3.409 \(\int \frac{\sec ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

Optimal. Leaf size=175 \[ \frac{b^{3/4} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}+\sqrt{b}\right )^2}+\frac{b^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}-\sqrt{b}\right )^2}+\frac{1}{4 d (a-b) (1-\sin (c+d x))}-\frac{1}{4 d (a-b) (\sin (c+d x)+1)}+\frac{(a-5 b) \tanh ^{-1}(\sin (c+d x))}{2 d (a-b)^2} \]

[Out]

(b^(3/4)*ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] + Sqrt[b])^2*d) + ((a - 5*b)*ArcTanh[Sin[
c + d*x]])/(2*(a - b)^2*d) + (b^(3/4)*ArcTanh[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] - Sqrt[b])^
2*d) + 1/(4*(a - b)*d*(1 - Sin[c + d*x])) - 1/(4*(a - b)*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.213722, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3223, 1171, 207, 1167, 205, 208} \[ \frac{b^{3/4} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}+\sqrt{b}\right )^2}+\frac{b^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \left (\sqrt{a}-\sqrt{b}\right )^2}+\frac{1}{4 d (a-b) (1-\sin (c+d x))}-\frac{1}{4 d (a-b) (\sin (c+d x)+1)}+\frac{(a-5 b) \tanh ^{-1}(\sin (c+d x))}{2 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a - b*Sin[c + d*x]^4),x]

[Out]

(b^(3/4)*ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] + Sqrt[b])^2*d) + ((a - 5*b)*ArcTanh[Sin[
c + d*x]])/(2*(a - b)^2*d) + (b^(3/4)*ArcTanh[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] - Sqrt[b])^
2*d) + 1/(4*(a - b)*d*(1 - Sin[c + d*x])) - 1/(4*(a - b)*d*(1 + Sin[c + d*x]))

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 ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2 \left (a-b x^4\right )} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{4 (a-b) (-1+x)^2}+\frac{1}{4 (a-b) (1+x)^2}+\frac{-a+5 b}{2 (a-b)^2 \left (-1+x^2\right )}+\frac{b \left (a+b+2 b x^2\right )}{(a-b)^2 \left (a-b x^4\right )}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{1}{4 (a-b) d (1-\sin (c+d x))}-\frac{1}{4 (a-b) d (1+\sin (c+d x))}-\frac{(a-5 b) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sin (c+d x)\right )}{2 (a-b)^2 d}+\frac{b \operatorname{Subst}\left (\int \frac{a+b+2 b x^2}{a-b x^4} \, dx,x,\sin (c+d x)\right )}{(a-b)^2 d}\\ &=\frac{(a-5 b) \tanh ^{-1}(\sin (c+d x))}{2 (a-b)^2 d}+\frac{1}{4 (a-b) d (1-\sin (c+d x))}-\frac{1}{4 (a-b) d (1+\sin (c+d x))}+\frac{b^{3/2} \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 )^2 d}-\frac{b^{3/2} \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 )^2 d}\\ &=\frac{b^{3/4} \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 )^2 d}+\frac{(a-5 b) \tanh ^{-1}(\sin (c+d x))}{2 (a-b)^2 d}+\frac{b^{3/4} \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 )^2 d}+\frac{1}{4 (a-b) d (1-\sin (c+d x))}-\frac{1}{4 (a-b) d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [C]  time = 1.00978, size = 255, normalized size = 1.46 \[ -\frac{\frac{b^{3/4} \log \left (\sqrt [4]{a}-\sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt{a}-\sqrt{b}\right )^2}-\frac{i b^{3/4} \log \left (\sqrt [4]{a}-i \sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt{a}+\sqrt{b}\right )^2}+\frac{i b^{3/4} \log \left (\sqrt [4]{a}+i \sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt{a}+\sqrt{b}\right )^2}-\frac{b^{3/4} \log \left (\sqrt [4]{a}+\sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt{a}-\sqrt{b}\right )^2}+\frac{1}{(a-b) (\sin (c+d x)-1)}+\frac{1}{(a-b) (\sin (c+d x)+1)}-\frac{2 (a-5 b) \tanh ^{-1}(\sin (c+d x))}{(a-b)^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a - b*Sin[c + d*x]^4),x]

[Out]

