3.416 \(\int \frac{\sec ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

Optimal. Leaf size=142 \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}-\sqrt{b}\right )^{3/2}}+\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}+\sqrt{b}\right )^{3/2}}+\frac{\tan (c+d x)}{d (a-b)} \]

[Out]

-(Sqrt[b]*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] - Sqrt[b])^(3/2)*d) + (S
qrt[b]*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] + Sqrt[b])^(3/2)*d) + Tan[c
+ d*x]/((a - b)*d)

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Rubi [A]  time = 0.23203, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3224, 1170, 1166, 205} \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}-\sqrt{b}\right )^{3/2}}+\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}+\sqrt{b}\right )^{3/2}}+\frac{\tan (c+d x)}{d (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a - b*Sin[c + d*x]^4),x]

[Out]

-(Sqrt[b]*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] - Sqrt[b])^(3/2)*d) + (S
qrt[b]*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] + Sqrt[b])^(3/2)*d) + Tan[c
+ d*x]/((a - b)*d)

Rule 3224

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(m/2 + 2
*p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 1170

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(d + e*x
^2)^q/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && IntegerQ[q]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 0.497502, size = 175, normalized size = 1.23 \[ \frac{\frac{\left (\sqrt{a} \sqrt{b}-b\right ) \tan ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\sqrt{a} \sqrt{\sqrt{a} \sqrt{b}+a}}+\frac{\left (\sqrt{a} \sqrt{b}+b\right ) \tanh ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\sqrt{a} \sqrt{\sqrt{a} \sqrt{b}-a}}+2 \tan (c+d x)}{2 d (a-b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a - b*Sin[c + d*x]^4),x]

[Out]

(((Sqrt[a]*Sqrt[b] - b)*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[a
+ Sqrt[a]*Sqrt[b]]) + ((Sqrt[a]*Sqrt[b] + b)*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt
[b]]])/(Sqrt[a]*Sqrt[-a + Sqrt[a]*Sqrt[b]]) + 2*Tan[c + d*x])/(2*(a - b)*d)

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Maple [B]  time = 0.128, size = 393, normalized size = 2.8 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{ \left ( a-b \right ) d}}-{\frac{ab}{2\, \left ( a-b \right ) d}\arctan \left ({ \left ( a-b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}+a \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ab}}}{\frac{1}{\sqrt{ \left ( \sqrt{ab}+a \right ) \left ( a-b \right ) }}}}-{\frac{{b}^{2}}{2\, \left ( a-b \right ) d}\arctan \left ({ \left ( a-b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}+a \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ab}}}{\frac{1}{\sqrt{ \left ( \sqrt{ab}+a \right ) \left ( a-b \right ) }}}}-{\frac{b}{ \left ( a-b \right ) d}\arctan \left ({ \left ( a-b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}+a \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}+a \right ) \left ( a-b \right ) }}}}+{\frac{ab}{2\, \left ( a-b \right ) d}{\it Artanh} \left ({ \left ( -a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}-a \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ab}}}{\frac{1}{\sqrt{ \left ( \sqrt{ab}-a \right ) \left ( a-b \right ) }}}}+{\frac{{b}^{2}}{2\, \left ( a-b \right ) d}{\it Artanh} \left ({ \left ( -a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}-a \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ab}}}{\frac{1}{\sqrt{ \left ( \sqrt{ab}-a \right ) \left ( a-b \right ) }}}}-{\frac{b}{ \left ( a-b \right ) d}{\it Artanh} \left ({ \left ( -a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}-a \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}-a \right ) \left ( a-b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a-b*sin(d*x+c)^4),x)

[Out]

tan(d*x+c)/(a-b)/d-1/2/d*a/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/
2)+a)*(a-b))^(1/2))*b-1/2/d/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1
/2)+a)*(a-b))^(1/2))*b^2-1/d*b/(a-b)/(((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a
-b))^(1/2))+1/2/d*a/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)-a)
*(a-b))^(1/2))*b+1/2/d/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)
-a)*(a-b))^(1/2))*b^2-1/d*b/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)-a)*(a-
b))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \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)^2/(a-b*sin(d*x+c)^4),x, algorithm="maxima")

[Out]

