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

Optimal. Leaf size=123 \[ -\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right )}+\frac {b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac {(a+2 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac {(a-2 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

[Out]

-1/4*(a+2*b)*ln(1-sin(d*x+c))/(a+b)^2/d+1/4*(a-2*b)*ln(1+sin(d*x+c))/(a-b)^2/d+b^3*ln(a+b*sin(d*x+c))/(a^2-b^2
)^2/d-1/2*sec(d*x+c)^2*(b-a*sin(d*x+c))/d/(a^2-b^2)

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Rubi [A]  time = 0.16, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2668, 741, 801} \[ \frac {b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right )}-\frac {(a+2 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac {(a-2 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 2*b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^2*d) + ((a - 2*b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^2*d) + (b^3*
Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^2*d) - (Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/(2*(a^2 - b^2)*d)

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.58, size = 170, normalized size = 1.38 \[ \frac {\frac {4 b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {1}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {1}{(a-b) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {2 (a+2 b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{(a+b)^2}+\frac {2 (a-2 b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{(a-b)^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*(a + 2*b)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(a + b)^2 + (2*(a - 2*b)*Log[Cos[(c + d*x)/2] + Sin[(
c + d*x)/2]])/(a - b)^2 + (4*b^3*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^2 + 1/((a + b)*(Cos[(c + d*x)/2] - Sin[(
c + d*x)/2])^2) - 1/((a - b)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2))/(4*d)

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fricas [A]  time = 0.58, size = 153, normalized size = 1.24 \[ \frac {4 \, b^{3} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.43, size = 177, normalized size = 1.44 \[ \frac {\frac {4 \, b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} + \frac {{\left (a - 2 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a + 2 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (b^{3} \sin \left (d x + c\right )^{2} - a^{3} \sin \left (d x + c\right ) + a b^{2} \sin \left (d x + c\right ) + a^{2} b - 2 \, b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.18, size = 164, normalized size = 1.33 \[ -\frac {1}{d \left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a}{4 d \left (a +b \right )^{2}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) b}{2 d \left (a +b \right )^{2}}+\frac {b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{d \left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) a}{4 d \left (a -b \right )^{2}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right ) b}{2 d \left (a -b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+b*sin(d*x+c)),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-1/2/d/(a+b)^2*ln(sin(d*x+c)-1)*b+1/d*b^3/(a+b)^
2/(a-b)^2*ln(a+b*sin(d*x+c))-1/d/(4*a-4*b)/(1+sin(d*x+c))+1/4/d/(a-b)^2*ln(1+sin(d*x+c))*a-1/2/d/(a-b)^2*ln(1+
sin(d*x+c))*b

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maxima [A]  time = 0.33, size = 139, normalized size = 1.13 \[ \frac {\frac {4 \, b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a - 2 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a + 2 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (a \sin \left (d x + c\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.39, size = 148, normalized size = 1.20 \[ \frac {\frac {b}{2\,\left (a^2-b^2\right )}-\frac {a\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {1}{4\,\left (a+b\right )}\right )}{d}+\frac {b^3\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-2\,b\right )}{4\,d\,{\left (a-b\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a + b*sin(c + d*x))),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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