3.442 \(\int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=177 \[ -\frac {b \left (a^2+3 b^2\right )}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {4 a b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {(a+3 b) \log (1-\sin (c+d x))}{4 d (a+b)^3}+\frac {(a-3 b) \log (\sin (c+d x)+1)}{4 d (a-b)^3} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 177, 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} \[ -\frac {b \left (a^2+3 b^2\right )}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {4 a b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {(a+3 b) \log (1-\sin (c+d x))}{4 d (a+b)^3}+\frac {(a-3 b) \log (\sin (c+d x)+1)}{4 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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))^2} \, dx &=\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{(a+x)^2 \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 (a+b \sin (c+d x))}+\frac {b \operatorname {Subst}\left (\int \frac {a^2-3 b^2+2 a x}{(a+x)^2 \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 (a+b \sin (c+d x))}+\frac {b \operatorname {Subst}\left (\int \left (\frac {(a-b) (a+3 b)}{2 b (a+b)^2 (b-x)}+\frac {a^2+3 b^2}{(a-b) (a+b) (a+x)^2}+\frac {8 a b^2}{(a-b)^2 (a+b)^2 (a+x)}+\frac {(a-3 b) (a+b)}{2 (a-b)^2 b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\frac {(a+3 b) \log (1-\sin (c+d x))}{4 (a+b)^3 d}+\frac {(a-3 b) \log (1+\sin (c+d x))}{4 (a-b)^3 d}+\frac {4 a b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac {b \left (a^2+3 b^2\right )}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.75, size = 222, normalized size = 1.25 \[ \frac {-b \left (-a^2-3 b^2\right ) \left (\frac {1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac {\log (\sin (c+d x)+1)}{2 b (a-b)^2}-\frac {2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}\right )+\frac {a ((a-b) \log (1-\sin (c+d x))-(a+b) \log (\sin (c+d x)+1)+2 b \log (a+b \sin (c+d x)))}{(a-b) (a+b)}+\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{a+b \sin (c+d x)}}{2 d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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fricas [B]  time = 0.58, size = 381, normalized size = 2.15 \[ -\frac {2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} - 16 \, {\left (a b^{4} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a^{2} b^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left ({\left (a^{4} b - 6 \, a^{2} b^{3} - 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{5} - 6 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (a^{4} b - 6 \, a^{2} b^{3} + 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{5} - 6 \, a^{3} b^{2} + 8 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/4*(2*a^4*b - 4*a^2*b^3 + 2*b^5 + 2*(a^4*b + 2*a^2*b^3 - 3*b^5)*cos(d*x + c)^2 - 16*(a*b^4*cos(d*x + c)^2*si
n(d*x + c) + a^2*b^3*cos(d*x + c)^2)*log(b*sin(d*x + c) + a) - ((a^4*b - 6*a^2*b^3 - 8*a*b^4 - 3*b^5)*cos(d*x
+ c)^2*sin(d*x + c) + (a^5 - 6*a^3*b^2 - 8*a^2*b^3 - 3*a*b^4)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((a^4*b
- 6*a^2*b^3 + 8*a*b^4 - 3*b^5)*cos(d*x + c)^2*sin(d*x + c) + (a^5 - 6*a^3*b^2 + 8*a^2*b^3 - 3*a*b^4)*cos(d*x +
c)^2)*log(-sin(d*x + c) + 1) - 2*(a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c))/((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^
7)*d*cos(d*x + c)^2*sin(d*x + c) + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2)

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giac [A]  time = 0.99, size = 244, normalized size = 1.38 \[ \frac {\frac {16 \, a b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} + \frac {{\left (a - 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a + 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left (a^{2} b \sin \left (d x + c\right )^{2} + 3 \, b^{3} \sin \left (d x + c\right )^{2} + a^{3} \sin \left (d x + c\right ) - a b^{2} \sin \left (d x + c\right ) - 2 \, a^{2} b - 2 \, b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - a\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/4*(16*a*b^4*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) + (a - 3*b)*log(abs(sin(d*x +
c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (a + 3*b)*log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^
3) - 2*(a^2*b*sin(d*x + c)^2 + 3*b^3*sin(d*x + c)^2 + a^3*sin(d*x + c) - a*b^2*sin(d*x + c) - 2*a^2*b - 2*b^3)
/((a^4 - 2*a^2*b^2 + b^4)*(b*sin(d*x + c)^3 + a*sin(d*x + c)^2 - b*sin(d*x + c) - a)))/d

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maple [A]  time = 0.32, size = 192, normalized size = 1.08 \[ -\frac {1}{4 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a}{4 d \left (a +b \right )^{3}}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) b}{4 d \left (a +b \right )^{3}}-\frac {b^{3}}{d \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {4 b^{3} a \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {1}{4 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) a}{4 d \left (a -b \right )^{3}}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) b}{4 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 275, normalized size = 1.55 \[ \frac {\frac {16 \, a b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left (2 \, a^{2} b + 2 \, b^{3} - {\left (a^{2} b + 3 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )\right )}}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{3} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/4*(16*a*b^3*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + (a - 3*b)*log(sin(d*x + c) + 1)/(a
^3 - 3*a^2*b + 3*a*b^2 - b^3) - (a + 3*b)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*(2*a^2*b +
2*b^3 - (a^2*b + 3*b^3)*sin(d*x + c)^2 - (a^3 - a*b^2)*sin(d*x + c))/(a^5 - 2*a^3*b^2 + a*b^4 - (a^4*b - 2*a^
2*b^3 + b^5)*sin(d*x + c)^3 - (a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c)^2 + (a^4*b - 2*a^2*b^3 + b^5)*sin(d*x + c
)))/d

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mupad [B]  time = 5.47, size = 227, normalized size = 1.28 \[ \frac {\frac {{\sin \left (c+d\,x\right )}^2\,\left (a^2\,b+3\,b^3\right )}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {a^2\,b+b^3}{{\left (a^2-b^2\right )}^2}+\frac {a\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}}{d\,\left (-b\,{\sin \left (c+d\,x\right )}^3-a\,{\sin \left (c+d\,x\right )}^2+b\,\sin \left (c+d\,x\right )+a\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{2\,{\left (a+b\right )}^3}+\frac {1}{4\,{\left (a+b\right )}^2}\right )}{d}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-3\,b\right )}{4\,d\,{\left (a-b\right )}^3}+\frac {4\,a\,b^3\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a + b*sin(c + d*x))^2),x)

[Out]

((sin(c + d*x)^2*(a^2*b + 3*b^3))/(2*(a^4 + b^4 - 2*a^2*b^2)) - (a^2*b + b^3)/(a^2 - b^2)^2 + (a*sin(c + d*x))
/(2*(a^2 - b^2)))/(d*(a + b*sin(c + d*x) - a*sin(c + d*x)^2 - b*sin(c + d*x)^3)) - (log(sin(c + d*x) - 1)*(b/(
2*(a + b)^3) + 1/(4*(a + b)^2)))/d + (log(sin(c + d*x) + 1)*(a - 3*b))/(4*d*(a - b)^3) + (4*a*b^3*log(a + b*si
n(c + d*x)))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+b*sin(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)**3/(a + b*sin(c + d*x))**2, x)

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