3.5.63 \(\int \frac {\sec ^5(c+d x)}{(a+b \tan ^2(c+d x))^2} \, dx\) [463]

Optimal. Leaf size=109 \[ \frac {\tanh ^{-1}(\sin (c+d x))}{b^2 d}-\frac {\sqrt {a-b} (2 a+b) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^2 d}-\frac {(a-b) \sin (c+d x)}{2 a b d \left (a-(a-b) \sin ^2(c+d x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3757, 425, 536, 212, 214} \begin {gather*} -\frac {\sqrt {a-b} (2 a+b) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^2 d}-\frac {(a-b) \sin (c+d x)}{2 a b d \left (a-(a-b) \sin ^2(c+d x)\right )}+\frac {\tanh ^{-1}(\sin (c+d x))}{b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

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

Rule 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 3757

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 {\sec ^5(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \left (a-(a-b) x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {(a-b) \sin (c+d x)}{2 a b d \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {-a-b+(-a+b) x^2}{\left (1-x^2\right ) \left (a+(-a+b) x^2\right )} \, dx,x,\sin (c+d x)\right )}{2 a b d}\\ &=-\frac {(a-b) \sin (c+d x)}{2 a b d \left (a-(a-b) \sin ^2(c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{b^2 d}-\frac {((a-b) (2 a+b)) \text {Subst}\left (\int \frac {1}{a+(-a+b) x^2} \, dx,x,\sin (c+d x)\right )}{2 a b^2 d}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{b^2 d}-\frac {\sqrt {a-b} (2 a+b) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^2 d}-\frac {(a-b) \sin (c+d x)}{2 a b d \left (a-(a-b) \sin ^2(c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.55, size = 191, normalized size = 1.75 \begin {gather*} \frac {-4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\sqrt {a-b} (2 a+b) \log \left (\sqrt {a}-\sqrt {a-b} \sin (c+d x)\right )}{a^{3/2}}+\frac {\left (-2 a^2+a b+b^2\right ) \log \left (\sqrt {a}+\sqrt {a-b} \sin (c+d x)\right )}{a^{3/2} \sqrt {a-b}}+\frac {4 b (-a+b) \sin (c+d x)}{a (a+b+(a-b) \cos (2 (c+d x)))}}{4 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

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Maple [A]
time = 0.43, size = 124, normalized size = 1.14

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 b^{2}}+\frac {\ln \left (\sin \left (d x +c \right )+1\right )}{2 b^{2}}+\frac {\left (a -b \right ) \left (\frac {b \sin \left (d x +c \right )}{2 a \left (a \left (\sin ^{2}\left (d x +c \right )\right )-b \left (\sin ^{2}\left (d x +c \right )\right )-a \right )}-\frac {\left (2 a +b \right ) \arctanh \left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{2 a \sqrt {a \left (a -b \right )}}\right )}{b^{2}}}{d}\)	\(124\)
default	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 b^{2}}+\frac {\ln \left (\sin \left (d x +c \right )+1\right )}{2 b^{2}}+\frac {\left (a -b \right ) \left (\frac {b \sin \left (d x +c \right )}{2 a \left (a \left (\sin ^{2}\left (d x +c \right )\right )-b \left (\sin ^{2}\left (d x +c \right )\right )-a \right )}-\frac {\left (2 a +b \right ) \arctanh \left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{2 a \sqrt {a \left (a -b \right )}}\right )}{b^{2}}}{d}\)	\(124\)
risch	\(\frac {i \left (a -b \right ) \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{b d a \left (a \,{\mathrm e}^{4 i \left (d x +c \right )}-b \,{\mathrm e}^{4 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,b^{2}}+\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{2 a d \,b^{2}}+\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{4 a^{2} d b}-\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{2 a d \,b^{2}}-\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{4 a^{2} d b}\)	\(375\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+b*tan(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/2/b^2*ln(sin(d*x+c)-1)+1/2/b^2*ln(sin(d*x+c)+1)+(a-b)/b^2*(1/2*b/a*sin(d*x+c)/(a*sin(d*x+c)^2-b*sin(d*
x+c)^2-a)-1/2*(2*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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(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

