3.391 \(\int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=59 \[ \frac {\left (a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2668, 723, 206} \[ \frac {\left (a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 723

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[((2*p + 3)*(c*d^2 + a*e^2))/(2*a*c*(p + 1)), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

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

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Mathematica [A]  time = 0.92, size = 113, normalized size = 1.92 \[ \frac {-2 \left (a^4-b^4\right ) \tan (c+d x) \sec (c+d x)+\left (4 a b^3-6 a^3 b\right ) \tan ^2(c+d x)+2 a^3 b \sec ^2(c+d x)+\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (\sin (c+d x)+1))}{4 d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2 - b^2)^2*(Log[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]]) + 2*a^3*b*Sec[c + d*x]^2 - 2*(a^4 - b^4)*Sec[c
+ d*x]*Tan[c + d*x] + (-6*a^3*b + 4*a*b^3)*Tan[c + d*x]^2)/(4*(-a^2 + b^2)*d)

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fricas [A]  time = 0.47, size = 90, normalized size = 1.53 \[ \frac {{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a b + 2 \, {\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.15, size = 86, normalized size = 1.46 \[ \frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (a^{2} \sin \left (d x + c\right ) + b^{2} \sin \left (d x + c\right ) + 2 \, a b\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.25, size = 118, normalized size = 2.00 \[ \frac {a^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a b}{d \cos \left (d x +c \right )^{2}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {b^{2} \sin \left (d x +c \right )}{2 d}-\frac {b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

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]

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

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maxima [A]  time = 0.32, size = 78, normalized size = 1.32 \[ \frac {{\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (2 \, a b + {\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 62, normalized size = 1.05 \[ \frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )}{d}-\frac {a\,b+\sin \left (c+d\,x\right )\,\left (\frac {a^2}{2}+\frac {b^2}{2}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh(sin(c + d*x))*(a^2/2 - b^2/2))/d - (a*b + sin(c + d*x)*(a^2/2 + b^2/2))/(d*(sin(c + d*x)^2 - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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