(((a - b)*d*cos(2*d*x + 2*c)^2 + (a - b)*d*sin(2*d*x + 2*c)^2 + 2*(a - b)*d*cos(2*d*x + 2*c) + (a - b)*d)*inte
grate(4*(4*b^2*cos(6*d*x + 6*c)^2 + 4*b^2*cos(2*d*x + 2*c)^2 + 4*b^2*sin(6*d*x + 6*c)^2 + 4*b^2*sin(2*d*x + 2*
c)^2 - 12*(8*a*b - 3*b^2)*cos(4*d*x + 4*c)^2 - b^2*cos(2*d*x + 2*c) - 12*(8*a*b - 3*b^2)*sin(4*d*x + 4*c)^2 +
2*(8*a*b - 15*b^2)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) - (b^2*cos(6*d*x + 6*c) - 6*b^2*cos(4*d*x + 4*c) + b^2*co
s(2*d*x + 2*c))*cos(8*d*x + 8*c) + (8*b^2*cos(2*d*x + 2*c) - b^2 + 2*(8*a*b - 15*b^2)*cos(4*d*x + 4*c))*cos(6*
d*x + 6*c) + 2*(3*b^2 + (8*a*b - 15*b^2)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - (b^2*sin(6*d*x + 6*c) - 6*b^2*si
n(4*d*x + 4*c) + b^2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 2*(4*b^2*sin(2*d*x + 2*c) + (8*a*b - 15*b^2)*sin(4*d
*x + 4*c))*sin(6*d*x + 6*c))/(a*b^2 - b^3 + (a*b^2 - b^3)*cos(8*d*x + 8*c)^2 + 16*(a*b^2 - b^3)*cos(6*d*x + 6*
c)^2 + 4*(64*a^3 - 112*a^2*b + 57*a*b^2 - 9*b^3)*cos(4*d*x + 4*c)^2 + 16*(a*b^2 - b^3)*cos(2*d*x + 2*c)^2 + (a
*b^2 - b^3)*sin(8*d*x + 8*c)^2 + 16*(a*b^2 - b^3)*sin(6*d*x + 6*c)^2 + 4*(64*a^3 - 112*a^2*b + 57*a*b^2 - 9*b^
3)*sin(4*d*x + 4*c)^2 + 16*(8*a^2*b - 11*a*b^2 + 3*b^3)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*(a*b^2 - b^3)*s
in(2*d*x + 2*c)^2 + 2*(a*b^2 - b^3 - 4*(a*b^2 - b^3)*cos(6*d*x + 6*c) - 2*(8*a^2*b - 11*a*b^2 + 3*b^3)*cos(4*d
*x + 4*c) - 4*(a*b^2 - b^3)*cos(2*d*x + 2*c))*cos(8*d*x + 8*c) - 8*(a*b^2 - b^3 - 2*(8*a^2*b - 11*a*b^2 + 3*b^
3)*cos(4*d*x + 4*c) - 4*(a*b^2 - b^3)*cos(2*d*x + 2*c))*cos(6*d*x + 6*c) - 4*(8*a^2*b - 11*a*b^2 + 3*b^3 - 4*(
8*a^2*b - 11*a*b^2 + 3*b^3)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - 8*(a*b^2 - b^3)*cos(2*d*x + 2*c) - 4*(2*(a*b^
2 - b^3)*sin(6*d*x + 6*c) + (8*a^2*b - 11*a*b^2 + 3*b^3)*sin(4*d*x + 4*c) + 2*(a*b^2 - b^3)*sin(2*d*x + 2*c))*
sin(8*d*x + 8*c) + 16*((8*a^2*b - 11*a*b^2 + 3*b^3)*sin(4*d*x + 4*c) + 2*(a*b^2 - b^3)*sin(2*d*x + 2*c))*sin(6
*d*x + 6*c)), x) + 2*sin(2*d*x + 2*c))/((a - b)*d*cos(2*d*x + 2*c)^2 + (a - b)*d*sin(2*d*x + 2*c)^2 + 2*(a - b
)*d*cos(2*d*x + 2*c) + (a - b)*d)

