3.404 \(\int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=111 \[ \frac {a b^2 \sin (c+d x)}{2 d}+\frac {(a+2 b) (a-b)^2 \log (\sin (c+d x)+1)}{4 d}-\frac {(a-2 b) (a+b)^2 \log (1-\sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2668, 739, 774, 633, 31} \[ \frac {a b^2 \sin (c+d x)}{2 d}+\frac {(a+2 b) (a-b)^2 \log (\sin (c+d x)+1)}{4 d}-\frac {(a-2 b) (a+b)^2 \log (1-\sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 739

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*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 774

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/c, x] + Dist[1
/c, Int[(c*d*f - a*e*g + c*(e*f + d*g)*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x]

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

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Mathematica [A]  time = 1.34, size = 176, normalized size = 1.59 \[ \frac {2 a^4 b \sec ^2(c+d x)+\left (a^2-b^2\right ) \left ((a-2 b) (a+b)^2 \log (1-\sin (c+d x))-(a-b)^2 (a+2 b) \log (\sin (c+d x)+1)\right )-a \tan (c+d x) \sec (c+d x) \left (2 a^4+4 a^2 b^2+b^4 \cos (2 (c+d x))-7 b^4\right )+\tan ^2(c+d x) \left (-8 a^4 b+4 a^2 b^3-2 a b^4 \sin (c+d x)+2 b^5\right )}{4 d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2 - b^2)*((a - 2*b)*(a + b)^2*Log[1 - Sin[c + d*x]] - (a - b)^2*(a + 2*b)*Log[1 + Sin[c + d*x]]) + 2*a^4*b
*Sec[c + d*x]^2 - a*(2*a^4 + 4*a^2*b^2 - 7*b^4 + b^4*Cos[2*(c + d*x)])*Sec[c + d*x]*Tan[c + d*x] + (-8*a^4*b +
4*a^2*b^3 + 2*b^5 - 2*a*b^4*Sin[c + d*x])*Tan[c + d*x]^2)/(4*(-a^2 + b^2)*d)

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fricas [A]  time = 0.47, size = 112, normalized size = 1.01 \[ \frac {{\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((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) + 6*a^2*b + 2*b^3 + 2*(a^3 + 3*a*b^2)*sin(d*x + c))/(d*cos(d*x + c)^2)

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giac [A]  time = 1.13, size = 114, normalized size = 1.03 \[ \frac {{\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (b^{3} \sin \left (d x + c\right )^{2} + a^{3} \sin \left (d x + c\right ) + 3 \, a b^{2} \sin \left (d x + c\right ) + 3 \, a^{2} b\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.28, size = 154, normalized size = 1.39 \[ \frac {a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 a^{2} b}{2 d \cos \left (d x +c \right )^{2}}+\frac {3 a \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {3 a \,b^{2} \sin \left (d x +c \right )}{2 d}-\frac {3 a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 98, normalized size = 0.88 \[ \frac {{\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, a^{2} b + b^{3} + {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.21, size = 99, normalized size = 0.89 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,{\left (a-b\right )}^2\,\left (a+2\,b\right )}{4\,d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^2\,\left (a-2\,b\right )}{4\,d}-\frac {\frac {3\,a^2\,b}{2}+\frac {b^3}{2}+\sin \left (c+d\,x\right )\,\left (\frac {a^3}{2}+\frac {3\,a\,b^2}{2}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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