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Fricas [A]
time = 2.54, size = 407, normalized size = 3.73 \begin {gather*} \left [\frac {{\left ({\left (2 \, a^{2} - a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a b + b^{2}\right )} \sqrt {\frac {a - b}{a}} \log \left (-\frac {{\left (a - b\right )} \cos \left (d x + c\right )^{2} + 2 \, a \sqrt {\frac {a - b}{a}} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) + 2 \, {\left ({\left (a^{2} - a b\right )} \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (a^{2} - a b\right )} \cos \left (d x + c\right )^{2} + a b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a b - b^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a b^{3} d + {\left (a^{2} b^{2} - a b^{3}\right )} d \cos \left (d x + c\right )^{2}\right )}}, \frac {{\left ({\left (2 \, a^{2} - a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a b + b^{2}\right )} \sqrt {-\frac {a - b}{a}} \arctan \left (\sqrt {-\frac {a - b}{a}} \sin \left (d x + c\right )\right ) + {\left ({\left (a^{2} - a b\right )} \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (a^{2} - a b\right )} \cos \left (d x + c\right )^{2} + a b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (a b - b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a b^{3} d + {\left (a^{2} b^{2} - a b^{3}\right )} d \cos \left (d x + c\right )^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/4*(((2*a^2 - a*b - b^2)*cos(d*x + c)^2 + 2*a*b + b^2)*sqrt((a - b)/a)*log(-((a - b)*cos(d*x + c)^2 + 2*a*sq
rt((a - b)/a)*sin(d*x + c) - 2*a + b)/((a - b)*cos(d*x + c)^2 + b)) + 2*((a^2 - a*b)*cos(d*x + c)^2 + a*b)*log
(sin(d*x + c) + 1) - 2*((a^2 - a*b)*cos(d*x + c)^2 + a*b)*log(-sin(d*x + c) + 1) - 2*(a*b - b^2)*sin(d*x + c))
/(a*b^3*d + (a^2*b^2 - a*b^3)*d*cos(d*x + c)^2), 1/2*(((2*a^2 - a*b - b^2)*cos(d*x + c)^2 + 2*a*b + b^2)*sqrt(
-(a - b)/a)*arctan(sqrt(-(a - b)/a)*sin(d*x + c)) + ((a^2 - a*b)*cos(d*x + c)^2 + a*b)*log(sin(d*x + c) + 1) -
((a^2 - a*b)*cos(d*x + c)^2 + a*b)*log(-sin(d*x + c) + 1) - (a*b - b^2)*sin(d*x + c))/(a*b^3*d + (a^2*b^2 - a
*b^3)*d*cos(d*x + c)^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+b*tan(d*x+c)**2)**2,x)

[Out]

Integral(sec(c + d*x)**5/(a + b*tan(c + d*x)**2)**2, x)

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Giac [A]
time = 0.74, size = 153, normalized size = 1.40 \begin {gather*} \frac {\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{b^{2}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{b^{2}} - \frac {{\left (2 \, a^{2} - a b - b^{2}\right )} \arctan \left (-\frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt {-a^{2} + a b}}\right )}{\sqrt {-a^{2} + a b} a b^{2}} + \frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right )^{2} - a\right )} a b}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/2*(log(abs(sin(d*x + c) + 1))/b^2 - log(abs(sin(d*x + c) - 1))/b^2 - (2*a^2 - a*b - b^2)*arctan(-(a*sin(d*x
+ c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/(sqrt(-a^2 + a*b)*a*b^2) + (a*sin(d*x + c) - b*sin(d*x + c))/((a*sin(
d*x + c)^2 - b*sin(d*x + c)^2 - a)*a*b))/d

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Mupad [B]
time = 14.37, size = 946, normalized size = 8.68 \begin {gather*} \frac {\left (-a^{5/2}\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}+a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a-2\,b+2\,a\,\cos \left (c+d\,x\right )-2\,b\,\cos \left (c+d\,x\right )\right )}{2\,\sqrt {a}\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\sqrt {b-a}}\right )\,\sqrt {b-a}\,1{}\mathrm {i}+\frac {b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a-2\,b+2\,a\,\cos \left (c+d\,x\right )-2\,b\,\cos \left (c+d\,x\right )\right )}{2\,\sqrt {a}\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\sqrt {b-a}}\right )\,\sqrt {b-a}\,1{}\mathrm {i}}{2}-a^{3/2}\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}-\sqrt {a}\,b^2\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}+a^2\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b-a}}\right )\,\sqrt {b-a}\,1{}\mathrm {i}+\frac {b^2\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b-a}}\right )\,\sqrt {b-a}\,1{}\mathrm {i}}{2}-a^{5/2}\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )\,2{}\mathrm {i}+a^{3/2}\,b\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}+a^2\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b-a}}\right )\,\cos \left (2\,c+2\,d\,x\right )\,\sqrt {b-a}\,1{}\mathrm {i}-\frac {b^2\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b-a}}\right )\,\cos \left (2\,c+2\,d\,x\right )\,\sqrt {b-a}\,1{}\mathrm {i}}{2}+\frac {a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a-2\,b+2\,a\,\cos \left (c+d\,x\right )-2\,b\,\cos \left (c+d\,x\right )\right )}{2\,\sqrt {a}\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\sqrt {b-a}}\right )\,\sqrt {b-a}\,3{}\mathrm {i}}{2}+\frac {a\,b\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b-a}}\right )\,\sqrt {b-a}\,3{}\mathrm {i}}{2}+a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a-2\,b+2\,a\,\cos \left (c+d\,x\right )-2\,b\,\cos \left (c+d\,x\right )\right )}{2\,\sqrt {a}\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\sqrt {b-a}}\right )\,\cos \left (2\,c+2\,d\,x\right )\,\sqrt {b-a}\,1{}\mathrm {i}-\frac {b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a-2\,b+2\,a\,\cos \left (c+d\,x\right )-2\,b\,\cos \left (c+d\,x\right )\right )}{2\,\sqrt {a}\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\sqrt {b-a}}\right )\,\cos \left (2\,c+2\,d\,x\right )\,\sqrt {b-a}\,1{}\mathrm {i}}{2}+a^{3/2}\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )\,2{}\mathrm {i}-\frac {a\,b\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b-a}}\right )\,\cos \left (2\,c+2\,d\,x\right )\,\sqrt {b-a}\,1{}\mathrm {i}}{2}-\frac {a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a-2\,b+2\,a\,\cos \left (c+d\,x\right )-2\,b\,\cos \left (c+d\,x\right )\right )}{2\,\sqrt {a}\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\sqrt {b-a}}\right )\,\cos \left (2\,c+2\,d\,x\right )\,\sqrt {b-a}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{2\,a^{3/2}\,b^2\,d\,\left (\frac {a}{2}+\frac {b}{2}+\frac {a\,\cos \left (2\,c+2\,d\,x\right )}{2}-\frac {b\,\cos \left (2\,c+2\,d\,x\right )}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + b*tan(c + d*x)^2)^2),x)

