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

Optimal. Leaf size=94 \[ \frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} d (a-b)^{3/2}}-\frac {b \sin (c+d x)}{2 a d (a-b) \left (a-(a-b) \sin ^2(c+d x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3676, 385, 208} \[ \frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} d (a-b)^{3/2}}-\frac {b \sin (c+d x)}{2 a d (a-b) \left (a-(a-b) \sin ^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.25, size = 92, normalized size = 0.98 \[ \frac {\frac {\sqrt {a} b \sin (c+d x)}{(a-b) \left ((a-b) \sin ^2(c+d x)-a\right )}+\frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{(a-b)^{3/2}}}{2 a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 337, normalized size = 3.59 \[ \left [\frac {{\left ({\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a b - b^{2}\right )} \sqrt {a^{2} - a b} \log \left (-\frac {{\left (a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - a b} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) - 2 \, {\left (a^{2} b - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} d\right )}}, -\frac {{\left ({\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a b - b^{2}\right )} \sqrt {-a^{2} + a b} \arctan \left (\frac {\sqrt {-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) + {\left (a^{2} b - a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(((2*a^2 - 3*a*b + b^2)*cos(d*x + c)^2 + 2*a*b - b^2)*sqrt(a^2 - a*b)*log(-((a - b)*cos(d*x + c)^2 - 2*sq
rt(a^2 - a*b)*sin(d*x + c) - 2*a + b)/((a - b)*cos(d*x + c)^2 + b)) - 2*(a^2*b - a*b^2)*sin(d*x + c))/((a^5 -
3*a^4*b + 3*a^3*b^2 - a^2*b^3)*d*cos(d*x + c)^2 + (a^4*b - 2*a^3*b^2 + a^2*b^3)*d), -1/2*(((2*a^2 - 3*a*b + b^
2)*cos(d*x + c)^2 + 2*a*b - b^2)*sqrt(-a^2 + a*b)*arctan(sqrt(-a^2 + a*b)*sin(d*x + c)/a) + (a^2*b - a*b^2)*si
n(d*x + c))/((a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*d*cos(d*x + c)^2 + (a^4*b - 2*a^3*b^2 + a^2*b^3)*d)]

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giac [A]  time = 1.85, size = 112, normalized size = 1.19 \[ -\frac {\frac {{\left (2 \, a - b\right )} \arctan \left (\frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt {-a^{2} + a b}}\right )}{{\left (a^{2} - a b\right )} \sqrt {-a^{2} + a b}} - \frac {b \sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right )^{2} - a\right )} {\left (a^{2} - a b\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.59, size = 102, normalized size = 1.09 \[ \frac {\frac {b \sin \left (d x +c \right )}{2 a \left (a -b \right ) \left (a \left (\sin ^{2}\left (d x +c \right )\right )-b \left (\sin ^{2}\left (d x +c \right )\right )-a \right )}+\frac {\left (2 a -b \right ) \arctanh \left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{2 a \left (a -b \right ) \sqrt {a \left (a -b \right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/2*b/a/(a-b)*sin(d*x+c)/(a*sin(d*x+c)^2-b*sin(d*x+c)^2-a)+1/2*(2*a-b)/a/(a-b)/(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(sec(d*x+c)/(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 = 12.78, size = 239, normalized size = 2.54 \[ -\frac {\left (a^2\,\mathrm {atanh}\left (\frac {\sin \left (c+d\,x\right )\,\sqrt {a-b}}{\sqrt {a}}\right )\,1{}\mathrm {i}-\frac {b^2\,\mathrm {atanh}\left (\frac {\sin \left (c+d\,x\right )\,\sqrt {a-b}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{2}+a^2\,\cos \left (2\,c+2\,d\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (c+d\,x\right )\,\sqrt {a-b}}{\sqrt {a}}\right )\,1{}\mathrm {i}+\frac {b^2\,\cos \left (2\,c+2\,d\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (c+d\,x\right )\,\sqrt {a-b}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{2}+\frac {a\,b\,\mathrm {atanh}\left (\frac {\sin \left (c+d\,x\right )\,\sqrt {a-b}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{2}-\frac {a\,b\,\cos \left (2\,c+2\,d\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (c+d\,x\right )\,\sqrt {a-b}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{2}-\sqrt {a}\,b\,\sin \left (c+d\,x\right )\,\sqrt {a-b}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a^{3/2}\,d\,{\left (a-b\right )}^{3/2}\,\left (\frac {a}{2}+\frac {b}{2}+\frac {a\,\cos \left (2\,c+2\,d\,x\right )}{2}-\frac {b\,\cos \left (2\,c+2\,d\,x\right )}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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