-((-2*(a - 5*b)*ArcTanh[Sin[c + d*x]])/(a - b)^2 + (b^(3/4)*Log[a^(1/4) - b^(1/4)*Sin[c + d*x]])/(a^(3/4)*(Sqr
t[a] - Sqrt[b])^2) - (I*b^(3/4)*Log[a^(1/4) - I*b^(1/4)*Sin[c + d*x]])/(a^(3/4)*(Sqrt[a] + Sqrt[b])^2) + (I*b^
(3/4)*Log[a^(1/4) + I*b^(1/4)*Sin[c + d*x]])/(a^(3/4)*(Sqrt[a] + Sqrt[b])^2) - (b^(3/4)*Log[a^(1/4) + b^(1/4)*
Sin[c + d*x]])/(a^(3/4)*(Sqrt[a] - Sqrt[b])^2) + 1/((a - b)*(-1 + Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x
])))/(4*d)

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Maple [B]  time = 0.108, size = 415, normalized size = 2.4 \begin{align*} -{\frac{1}{d \left ( 4\,a-4\,b \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) a}{4\,d \left ( a-b \right ) ^{2}}}+{\frac{5\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) b}{4\,d \left ( a-b \right ) ^{2}}}-{\frac{1}{d \left ( 4\,a-4\,b \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) a}{4\,d \left ( a-b \right ) ^{2}}}-{\frac{5\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) b}{4\,d \left ( a-b \right ) ^{2}}}+{\frac{b}{2\,d \left ( a-b \right ) ^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sin \left ( dx+c \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ) }+{\frac{{b}^{2}}{2\,d \left ( a-b \right ) ^{2}a}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sin \left ( dx+c \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ) }+{\frac{b}{4\,d \left ( a-b \right ) ^{2}}\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}}{4\,d \left ( a-b \right ) ^{2}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}{d \left ( a-b \right ) ^{2}}\arctan \left ({\sin \left ( dx+c \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{b}{2\,d \left ( a-b \right ) ^{2}}\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)^3/(a-b*sin(d*x+c)^4),x)

[Out]

-1/d/(4*a-4*b)/(sin(d*x+c)-1)-1/4/d/(a-b)^2*ln(sin(d*x+c)-1)*a+5/4/d/(a-b)^2*ln(sin(d*x+c)-1)*b-1/d/(4*a-4*b)/
(1+sin(d*x+c))+1/4/d/(a-b)^2*ln(1+sin(d*x+c))*a-5/4/d/(a-b)^2*ln(1+sin(d*x+c))*b+1/2/d*b/(a-b)^2*(a/b)^(1/4)*a
rctan(sin(d*x+c)/(a/b)^(1/4))+1/2/d*b^2/(a-b)^2*(a/b)^(1/4)/a*arctan(sin(d*x+c)/(a/b)^(1/4))+1/4/d*b/(a-b)^2*(
a/b)^(1/4)*ln((sin(d*x+c)+(a/b)^(1/4))/(sin(d*x+c)-(a/b)^(1/4)))+1/4/d*b^2/(a-b)^2*(a/b)^(1/4)/a*ln((sin(d*x+c
)+(a/b)^(1/4))/(sin(d*x+c)-(a/b)^(1/4)))-1/d*b/(a-b)^2/(a/b)^(1/4)*arctan(sin(d*x+c)/(a/b)^(1/4))+1/2/d*b/(a-b
)^2/(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)^3/(a-b*sin(d*x+c)^4),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 9.82203, size = 5469, normalized size = 31.25 \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)^3/(a-b*sin(d*x+c)^4),x, algorithm="fricas")

[Out]

