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

Optimal. Leaf size=117 \[ \frac{\sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}+\sqrt{b}\right )}-\frac{\sqrt [4]{b} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}-\sqrt{b}\right )}+\frac{\tanh ^{-1}(\sin (c+d x))}{d (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.153484, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3223, 1171, 207, 1167, 205, 208} \[ \frac{\sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}+\sqrt{b}\right )}-\frac{\sqrt [4]{b} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}-\sqrt{b}\right )}+\frac{\tanh ^{-1}(\sin (c+d x))}{d (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3223

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x]
, x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0
] || IGtQ[p, 0] || IntegersQ[m, p])

Rule 1171

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

Rule 207

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

Rule 1167

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

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]

Rubi steps

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

Mathematica [C]  time = 0.181685, size = 184, normalized size = 1.57 \[ \frac{4 a^{3/4} \tanh ^{-1}(\sin (c+d x))+\sqrt [4]{b} \left (\left (\sqrt{a}+\sqrt{b}\right ) \log \left (\sqrt [4]{a}-\sqrt [4]{b} \sin (c+d x)\right )+i \left (\left (\sqrt{a}-\sqrt{b}\right ) \log \left (\sqrt [4]{a}-i \sqrt [4]{b} \sin (c+d x)\right )+\left (\sqrt{b}-\sqrt{a}\right ) \log \left (\sqrt [4]{a}+i \sqrt [4]{b} \sin (c+d x)\right )+i \left (\sqrt{a}+\sqrt{b}\right ) \log \left (\sqrt [4]{a}+\sqrt [4]{b} \sin (c+d x)\right )\right )\right )}{4 a^{3/4} d (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^(3/4)*ArcTanh[Sin[c + d*x]] + b^(1/4)*((Sqrt[a] + Sqrt[b])*Log[a^(1/4) - b^(1/4)*Sin[c + d*x]] + I*((Sqrt
[a] - Sqrt[b])*Log[a^(1/4) - I*b^(1/4)*Sin[c + d*x]] + (-Sqrt[a] + Sqrt[b])*Log[a^(1/4) + I*b^(1/4)*Sin[c + d*
x]] + I*(Sqrt[a] + Sqrt[b])*Log[a^(1/4) + b^(1/4)*Sin[c + d*x]])))/(4*a^(3/4)*(a - b)*d)

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Maple [B]  time = 0.095, size = 229, normalized size = 2. \begin{align*} -{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{d \left ( 2\,a-2\,b \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d \left ( 2\,a-2\,b \right ) }}-{\frac{b}{4\,d \left ( a-b \right ) a}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( \sin \left ( dx+c \right ) +\sqrt [4]{{\frac{a}{b}}} \right ) \left ( \sin \left ( dx+c \right ) -\sqrt [4]{{\frac{a}{b}}} \right ) ^{-1}} \right ) }-{\frac{b}{2\,d \left ( a-b \right ) a}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sin \left ( dx+c \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ) }+{\frac{1}{2\,d \left ( a-b \right ) }\arctan \left ({\sin \left ( dx+c \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{1}{4\,d \left ( a-b \right ) }\ln \left ({ \left ( \sin \left ( dx+c \right ) +\sqrt [4]{{\frac{a}{b}}} \right ) \left ( \sin \left ( dx+c \right ) -\sqrt [4]{{\frac{a}{b}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/(2*a-2*b)*ln(sin(d*x+c)-1)+1/d/(2*a-2*b)*ln(1+sin(d*x+c))-1/4/d*b/(a-b)*(a/b)^(1/4)/a*ln((sin(d*x+c)+(a/b
)^(1/4))/(sin(d*x+c)-(a/b)^(1/4)))-1/2/d*b/(a-b)*(a/b)^(1/4)/a*arctan(sin(d*x+c)/(a/b)^(1/4))+1/2/d/(a-b)/(a/b
)^(1/4)*arctan(sin(d*x+c)/(a/b)^(1/4))-1/4/d/(a-b)/(a/b)^(1/4)*ln((sin(d*x+c)+(a/b)^(1/4))/(sin(d*x+c)-(a/b)^(
1/4)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.22071, size = 2836, normalized size = 24.24 \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)/(a-b*sin(d*x+c)^4),x, algorithm="fricas")

[Out]

