3.1549 \(\int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=159 \[ \frac {b^2 (A b-a B) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac {\sec ^2(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{2 d \left (a^2-b^2\right )}-\frac {(a A+b (2 A+B)) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac {(a A-b (2 A-B)) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

[Out]

-1/4*(a*A+b*(2*A+B))*ln(1-sin(d*x+c))/(a+b)^2/d+1/4*(a*A-b*(2*A-B))*ln(1+sin(d*x+c))/(a-b)^2/d+b^2*(A*b-B*a)*l
n(a+b*sin(d*x+c))/(a^2-b^2)^2/d-1/2*sec(d*x+c)^2*(A*b-a*B-(A*a-B*b)*sin(d*x+c))/(a^2-b^2)/d

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Rubi [A]  time = 0.29, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2837, 823, 801} \[ \frac {b^2 (A b-a B) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac {\sec ^2(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{2 d \left (a^2-b^2\right )}-\frac {(a A+b (2 A+B)) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac {(a A-b (2 A-B)) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(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))}{a+b \sin (c+d x)} \, dx &=\frac {b^3 \operatorname {Subst}\left (\int \frac {A+\frac {B x}{b}}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\sec ^2(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{2 \left (a^2-b^2\right ) d}-\frac {b \operatorname {Subst}\left (\int \frac {-a^2 A+2 A b^2-a b B-(a A-b B) 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) (A b-a B-(a A-b B) \sin (c+d x))}{2 \left (a^2-b^2\right ) d}-\frac {b \operatorname {Subst}\left (\int \left (\frac {(a-b) (-a A-b (2 A+B))}{2 b (a+b) (b-x)}+\frac {2 b (-A b+a B)}{(a-b) (a+b) (a+x)}+\frac {(a+b) (-a A+b (2 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 A+b (2 A+B)) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {(a A-b (2 A-B)) \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac {b^2 (A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {\sec ^2(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{2 \left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [A]  time = 0.77, size = 197, normalized size = 1.24 \[ \frac {\frac {4 b^2 (A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {A+B}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {B-A}{(a-b) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {2 (a A+b (2 A+B)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{(a+b)^2}+\frac {2 (a A+b (B-2 A)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{(a-b)^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 234, normalized size = 1.47 \[ \frac {2 \, B a^{3} - 2 \, A a^{2} b - 2 \, B a b^{2} + 2 \, A b^{3} - 4 \, {\left (B a b^{2} - A b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (A a^{3} + B a^{2} b - {\left (3 \, A - 2 \, B\right )} a b^{2} - {\left (2 \, A - B\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{3} + B a^{2} b - {\left (3 \, A + 2 \, B\right )} a b^{2} + {\left (2 \, A + B\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{3} - B a^{2} b - A a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*B*a^3 - 2*A*a^2*b - 2*B*a*b^2 + 2*A*b^3 - 4*(B*a*b^2 - A*b^3)*cos(d*x + c)^2*log(b*sin(d*x + c) + a) +
(A*a^3 + B*a^2*b - (3*A - 2*B)*a*b^2 - (2*A - B)*b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (A*a^3 + B*a^2*b
- (3*A + 2*B)*a*b^2 + (2*A + B)*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^3 - B*a^2*b - A*a*b^2 + B*
b^3)*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 = 0.32, size = 260, normalized size = 1.64 \[ -\frac {\frac {4 \, {\left (B a b^{3} - A b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} + \frac {{\left (A a + 2 \, A b + B b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {{\left (A a - 2 \, A b + B b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {2 \, {\left (B a b^{2} \sin \left (d x + c\right )^{2} - A b^{3} \sin \left (d x + c\right )^{2} + A a^{3} \sin \left (d x + c\right ) - B a^{2} b \sin \left (d x + c\right ) - A a b^{2} \sin \left (d x + c\right ) + B b^{3} \sin \left (d x + c\right ) + B a^{3} - A a^{2} b - 2 \, B a b^{2} + 2 \, A b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*(B*a*b^3 - A*b^4)*log(abs(b*sin(d*x + c) + a))/(a^4*b - 2*a^2*b^3 + b^5) + (A*a + 2*A*b + B*b)*log(abs
(-sin(d*x + c) + 1))/(a^2 + 2*a*b + b^2) - (A*a - 2*A*b + B*b)*log(abs(-sin(d*x + c) - 1))/(a^2 - 2*a*b + b^2)
+ 2*(B*a*b^2*sin(d*x + c)^2 - A*b^3*sin(d*x + c)^2 + A*a^3*sin(d*x + c) - B*a^2*b*sin(d*x + c) - A*a*b^2*sin(
d*x + c) + B*b^3*sin(d*x + c) + B*a^3 - A*a^2*b - 2*B*a*b^2 + 2*A*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 = 0.50, size = 297, normalized size = 1.87 \[ -\frac {A}{d \left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}-\frac {B}{d \left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a A}{4 d \left (a +b \right )^{2}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) A b}{2 d \left (a +b \right )^{2}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) B b}{4 d \left (a +b \right )^{2}}+\frac {b^{3} \ln \left (a +b \sin \left (d x +c \right )\right ) A}{d \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {b^{2} \ln \left (a +b \sin \left (d x +c \right )\right ) a B}{d \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {A}{d \left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {B}{d \left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) a A}{4 d \left (a -b \right )^{2}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right ) A b}{2 d \left (a -b \right )^{2}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) B b}{4 d \left (a -b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.73, size = 175, normalized size = 1.10 \[ -\frac {\frac {4 \, {\left (B a b^{2} - A b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (A a - {\left (2 \, A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (A a + {\left (2 \, A + B\right )} b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (B a - A b + {\left (A a - B b\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.52, size = 197, normalized size = 1.24 \[ \frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (A\,b^3-B\,a\,b^2\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {\frac {A\,b-B\,a}{2\,\left (a^2-b^2\right )}-\frac {\sin \left (c+d\,x\right )\,\left (A\,a-B\,b\right )}{2\,\left (a^2-b^2\right )}}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (A\,a-b\,\left (2\,A-B\right )\right )}{d\,\left (4\,a^2-8\,a\,b+4\,b^2\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (A\,a+b\,\left (2\,A+B\right )\right )}{d\,\left (4\,a^2+8\,a\,b+4\,b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(a + b*sin(c + d*x))*(A*b^3 - B*a*b^2))/(d*(a^4 + b^4 - 2*a^2*b^2)) - ((A*b - B*a)/(2*(a^2 - b^2)) - (sin(
c + d*x)*(A*a - B*b))/(2*(a^2 - b^2)))/(d*cos(c + d*x)^2) + (log(sin(c + d*x) + 1)*(A*a - b*(2*A - B)))/(d*(4*
a^2 - 8*a*b + 4*b^2)) - (log(sin(c + d*x) - 1)*(A*a + b*(2*A + B)))/(d*(8*a*b + 4*a^2 + 4*b^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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