-1/4*((a^2 - 2*a*b + b^2)*d*sqrt(((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2*sqrt((a^4*b^3 + 12*a^3*b
^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a^11 - 8*a^10*b + 28*a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a
^5*b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) + 4*a*b^2 + 4*b^3)/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2))*
cos(d*x + c)^2*log(1/2*(a^2*b^2 + 6*a*b^3 + b^4)*sin(d*x + c) + 1/2*(2*(a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3
+ a^3*b^4)*d^3*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a^11 - 8*a^10*b + 28*a^9*b^2 - 56*a
^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5*b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) - (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 +
a*b^4)*d)*sqrt(((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 +
12*a*b^6 + b^7)/((a^11 - 8*a^10*b + 28*a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5*b^6 - 8*a^4*b^7
+ a^3*b^8)*d^4)) + 4*a*b^2 + 4*b^3)/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2))) - (a^2 - 2*a*b +
b^2)*d*sqrt(-((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12
*a*b^6 + b^7)/((a^11 - 8*a^10*b + 28*a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5*b^6 - 8*a^4*b^7 +
a^3*b^8)*d^4)) - 4*a*b^2 - 4*b^3)/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2))*cos(d*x + c)^2*log(1
/2*(a^2*b^2 + 6*a*b^3 + b^4)*sin(d*x + c) + 1/2*(2*(a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*d^3*sqrt(
(a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a^11 - 8*a^10*b + 28*a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4
- 56*a^6*b^5 + 28*a^5*b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) + (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 + a*b^4)*d)*sqrt(-((a^
5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a
^11 - 8*a^10*b + 28*a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5*b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) -
4*a*b^2 - 4*b^3)/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2))) - (a^2 - 2*a*b + b^2)*d*sqrt(((a^5 -
4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a^11
- 8*a^10*b + 28*a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5*b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) + 4*
a*b^2 + 4*b^3)/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2))*cos(d*x + c)^2*log(-1/2*(a^2*b^2 + 6*a*b
^3 + b^4)*sin(d*x + c) + 1/2*(2*(a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*d^3*sqrt((a^4*b^3 + 12*a^3*b
^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a^11 - 8*a^10*b + 28*a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a
^5*b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) - (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 + a*b^4)*d)*sqrt(((a^5 - 4*a^4*b + 6*a^3*
b^2 - 4*a^2*b^3 + a*b^4)*d^2*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a^11 - 8*a^10*b + 28*
a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5*b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) + 4*a*b^2 + 4*b^3)/((
a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2))) + (a^2 - 2*a*b + b^2)*d*sqrt(-((a^5 - 4*a^4*b + 6*a^3*b^
2 - 4*a^2*b^3 + a*b^4)*d^2*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a^11 - 8*a^10*b + 28*a^
9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5*b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) - 4*a*b^2 - 4*b^3)/((a^
5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2))*cos(d*x + c)^2*log(-1/2*(a^2*b^2 + 6*a*b^3 + b^4)*sin(d*x +
c) + 1/2*(2*(a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*d^3*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 1
2*a*b^6 + b^7)/((a^11 - 8*a^10*b + 28*a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5*b^6 - 8*a^4*b^7
+ a^3*b^8)*d^4)) + (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 + a*b^4)*d)*sqrt(-((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 +
a*b^4)*d^2*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a^11 - 8*a^10*b + 28*a^9*b^2 - 56*a^8*b
^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5*b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) - 4*a*b^2 - 4*b^3)/((a^5 - 4*a^4*b + 6*
a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2))) - (a - 5*b)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (a - 5*b)*cos(d*x + c)^
2*log(-sin(d*x + c) + 1) - 2*(a - b)*sin(d*x + c))/((a^2 - 2*a*b + b^2)*d*cos(d*x + c)^2)

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

[Out]

Timed out

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Giac [B]  time = 5.73722, size = 635, normalized size = 3.63 \begin{align*} \frac{\frac{2 \,{\left (\left (-a b^{3}\right )^{\frac{1}{4}}{\left (a b + b^{2}\right )} + 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^{3} b - 2 \, \sqrt{2} a^{2} b^{2} + \sqrt{2} a b^{3}} + \frac{2 \,{\left (\left (-a b^{3}\right )^{\frac{1}{4}}{\left (a b + b^{2}\right )} + 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^{3} b - 2 \, \sqrt{2} a^{2} b^{2} + \sqrt{2} a b^{3}} + \frac{{\left (\left (-a b^{3}\right )^{\frac{1}{4}}{\left (a b + b^{2}\right )} - 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^{3} b - 2 \, \sqrt{2} a^{2} b^{2} + \sqrt{2} a b^{3}} - \frac{{\left (\left (-a b^{3}\right )^{\frac{1}{4}}{\left (a b + b^{2}\right )} - 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^{3} b - 2 \, \sqrt{2} a^{2} b^{2} + \sqrt{2} a b^{3}} + \frac{{\left (a - 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{{\left (a - 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{2 \, \sin \left (d x + c\right )}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}{\left (a - b\right )}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a-b*sin(d*x+c)^4),x, algorithm="giac")

[Out]

1/4*(2*((-a*b^3)^(1/4)*(a*b + b^2) + 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^3*b - 2*sqrt(2)*a^2*b^2 + sqrt(2)*a*b^3) + 2*((-a*b^3)^(1/4)*(a*b + b^2) + 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^3*b - 2*sqrt(
2)*a^2*b^2 + sqrt(2)*a*b^3) + ((-a*b^3)^(1/4)*(a*b + b^2) - 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^3*b - 2*sqrt(2)*a^2*b^2 + sqrt(2)*a*b^3) - ((-a*b^3)^(1/4)*(a*
b + b^2) - 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^3
*b - 2*sqrt(2)*a^2*b^2 + sqrt(2)*a*b^3) + (a - 5*b)*log(abs(sin(d*x + c) + 1))/(a^2 - 2*a*b + b^2) - (a - 5*b)
*log(abs(sin(d*x + c) - 1))/(a^2 - 2*a*b + b^2) - 2*sin(d*x + c)/((sin(d*x + c)^2 - 1)*(a - b)))/d