3.5.25 \(\int (a+b \cos (c+d x))^2 \sec ^4(c+d x) \, dx\) [425]

Optimal. Leaf size=80 \[ \frac {a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {a^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2868, 3853, 3855, 3091, 3852, 8} \begin {gather*} \frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a b \tan (c+d x) \sec (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2868

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[2*c*(d/b)
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 3091

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A*Cos[e +
f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]
time = 0.25, size = 71, normalized size = 0.89 \begin {gather*} \frac {a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d}+\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {a^2 \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4,x]

[Out]

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

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Maple [A]
time = 0.18, size = 74, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} \tan \left (d x +c \right )}{d}\)	\(74\)
default	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} \tan \left (d x +c \right )}{d}\)	\(74\)
risch	\(-\frac {2 i \left (3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}-2 a^{2}-3 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(139\)
norman	\(\frac {-\frac {4 \left (a^{2}-3 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (a^{2}-a b +b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (a^{2}+a b +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 a \left (2 a -3 b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a \left (2 a +3 b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(195\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*sec(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 84, normalized size = 1.05 \begin {gather*} \frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} - 3 \, a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, b^{2} \tan \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.45, size = 100, normalized size = 1.25 \begin {gather*} \frac {3 \, a b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \cos {\left (c + d x \right )}\right )^{2} \sec ^{4}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**2*sec(d*x+c)**4,x)

[Out]

Integral((a + b*cos(c + d*x))**2*sec(c + d*x)**4, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (76) = 152\).
time = 0.43, size = 178, normalized size = 2.22 \begin {gather*} \frac {3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/3*(3*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*a^2*tan(1/2*d*
x + 1/2*c)^5 - 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*b^2*tan(1/2*d*x + 1/2*c)^5 - 2*a^2*tan(1/2*d*x + 1/2*c)^3 - 6*
b^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*tan(1/2*d*x + 1/2*c) + 3*a*b*tan(1/2*d*x + 1/2*c) + 3*b^2*tan(1/2*d*x + 1/2
*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

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Mupad [B]
time = 2.59, size = 141, normalized size = 1.76 \begin {gather*} \frac {2\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\left (2\,a^2-2\,a\,b+2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,a^2}{3}-4\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^2+2\,a\,b+2\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^2/cos(c + d*x)^4,x)

[Out]

(2*a*b*atanh(tan(c/2 + (d*x)/2)))/d - (tan(c/2 + (d*x)/2)^5*(2*a^2 - 2*a*b + 2*b^2) - tan(c/2 + (d*x)/2)^3*((4
*a^2)/3 + 4*b^2) + tan(c/2 + (d*x)/2)*(2*a*b + 2*a^2 + 2*b^2))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/
2)^4 + tan(c/2 + (d*x)/2)^6 - 1))

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