3.5.57 \(\int \sec ^3(c+d x) (a+b \sec (c+d x))^2 \, dx\) [457]

Optimal. Leaf size=110 \[ \frac {\left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {\left (4 a^2+3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a b \tan ^3(c+d x)}{3 d} \]

[Out]

1/8*(4*a^2+3*b^2)*arctanh(sin(d*x+c))/d+2*a*b*tan(d*x+c)/d+1/8*(4*a^2+3*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/4*b^2*s
ec(d*x+c)^3*tan(d*x+c)/d+2/3*a*b*tan(d*x+c)^3/d

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Rubi [A]
time = 0.07, antiderivative size = 110, 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 = {3873, 3852, 4131, 3853, 3855} \begin {gather*} \frac {\left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2+3 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {2 a b \tan ^3(c+d x)}{3 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {b^2 \tan (c+d x) \sec ^3(c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^2 + 3*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (2*a*b*Tan[c + d*x])/d + ((4*a^2 + 3*b^2)*Sec[c + d*x]*Tan[c +
d*x])/(8*d) + (b^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (2*a*b*Tan[c + d*x]^3)/(3*d)

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]

Rule 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.30, size = 82, normalized size = 0.75 \begin {gather*} \frac {3 \left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 \left (4 a^2+3 b^2\right ) \sec (c+d x)+6 b^2 \sec ^3(c+d x)+16 a b \left (3+\tan ^2(c+d x)\right )\right )}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(4*a^2 + 3*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*(4*a^2 + 3*b^2)*Sec[c + d*x] + 6*b^2*Sec[c + d*x]^3
+ 16*a*b*(3 + Tan[c + d*x]^2)))/(24*d)

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Maple [A]
time = 0.10, size = 111, normalized size = 1.01

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 b a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(111\)
default	\(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 b a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(111\)
norman	\(\frac {\frac {\left (4 a^{2}-16 b a +5 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (4 a^{2}+16 b a +5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 a^{2}-80 b a -9 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {\left (12 a^{2}+80 b a -9 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (4 a^{2}+3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (4 a^{2}+3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(197\)
risch	\(-\frac {i \left (12 a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+9 b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+12 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+33 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-96 b a \,{\mathrm e}^{4 i \left (d x +c \right )}-12 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-33 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-128 a b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 a^{2} {\mathrm e}^{i \left (d x +c \right )}-9 b^{2} {\mathrm e}^{i \left (d x +c \right )}-32 b a \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d}\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-2*b*a*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+b^
2*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c))))

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Maxima [A]
time = 0.26, size = 144, normalized size = 1.31 \begin {gather*} \frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b - 3 \, b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{2} {\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 )}}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*a*b - 3*b^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4
- 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*a^2*(2*sin(d*x + c)/(sin(d*x
+ c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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Fricas [A]
time = 2.93, size = 133, normalized size = 1.21 \begin {gather*} \frac {3 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, a b \cos \left (d x + c\right )^{3} + 16 \, a b \cos \left (d x + c\right ) + 3 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, b^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(4*a^2 + 3*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*a^2 + 3*b^2)*cos(d*x + c)^4*log(-sin(d*x +
c) + 1) + 2*(32*a*b*cos(d*x + c)^3 + 16*a*b*cos(d*x + c) + 3*(4*a^2 + 3*b^2)*cos(d*x + c)^2 + 6*b^2)*sin(d*x
+ c))/(d*cos(d*x + c)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (102) = 204\).
time = 0.48, size = 258, normalized size = 2.35 \begin {gather*} \frac {3 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 80 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, 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 )}^{4}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(4*a^2 + 3*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*a^2 + 3*b^2)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) + 2*(12*a^2*tan(1/2*d*x + 1/2*c)^7 - 48*a*b*tan(1/2*d*x + 1/2*c)^7 + 15*b^2*tan(1/2*d*x + 1/2*c)^7 - 12*a^
2*tan(1/2*d*x + 1/2*c)^5 + 80*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*b^2*tan(1/2*d*x + 1/2*c)^5 - 12*a^2*tan(1/2*d*x +
1/2*c)^3 - 80*a*b*tan(1/2*d*x + 1/2*c)^3 + 9*b^2*tan(1/2*d*x + 1/2*c)^3 + 12*a^2*tan(1/2*d*x + 1/2*c) + 48*a*
b*tan(1/2*d*x + 1/2*c) + 15*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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Mupad [B]
time = 3.66, size = 184, normalized size = 1.67 \begin {gather*} \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2+\frac {3\,b^2}{4}\right )}{d}+\frac {\left (a^2-4\,a\,b+\frac {5\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-a^2+\frac {20\,a\,b}{3}+\frac {3\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-a^2-\frac {20\,a\,b}{3}+\frac {3\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (a^2+4\,a\,b+\frac {5\,b^2}{4}\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 )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\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)^3,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*(a^2 + (3*b^2)/4))/d + (tan(c/2 + (d*x)/2)^5*((20*a*b)/3 - a^2 + (3*b^2)/4) + tan(c
/2 + (d*x)/2)*(4*a*b + a^2 + (5*b^2)/4) + tan(c/2 + (d*x)/2)^7*(a^2 - 4*a*b + (5*b^2)/4) - tan(c/2 + (d*x)/2)^
3*((20*a*b)/3 + a^2 - (3*b^2)/4))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6
+ tan(c/2 + (d*x)/2)^8 + 1))

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