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Fricas [B]  time = 5.37704, size = 5536, normalized size = 38.99 \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)^2/(a-b*sin(d*x+c)^4),x, algorithm="fricas")

[Out]

1/8*((a - b)*d*sqrt(((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b
+ 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4)) - a*b - 3*b^2)/((a^4 - 3*a^3*b + 3*a^2*b^2
- a*b^3)*d^2))*cos(d*x + c)*log(3/4*a*b^2 + 1/4*b^3 - 1/4*(3*a*b^2 + b^3)*cos(d*x + c)^2 + 1/2*(2*(a^6 - 3*a^
5*b + 3*a^4*b^2 - a^3*b^3)*d^3*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15
*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4))*cos(d*x + c)*sin(d*x + c) + (3*a^3*b + 4*a^2*b^2 + a*b^3)*d*cos(d*x + c)
*sin(d*x + c))*sqrt(((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b
+ 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4)) - a*b - 3*b^2)/((a^4 - 3*a^3*b + 3*a^2*b^2
- a*b^3)*d^2)) - 1/4*(2*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*d^2*cos(d*x + c)^2 - (a^5 - 3*a^4*b + 3*a^3*b^2
- a^2*b^3)*d^2)*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a
^4*b^5 + a^3*b^6)*d^4))) - (a - b)*d*sqrt(((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2*sqrt((9*a^2*b^3 + 6*a*b^4 +
b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4)) - a*b - 3*b^2)/((a^4
- 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2))*cos(d*x + c)*log(3/4*a*b^2 + 1/4*b^3 - 1/4*(3*a*b^2 + b^3)*cos(d*x + c)^
2 - 1/2*(2*(a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*d^3*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7
*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4))*cos(d*x + c)*sin(d*x + c) + (3*a^3*b + 4*a^2*b^2 +
a*b^3)*d*cos(d*x + c)*sin(d*x + c))*sqrt(((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2*sqrt((9*a^2*b^3 + 6*a*b^4 +
b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4)) - a*b - 3*b^2)/((a^4
- 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2)) - 1/4*(2*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*d^2*cos(d*x + c)^2 - (a^5
- 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*d^2)*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*
b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4))) + (a - b)*d*sqrt(-((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2*sqrt
((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4
)) + a*b + 3*b^2)/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2))*cos(d*x + c)*log(-3/4*a*b^2 - 1/4*b^3 + 1/4*(3*a*
b^2 + b^3)*cos(d*x + c)^2 + 1/2*(2*(a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*d^3*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/
((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4))*cos(d*x + c)*sin(d*x + c)
- (3*a^3*b + 4*a^2*b^2 + a*b^3)*d*cos(d*x + c)*sin(d*x + c))*sqrt(-((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2*sq
rt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d
^4)) + a*b + 3*b^2)/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2)) - 1/4*(2*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*
d^2*cos(d*x + c)^2 - (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*d^2)*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8
*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4))) - (a - b)*d*sqrt(-((a^4 - 3*a^3*b + 3*
a^2*b^2 - a*b^3)*d^2*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 -
6*a^4*b^5 + a^3*b^6)*d^4)) + a*b + 3*b^2)/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2))*cos(d*x + c)*log(-3/4*a*
b^2 - 1/4*b^3 + 1/4*(3*a*b^2 + b^3)*cos(d*x + c)^2 - 1/2*(2*(a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*d^3*sqrt((9*
a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4))*c
os(d*x + c)*sin(d*x + c) - (3*a^3*b + 4*a^2*b^2 + a*b^3)*d*cos(d*x + c)*sin(d*x + c))*sqrt(-((a^4 - 3*a^3*b +
3*a^2*b^2 - a*b^3)*d^2*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4
- 6*a^4*b^5 + a^3*b^6)*d^4)) + a*b + 3*b^2)/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2)) - 1/4*(2*(a^5 - 3*a^4*
b + 3*a^3*b^2 - a^2*b^3)*d^2*cos(d*x + c)^2 - (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*d^2)*sqrt((9*a^2*b^3 + 6*a
*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4))) + 8*sin(d*x +
c))/((a - b)*d*cos(d*x + c))

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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)**2/(a-b*sin(d*x+c)**4),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a-b*sin(d*x+c)^4),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError