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

Optimal. Leaf size=1093 \[ \text{result too large to display} \]

[Out]

(-2*(-1)^(2/3)*a^(2/3)*b^(8/3)*ArcTan[((-1)^(1/3)*b^(1/3) - a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - (-1)^(2/3
)*b^(2/3)]])/(Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*(a^2 - b^2)^2*d) - (2*b^2*(2*a^2 + b^2)*ArcTan[((-1)^(1/3)*b^
(1/3) - a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3*a^(2/3)*Sqrt[a^(2/3) - (-1)^(2/3)*b^
(2/3)]*(a^2 - b^2)^2*d) + (2*a^(2/3)*b^(8/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3
)]])/(Sqrt[a^(2/3) - b^(2/3)]*(a^2 - b^2)^2*d) + (2*b^2*(2*a^2 + b^2)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/
2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*a^(2/3)*Sqrt[a^(2/3) - b^(2/3)]*(a^2 - b^2)^2*d) + (2*b^(4/3)*(a^2 + 2*b^2)*A
rcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*Sqrt[a^(2/3) - b^(2/3)]*(a^2 - b^2)^2*
d) - (2*(-1)^(1/3)*a^(2/3)*b^(8/3)*ArcTan[((-1)^(2/3)*b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^
(1/3)*b^(2/3)]])/(Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*(a^2 - b^2)^2*d) + (2*b^2*(2*a^2 + b^2)*ArcTan[((-1)^(2/3
)*b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]])/(3*a^(2/3)*Sqrt[a^(2/3) + (-1)^(1/3
)*b^(2/3)]*(a^2 - b^2)^2*d) - (2*b^(4/3)*(a^2 + 2*b^2)*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^(1/3)*Tan[(c + d*x)/2])
/Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]])/(3*Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]*(a^2 - b^2)^2*d) - (2*b^(4/3
)*(a^2 + 2*b^2)*ArcTanh[(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x)/2])/Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]])/(
3*Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]*(a^2 - b^2)^2*d) + Cos[c + d*x]/(12*(a + b)*d*(1 - Sin[c + d*x])^2) + Cos
[c + d*x]/(12*(a + b)*d*(1 - Sin[c + d*x])) + ((a + 4*b)*Cos[c + d*x])/(4*(a + b)^2*d*(1 - Sin[c + d*x])) - Co
s[c + d*x]/(12*(a - b)*d*(1 + Sin[c + d*x])^2) - ((a - 4*b)*Cos[c + d*x])/(4*(a - b)^2*d*(1 + Sin[c + d*x])) -
Cos[c + d*x]/(12*(a - b)*d*(1 + Sin[c + d*x]))

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Rubi [F]  time = 0.0462909, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx &=\int \frac{\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx\\ \end{align*}