[Out]

((a^2*atan((sin(c/2 + (d*x)/2)*(a - 2*b + 2*a*cos(c + d*x) - 2*b*cos(c + d*x)))/(2*a^(1/2)*cos(c/2 + (d*x)/2)^
3*(b - a)^(1/2)))*(b - a)^(1/2)*1i - a^(5/2)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*2i + (b^2*atan((sin(
c/2 + (d*x)/2)*(a - 2*b + 2*a*cos(c + d*x) - 2*b*cos(c + d*x)))/(2*a^(1/2)*cos(c/2 + (d*x)/2)^3*(b - a)^(1/2))
)*(b - a)^(1/2)*1i)/2 - a^(3/2)*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*2i - a^(1/2)*b^2*sin(c + d*x)*1
i + a^2*atan((a^(1/2)*sin(c/2 + (d*x)/2))/(2*cos(c/2 + (d*x)/2)*(b - a)^(1/2)))*(b - a)^(1/2)*1i + (b^2*atan((
a^(1/2)*sin(c/2 + (d*x)/2))/(2*cos(c/2 + (d*x)/2)*(b - a)^(1/2)))*(b - a)^(1/2)*1i)/2 - a^(5/2)*atanh(sin(c/2
+ (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*2i + a^(3/2)*b*sin(c + d*x)*1i + a^2*atan((a^(1/2)*sin(c/2 + (
d*x)/2))/(2*cos(c/2 + (d*x)/2)*(b - a)^(1/2)))*cos(2*c + 2*d*x)*(b - a)^(1/2)*1i - (b^2*atan((a^(1/2)*sin(c/2
+ (d*x)/2))/(2*cos(c/2 + (d*x)/2)*(b - a)^(1/2)))*cos(2*c + 2*d*x)*(b - a)^(1/2)*1i)/2 + (a*b*atan((sin(c/2 +
(d*x)/2)*(a - 2*b + 2*a*cos(c + d*x) - 2*b*cos(c + d*x)))/(2*a^(1/2)*cos(c/2 + (d*x)/2)^3*(b - a)^(1/2)))*(b -
a)^(1/2)*3i)/2 + (a*b*atan((a^(1/2)*sin(c/2 + (d*x)/2))/(2*cos(c/2 + (d*x)/2)*(b - a)^(1/2)))*(b - a)^(1/2)*3
i)/2 + a^2*atan((sin(c/2 + (d*x)/2)*(a - 2*b + 2*a*cos(c + d*x) - 2*b*cos(c + d*x)))/(2*a^(1/2)*cos(c/2 + (d*x
)/2)^3*(b - a)^(1/2)))*cos(2*c + 2*d*x)*(b - a)^(1/2)*1i - (b^2*atan((sin(c/2 + (d*x)/2)*(a - 2*b + 2*a*cos(c
+ d*x) - 2*b*cos(c + d*x)))/(2*a^(1/2)*cos(c/2 + (d*x)/2)^3*(b - a)^(1/2)))*cos(2*c + 2*d*x)*(b - a)^(1/2)*1i)
/2 + a^(3/2)*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*2i - (a*b*atan((a^(1/2)*sin(c/2 +
(d*x)/2))/(2*cos(c/2 + (d*x)/2)*(b - a)^(1/2)))*cos(2*c + 2*d*x)*(b - a)^(1/2)*1i)/2 - (a*b*atan((sin(c/2 + (
d*x)/2)*(a - 2*b + 2*a*cos(c + d*x) - 2*b*cos(c + d*x)))/(2*a^(1/2)*cos(c/2 + (d*x)/2)^3*(b - a)^(1/2)))*cos(2
*c + 2*d*x)*(b - a)^(1/2)*1i)/2)*1i)/(2*a^(3/2)*b^2*d*(a/2 + b/2 + (a*cos(2*c + 2*d*x))/2 - (b*cos(2*c + 2*d*x
))/2))

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