3.8.77 \(\int (a+b \sec (c+d x))^2 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [777]

Optimal. Leaf size=116 \[ \frac {\left (2 a^2 B+b^2 B+2 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {2 \left (3 a b B+a^2 C+b^2 C\right ) \tan (c+d x)}{3 d}+\frac {b (3 b B+2 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d} \]

[Out]

1/2*(2*B*a^2+B*b^2+2*C*a*b)*arctanh(sin(d*x+c))/d+2/3*(3*B*a*b+C*a^2+C*b^2)*tan(d*x+c)/d+1/6*b*(3*B*b+2*C*a)*s
ec(d*x+c)*tan(d*x+c)/d+1/3*C*(a+b*sec(d*x+c))^2*tan(d*x+c)/d

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Rubi [A]
time = 0.11, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {4141, 4133, 3855, 3852, 8} \begin {gather*} \frac {2 \left (a^2 C+3 a b B+b^2 C\right ) \tan (c+d x)}{3 d}+\frac {\left (2 a^2 B+2 a b C+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b (2 a C+3 b B) \tan (c+d x) \sec (c+d x)}{6 d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*B + b^2*B + 2*a*b*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (2*(3*a*b*B + a^2*C + b^2*C)*Tan[c + d*x])/(3*d) +
(b*(3*b*B + 2*a*C)*Sec[c + d*x]*Tan[c + d*x])/(6*d) + (C*(a + b*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 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 3855

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

Rule 4133

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*(Cot[e + f*x]/(2*f)), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b
*(2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

Rule 4141

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

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+b \sec (c+d x)) \left ((3 a B+2 b C) \sec (c+d x)+(3 b B+2 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b (3 b B+2 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (3 \left (2 a^2 B+b^2 B+2 a b C\right ) \sec (c+d x)+4 \left (3 a b B+a^2 C+b^2 C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b (3 b B+2 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{2} \left (2 a^2 B+b^2 B+2 a b C\right ) \int \sec (c+d x) \, dx+\frac {1}{3} \left (2 \left (3 a b B+a^2 C+b^2 C\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {\left (2 a^2 B+b^2 B+2 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b (3 b B+2 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac {\left (2 \left (3 a b B+a^2 C+b^2 C\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {\left (2 a^2 B+b^2 B+2 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {2 \left (3 a b B+a^2 C+b^2 C\right ) \tan (c+d x)}{3 d}+\frac {b (3 b B+2 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.49, size = 92, normalized size = 0.79 \begin {gather*} \frac {3 \left (2 a^2 B+b^2 B+2 a b C\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 b (b B+2 a C) \sec (c+d x)+2 \left (6 a b B+3 a^2 C+3 b^2 C+b^2 C \tan ^2(c+d x)\right )\right )}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 143, normalized size = 1.23

method	result	size
derivativedivides	\(\frac {b^{2} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-b^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a b B \tan \left (d x +c \right )+2 a b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \tan \left (d x +c \right )}{d}\)	\(143\)
default	\(\frac {b^{2} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-b^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a b B \tan \left (d x +c \right )+2 a b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \tan \left (d x +c \right )}{d}\)	\(143\)
norman	\(\frac {\frac {4 \left (6 a b B +3 a^{2} C +b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (4 a b B -b^{2} B +2 a^{2} C -2 a b C +2 b^{2} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (4 a b B +b^{2} B +2 a^{2} C +2 a b C +2 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {\left (2 a^{2} B +b^{2} B +2 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 a^{2} B +b^{2} B +2 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(207\)
risch	\(-\frac {i \left (3 B \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+6 C a b \,{\mathrm e}^{5 i \left (d x +c \right )}-12 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-24 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-6 C b a \,{\mathrm e}^{i \left (d x +c \right )}-12 a b B -6 a^{2} C -4 b^{2} C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} B}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} B}{2 d}+\frac {b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} B}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} B}{2 d}-\frac {b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(298\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 165, normalized size = 1.42 \begin {gather*} \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{2} - 6 \, C 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 )} - 3 \, B b^{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 )} + 12 \, B a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, C a^{2} \tan \left (d x + c\right ) + 24 \, B a b \tan \left (d x + c\right )}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*b^2 - 6*C*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)) - 3*B*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log
(sin(d*x + c) - 1)) + 12*B*a^2*log(sec(d*x + c) + tan(d*x + c)) + 12*C*a^2*tan(d*x + c) + 24*B*a*b*tan(d*x + c
))/d

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Fricas [A]
time = 2.29, size = 150, normalized size = 1.29 \begin {gather*} \frac {3 \, {\left (2 \, B a^{2} + 2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, B a^{2} + 2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C b^{2} + 2 \, {\left (3 \, C a^{2} + 6 \, B a b + 2 \, C b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(2*B*a^2 + 2*C*a*b + B*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*B*a^2 + 2*C*a*b + B*b^2)*cos(d
*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*C*b^2 + 2*(3*C*a^2 + 6*B*a*b + 2*C*b^2)*cos(d*x + c)^2 + 3*(2*C*a*b +
B*b^2)*cos(d*x + c))*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 (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (108) = 216\).
time = 0.48, size = 294, normalized size = 2.53 \begin {gather*} \frac {3 \, {\left (2 \, B a^{2} + 2 \, C a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, B a^{2} + 2 \, C a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C 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}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(2*B*a^2 + 2*C*a*b + B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*B*a^2 + 2*C*a*b + B*b^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 12*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b*tan
(1/2*d*x + 1/2*c)^5 - 3*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^2*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^2*tan(1/2*d*x +
1/2*c)^3 - 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 4*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^2*tan(1/2*d*x + 1/2*c) +
12*B*a*b*tan(1/2*d*x + 1/2*c) + 6*C*a*b*tan(1/2*d*x + 1/2*c) + 3*B*b^2*tan(1/2*d*x + 1/2*c) + 6*C*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 = 7.38, size = 227, normalized size = 1.96 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B\,a^2+C\,a\,b+\frac {B\,b^2}{2}\right )}{4\,B\,a^2+4\,C\,a\,b+2\,B\,b^2}\right )\,\left (2\,B\,a^2+2\,C\,a\,b+B\,b^2\right )}{d}-\frac {\left (2\,C\,a^2-B\,b^2+2\,C\,b^2+4\,B\,a\,b-2\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,C\,a^2-8\,B\,a\,b-\frac {4\,C\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (B\,b^2+2\,C\,a^2+2\,C\,b^2+4\,B\,a\,b+2\,C\,a\,b\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((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^2,x)

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*(B*a^2 + (B*b^2)/2 + C*a*b))/(4*B*a^2 + 2*B*b^2 + 4*C*a*b))*(2*B*a^2 + B*b^2 + 2*
C*a*b))/d - (tan(c/2 + (d*x)/2)*(B*b^2 + 2*C*a^2 + 2*C*b^2 + 4*B*a*b + 2*C*a*b) - tan(c/2 + (d*x)/2)^3*(4*C*a^
2 + (4*C*b^2)/3 + 8*B*a*b) + tan(c/2 + (d*x)/2)^5*(2*C*a^2 - B*b^2 + 2*C*b^2 + 4*B*a*b - 2*C*a*b))/(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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