Mathematica [C]  time = 1.69115, size = 679, normalized size = 0.62 \[ \frac{\sec ^3(c+d x) \left (-3 b \left (5 a^2+13 b^2\right ) \cos (c+d x)+12 b \left (a^2+2 b^2\right ) \cos (2 (c+d x))-5 a^2 b \cos (3 (c+d x))+4 a^2 b+12 a^3 \sin (c+d x)+4 a^3 \sin (3 (c+d x))-30 a b^2 \sin (c+d x)-22 a b^2 \sin (3 (c+d x))-13 b^3 \cos (3 (c+d x))+32 b^3\right )+4 i b^2 \text{RootSum}\left [8 \text{$\#$1}^3 a+i \text{$\#$1}^6 b-3 i \text{$\#$1}^4 b+3 i \text{$\#$1}^2 b-i b\& ,\frac{-i \text{$\#$1}^4 a^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+10 i \text{$\#$1}^2 a^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-i a^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^4 a^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-20 \text{$\#$1}^2 a^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-6 \text{$\#$1}^3 a b \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+6 \text{$\#$1} a b \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-12 i \text{$\#$1}^3 a b \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-2 i \text{$\#$1}^4 b^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+8 i \text{$\#$1}^2 b^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-2 i b^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+4 \text{$\#$1}^4 b^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-16 \text{$\#$1}^2 b^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+2 a^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+12 i \text{$\#$1} a b \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+4 b^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{-4 i \text{$\#$1}^2 a+\text{$\#$1}^5 b-2 \text{$\#$1}^3 b+\text{$\#$1} b}\& \right ]}{24 d (a-b)^2 (a+b)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((4*I)*b^2*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^6 & , (2*a^2*ArcTan[Sin[c + d*x]/(
Cos[c + d*x] - #1)] + 4*b^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] - I*a^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]
- (2*I)*b^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + (12*I)*a*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1 + 6*a*
b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1 - 20*a^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - 16*b^2*ArcTan[
Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + (10*I)*a^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + (8*I)*b^2*Log[1 -
2*Cos[c + d*x]*#1 + #1^2]*#1^2 - (12*I)*a*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^3 - 6*a*b*Log[1 - 2*C
os[c + d*x]*#1 + #1^2]*#1^3 + 2*a^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 + 4*b^2*ArcTan[Sin[c + d*x]/
(Cos[c + d*x] - #1)]*#1^4 - I*a^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4 - (2*I)*b^2*Log[1 - 2*Cos[c + d*x]*#1
+ #1^2]*#1^4)/(b*#1 - (4*I)*a*#1^2 - 2*b*#1^3 + b*#1^5) & ] + Sec[c + d*x]^3*(4*a^2*b + 32*b^3 - 3*b*(5*a^2 +
13*b^2)*Cos[c + d*x] + 12*b*(a^2 + 2*b^2)*Cos[2*(c + d*x)] - 5*a^2*b*Cos[3*(c + d*x)] - 13*b^3*Cos[3*(c + d*x
)] + 12*a^3*Sin[c + d*x] - 30*a*b^2*Sin[c + d*x] + 4*a^3*Sin[3*(c + d*x)] - 22*a*b^2*Sin[3*(c + d*x)]))/(24*(a
- b)^2*(a + b)^2*d)

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Maple [C]  time = 0.241, size = 346, normalized size = 0.3 \begin{align*}{\frac{{b}^{2}}{3\,d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{6}+3\,a{{\it \_Z}}^{4}+8\,b{{\it \_Z}}^{3}+3\,a{{\it \_Z}}^{2}+a \right ) }{\frac{ \left ( 2\,{a}^{2}+{b}^{2} \right ){{\it \_R}}^{4}-6\,{{\it \_R}}^{3}ab+2\, \left ( 4\,{a}^{2}+5\,{b}^{2} \right ){{\it \_R}}^{2}-6\,{\it \_R}\,ab+2\,{a}^{2}+{b}^{2}}{{{\it \_R}}^{5}a+2\,{{\it \_R}}^{3}a+4\,{{\it \_R}}^{2}b+{\it \_R}\,a}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -{\it \_R} \right ) }}-{\frac{2}{3\,d \left ( 2\,a+2\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{d \left ( 2\,a+2\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{a}{d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{5\,b}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{2}{3\,d \left ( 2\,a-2\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{d \left ( 2\,a-2\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{a}{d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{5\,b}{2\,d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+b*sin(d*x+c)^3),x)

[Out]

1/3/d*b^2/(a-b)^2/(a+b)^2*sum(((2*a^2+b^2)*_R^4-6*_R^3*a*b+2*(4*a^2+5*b^2)*_R^2-6*_R*a*b+2*a^2+b^2)/(_R^5*a+2*
_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a))-2/3/d/(tan(1/2
*d*x+1/2*c)-1)^3/(2*a+2*b)-1/d/(2*a+2*b)/(tan(1/2*d*x+1/2*c)-1)^2-1/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)*a-5/2/d/(
a+b)^2/(tan(1/2*d*x+1/2*c)-1)*b-2/3/d/(tan(1/2*d*x+1/2*c)+1)^3/(2*a-2*b)+1/d/(2*a-2*b)/(tan(1/2*d*x+1/2*c)+1)^
2-1/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)*a+5/2/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)*b

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Maxima [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)^4/(a+b*sin(d*x+c)^3),x, algorithm="maxima")

[Out]

Timed out

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Fricas [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)^4/(a+b*sin(d*x+c)^3),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{4}}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(d*x + c)^4/(b*sin(d*x + c)^3 + a), x)