-1/4*((a - b)*d*sqrt(((a^3 - 2*a^2*b + a*b^2)*d^2*sqrt((a^2*b + 2*a*b^2 + b^3)/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4
*a^4*b^3 + a^3*b^4)*d^4)) + 2*b)/((a^3 - 2*a^2*b + a*b^2)*d^2))*log(1/2*(a*b + b^2)*sin(d*x + c) + 1/2*((a^5 -
2*a^4*b + a^3*b^2)*d^3*sqrt((a^2*b + 2*a*b^2 + b^3)/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*d^4))
- (a^2*b + a*b^2)*d)*sqrt(((a^3 - 2*a^2*b + a*b^2)*d^2*sqrt((a^2*b + 2*a*b^2 + b^3)/((a^7 - 4*a^6*b + 6*a^5*b^
2 - 4*a^4*b^3 + a^3*b^4)*d^4)) + 2*b)/((a^3 - 2*a^2*b + a*b^2)*d^2))) - (a - b)*d*sqrt(-((a^3 - 2*a^2*b + a*b^
2)*d^2*sqrt((a^2*b + 2*a*b^2 + b^3)/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*d^4)) - 2*b)/((a^3 - 2*
a^2*b + a*b^2)*d^2))*log(1/2*(a*b + b^2)*sin(d*x + c) + 1/2*((a^5 - 2*a^4*b + a^3*b^2)*d^3*sqrt((a^2*b + 2*a*b
^2 + b^3)/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*d^4)) + (a^2*b + a*b^2)*d)*sqrt(-((a^3 - 2*a^2*b
+ a*b^2)*d^2*sqrt((a^2*b + 2*a*b^2 + b^3)/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*d^4)) - 2*b)/((a^
3 - 2*a^2*b + a*b^2)*d^2))) - (a - b)*d*sqrt(((a^3 - 2*a^2*b + a*b^2)*d^2*sqrt((a^2*b + 2*a*b^2 + b^3)/((a^7 -
4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*d^4)) + 2*b)/((a^3 - 2*a^2*b + a*b^2)*d^2))*log(-1/2*(a*b + b^2)*s
in(d*x + c) + 1/2*((a^5 - 2*a^4*b + a^3*b^2)*d^3*sqrt((a^2*b + 2*a*b^2 + b^3)/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4*
a^4*b^3 + a^3*b^4)*d^4)) - (a^2*b + a*b^2)*d)*sqrt(((a^3 - 2*a^2*b + a*b^2)*d^2*sqrt((a^2*b + 2*a*b^2 + b^3)/(
(a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*d^4)) + 2*b)/((a^3 - 2*a^2*b + a*b^2)*d^2))) + (a - b)*d*sqr
t(-((a^3 - 2*a^2*b + a*b^2)*d^2*sqrt((a^2*b + 2*a*b^2 + b^3)/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4
)*d^4)) - 2*b)/((a^3 - 2*a^2*b + a*b^2)*d^2))*log(-1/2*(a*b + b^2)*sin(d*x + c) + 1/2*((a^5 - 2*a^4*b + a^3*b^
2)*d^3*sqrt((a^2*b + 2*a*b^2 + b^3)/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*d^4)) + (a^2*b + a*b^2)
*d)*sqrt(-((a^3 - 2*a^2*b + a*b^2)*d^2*sqrt((a^2*b + 2*a*b^2 + b^3)/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 +
a^3*b^4)*d^4)) - 2*b)/((a^3 - 2*a^2*b + a*b^2)*d^2))) - 2*log(sin(d*x + c) + 1) + 2*log(-sin(d*x + c) + 1))/((
a - b)*d)

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

[Out]

Timed out

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Giac [B]  time = 6.21232, size = 500, normalized size = 4.27 \begin{align*} -\frac{\frac{2 \,{\left (\left (-a b^{3}\right )^{\frac{1}{4}} b^{2} + \left (-a b^{3}\right )^{\frac{3}{4}}\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-\frac{a}{b}\right )^{\frac{1}{4}} + 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{\sqrt{2} a^{2} b^{2} - \sqrt{2} a b^{3}} + \frac{2 \,{\left (\left (-a b^{3}\right )^{\frac{1}{4}} b^{2} + \left (-a b^{3}\right )^{\frac{3}{4}}\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-\frac{a}{b}\right )^{\frac{1}{4}} - 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{\sqrt{2} a^{2} b^{2} - \sqrt{2} a b^{3}} + \frac{{\left (\left (-a b^{3}\right )^{\frac{1}{4}} b^{2} - \left (-a b^{3}\right )^{\frac{3}{4}}\right )} \log \left (\sin \left (d x + c\right )^{2} + \sqrt{2} \left (-\frac{a}{b}\right )^{\frac{1}{4}} \sin \left (d x + c\right ) + \sqrt{-\frac{a}{b}}\right )}{\sqrt{2} a^{2} b^{2} - \sqrt{2} a b^{3}} - \frac{{\left (\left (-a b^{3}\right )^{\frac{1}{4}} b^{2} - \left (-a b^{3}\right )^{\frac{3}{4}}\right )} \log \left (\sin \left (d x + c\right )^{2} - \sqrt{2} \left (-\frac{a}{b}\right )^{\frac{1}{4}} \sin \left (d x + c\right ) + \sqrt{-\frac{a}{b}}\right )}{\sqrt{2} a^{2} b^{2} - \sqrt{2} a b^{3}} - \frac{2 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} + \frac{2 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a - b}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*((-a*b^3)^(1/4)*b^2 + (-a*b^3)^(3/4))*arctan(1/2*sqrt(2)*(sqrt(2)*(-a/b)^(1/4) + 2*sin(d*x + c))/(-a/b
)^(1/4))/(sqrt(2)*a^2*b^2 - sqrt(2)*a*b^3) + 2*((-a*b^3)^(1/4)*b^2 + (-a*b^3)^(3/4))*arctan(-1/2*sqrt(2)*(sqrt
(2)*(-a/b)^(1/4) - 2*sin(d*x + c))/(-a/b)^(1/4))/(sqrt(2)*a^2*b^2 - sqrt(2)*a*b^3) + ((-a*b^3)^(1/4)*b^2 - (-a
*b^3)^(3/4))*log(sin(d*x + c)^2 + sqrt(2)*(-a/b)^(1/4)*sin(d*x + c) + sqrt(-a/b))/(sqrt(2)*a^2*b^2 - sqrt(2)*a
*b^3) - ((-a*b^3)^(1/4)*b^2 - (-a*b^3)^(3/4))*log(sin(d*x + c)^2 - sqrt(2)*(-a/b)^(1/4)*sin(d*x + c) + sqrt(-a
/b))/(sqrt(2)*a^2*b^2 - sqrt(2)*a*b^3) - 2*log(abs(sin(d*x + c) + 1))/(a - b) + 2*log(abs(sin(d*x + c) - 1))/(
a - b))/d