*sin(d*x+c)^3-1/3*b*sin(d*x+c)^3-a*sin(d*x+c)+3*b*sin(d*x+c))-b^2/(a-b)^3*
(-1/2/a*b*sin(d*x+c)/(a*sin(d*x+c)^2-b*sin(d*x+c)^2-a)-1/2*(6*a-b)/a/(a*(a-b))^(1/2)*arctanh((a-b)*sin(d*x+c)/
(a*(a-b))^(1/2))))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

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mupad [B]  time = 16.01, size = 1690, normalized size = 11.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a + b*tan(c + d*x)^2)^2,x)

[Out]

- ((tan(c/2 + (d*x)/2)*(6*a^2*b - 2*a^3 + b^3))/(a*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (tan(c/2 + (d*x)/2)^9*(6
*a^2*b - 2*a^3 + b^3))/(a*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (4*tan(c/2 + (d*x)/2)^3*(18*a*b^2 - 8*a^2*b + 2*a
^3 + 3*b^3))/(3*a*(a - b)*(a^2 - 2*a*b + b^2)) + (4*tan(c/2 + (d*x)/2)^7*(18*a*b^2 - 8*a^2*b + 2*a^3 + 3*b^3))
/(3*a*(a - b)*(a^2 - 2*a*b + b^2)) + (2*tan(c/2 + (d*x)/2)^5*(56*a*b^2 - 18*a^2*b - 2*a^3 + 9*b^3))/(3*a*(a -
b)*(a^2 - 2*a*b + b^2)))/(d*(a + tan(c/2 + (d*x)/2)^2*(a + 4*b) + tan(c/2 + (d*x)/2)^8*(a + 4*b) - tan(c/2 + (
d*x)/2)^4*(2*a - 12*b) - tan(c/2 + (d*x)/2)^6*(2*a - 12*b) + a*tan(c/2 + (d*x)/2)^10)) - (b^2*atan(((b^2*(tan(
c/2 + (d*x)/2)*(8*a^3*b^10 - 96*a^4*b^9 + 408*a^5*b^8 - 880*a^6*b^7 + 1080*a^7*b^6 - 768*a^8*b^5 + 296*a^9*b^4
- 48*a^10*b^3) - (b^2*(6*a - b)*(tan(c/2 + (d*x)/2)^2*(16*a^15 - 176*a^14*b + 32*a^5*b^10 - 304*a^6*b^9 + 129
6*a^7*b^8 - 3264*a^8*b^7 + 5376*a^9*b^6 - 6048*a^10*b^5 + 4704*a^11*b^4 - 2496*a^12*b^3 + 864*a^13*b^2) + 144*
a^14*b - 16*a^15 + 16*a^6*b^9 - 144*a^7*b^8 + 576*a^8*b^7 - 1344*a^9*b^6 + 2016*a^10*b^5 - 2016*a^11*b^4 + 134
4*a^12*b^3 - 576*a^13*b^2))/(4*a^(3/2)*(a - b)^(7/2)))*(6*a - b)*1i)/(4*a^(3/2)*(a - b)^(7/2)) + (b^2*(tan(c/2
+ (d*x)/2)*(8*a^3*b^10 - 96*a^4*b^9 + 408*a^5*b^8 - 880*a^6*b^7 + 1080*a^7*b^6 - 768*a^8*b^5 + 296*a^9*b^4 -
48*a^10*b^3) + (b^2*(6*a - b)*(tan(c/2 + (d*x)/2)^2*(16*a^15 - 176*a^14*b + 32*a^5*b^10 - 304*a^6*b^9 + 1296*a
^7*b^8 - 3264*a^8*b^7 + 5376*a^9*b^6 - 6048*a^10*b^5 + 4704*a^11*b^4 - 2496*a^12*b^3 + 864*a^13*b^2) + 144*a^1
4*b - 16*a^15 + 16*a^6*b^9 - 144*a^7*b^8 + 576*a^8*b^7 - 1344*a^9*b^6 + 2016*a^10*b^5 - 2016*a^11*b^4 + 1344*a
^12*b^3 - 576*a^13*b^2))/(4*a^(3/2)*(a - b)^(7/2)))*(6*a - b)*1i)/(4*a^(3/2)*(a - b)^(7/2)))/(2*tan(c/2 + (d*x
)/2)^2*(a^2*b^9 - 15*a^3*b^8 + 75*a^4*b^7 - 145*a^5*b^6 + 120*a^6*b^5 - 36*a^7*b^4) - 2*a^2*b^9 + 30*a^3*b^8 -
150*a^4*b^7 + 290*a^5*b^6 - 240*a^6*b^5 + 72*a^7*b^4 - (b^2*(tan(c/2 + (d*x)/2)*(8*a^3*b^10 - 96*a^4*b^9 + 40
8*a^5*b^8 - 880*a^6*b^7 + 1080*a^7*b^6 - 768*a^8*b^5 + 296*a^9*b^4 - 48*a^10*b^3) - (b^2*(6*a - b)*(tan(c/2 +
(d*x)/2)^2*(16*a^15 - 176*a^14*b + 32*a^5*b^10 - 304*a^6*b^9 + 1296*a^7*b^8 - 3264*a^8*b^7 + 5376*a^9*b^6 - 60
48*a^10*b^5 + 4704*a^11*b^4 - 2496*a^12*b^3 + 864*a^13*b^2) + 144*a^14*b - 16*a^15 + 16*a^6*b^9 - 144*a^7*b^8
+ 576*a^8*b^7 - 1344*a^9*b^6 + 2016*a^10*b^5 - 2016*a^11*b^4 + 1344*a^12*b^3 - 576*a^13*b^2))/(4*a^(3/2)*(a -
b)^(7/2)))*(6*a - b))/(4*a^(3/2)*(a - b)^(7/2)) + (b^2*(tan(c/2 + (d*x)/2)*(8*a^3*b^10 - 96*a^4*b^9 + 408*a^5*
b^8 - 880*a^6*b^7 + 1080*a^7*b^6 - 768*a^8*b^5 + 296*a^9*b^4 - 48*a^10*b^3) + (b^2*(6*a - b)*(tan(c/2 + (d*x)/
2)^2*(16*a^15 - 176*a^14*b + 32*a^5*b^10 - 304*a^6*b^9 + 1296*a^7*b^8 - 3264*a^8*b^7 + 5376*a^9*b^6 - 6048*a^1
0*b^5 + 4704*a^11*b^4 - 2496*a^12*b^3 + 864*a^13*b^2) + 144*a^14*b - 16*a^15 + 16*a^6*b^9 - 144*a^7*b^8 + 576*
a^8*b^7 - 1344*a^9*b^6 + 2016*a^10*b^5 - 2016*a^11*b^4 + 1344*a^12*b^3 - 576*a^13*b^2))/(4*a^(3/2)*(a - b)^(7/
2)))*(6*a - b))/(4*a^(3/2)*(a - b)^(7/2))))*(6*a - b)*1i)/(2*a^(3/2)*d*(a - b)^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+b*tan(d*x+c)**2)**2,x)

[Out]

Timed out

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