Optimal. Leaf size=249 \[ -\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 a b}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\tan (c+d x)}} \]
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Rubi [A]
time = 0.15, antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3623, 3610,
3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\tan (c+d x)}}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 a b}{3 d \tan ^{\frac {3}{2}}(c+d x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 210
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3610
Rule 3615
Rule 3623
Rubi steps
\begin {align*} \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\int \frac {2 a b-\left (a^2-b^2\right ) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 a b}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\int \frac {-a^2+b^2-2 a b \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 a b}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\tan (c+d x)}}+\int \frac {-2 a b+\left (a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 a b}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\tan (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {-2 a b+\left (a^2-b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 a b}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\tan (c+d x)}}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}-\frac {\left (a^2+2 a b-b^2\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 a b}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\tan (c+d x)}}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}+\frac {\left (a^2+2 a b-b^2\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}\\ &=\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 a b}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\tan (c+d x)}}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}\\ &=-\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 a b}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 0.22, size = 81, normalized size = 0.33 \begin {gather*} \frac {-6 \left (a^2-b^2\right ) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\tan ^2(c+d x)\right )-2 b \left (3 b+10 a \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\tan ^2(c+d x)\right ) \tan (c+d x)\right )}{15 d \tan ^{\frac {5}{2}}(c+d x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.08, size = 231, normalized size = 0.93
method | result | size |
derivativedivides | \(\frac {-\frac {2 a^{2}}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 \left (-a^{2}+b^{2}\right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {4 a b}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}+\frac {\left (a^{2}-b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(231\) |
default | \(\frac {-\frac {2 a^{2}}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 \left (-a^{2}+b^{2}\right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {4 a b}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}+\frac {\left (a^{2}-b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(231\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 206, normalized size = 0.83 \begin {gather*} \frac {30 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 30 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 15 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 15 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \frac {8 \, {\left (10 \, a b \tan \left (d x + c\right ) - 15 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, a^{2}\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{60 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 5441 vs.
\(2 (211) = 422\).
time = 1.85, size = 5441, normalized size = 21.85 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{2}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.69, size = 983, normalized size = 3.95 \begin {gather*} -2\,\mathrm {atanh}\left (\frac {32\,a^4\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3\,b}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}}{16\,a^6\,d^2+a^5\,b\,d^2\,32{}\mathrm {i}-112\,a^4\,b^2\,d^2-a^3\,b^3\,d^2\,192{}\mathrm {i}+112\,a^2\,b^4\,d^2+a\,b^5\,d^2\,32{}\mathrm {i}-16\,b^6\,d^2}+\frac {32\,b^4\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3\,b}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}}{16\,a^6\,d^2+a^5\,b\,d^2\,32{}\mathrm {i}-112\,a^4\,b^2\,d^2-a^3\,b^3\,d^2\,192{}\mathrm {i}+112\,a^2\,b^4\,d^2+a\,b^5\,d^2\,32{}\mathrm {i}-16\,b^6\,d^2}-\frac {192\,a^2\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3\,b}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}}{16\,a^6\,d^2+a^5\,b\,d^2\,32{}\mathrm {i}-112\,a^4\,b^2\,d^2-a^3\,b^3\,d^2\,192{}\mathrm {i}+112\,a^2\,b^4\,d^2+a\,b^5\,d^2\,32{}\mathrm {i}-16\,b^6\,d^2}\right )\,\sqrt {-\frac {a^4\,1{}\mathrm {i}-4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}+4\,a\,b^3+b^4\,1{}\mathrm {i}}{4\,d^2}}+2\,\mathrm {atanh}\left (\frac {32\,a^4\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}+\frac {a^3\,b}{d^2}-\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}}{-16\,a^6\,d^2+a^5\,b\,d^2\,32{}\mathrm {i}+112\,a^4\,b^2\,d^2-a^3\,b^3\,d^2\,192{}\mathrm {i}-112\,a^2\,b^4\,d^2+a\,b^5\,d^2\,32{}\mathrm {i}+16\,b^6\,d^2}+\frac {32\,b^4\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}+\frac {a^3\,b}{d^2}-\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}}{-16\,a^6\,d^2+a^5\,b\,d^2\,32{}\mathrm {i}+112\,a^4\,b^2\,d^2-a^3\,b^3\,d^2\,192{}\mathrm {i}-112\,a^2\,b^4\,d^2+a\,b^5\,d^2\,32{}\mathrm {i}+16\,b^6\,d^2}-\frac {192\,a^2\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}+\frac {a^3\,b}{d^2}-\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}}{-16\,a^6\,d^2+a^5\,b\,d^2\,32{}\mathrm {i}+112\,a^4\,b^2\,d^2-a^3\,b^3\,d^2\,192{}\mathrm {i}-112\,a^2\,b^4\,d^2+a\,b^5\,d^2\,32{}\mathrm {i}+16\,b^6\,d^2}\right )\,\sqrt {\frac {a^4\,1{}\mathrm {i}+4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}-4\,a\,b^3+b^4\,1{}\mathrm {i}}{4\,d^2}}-\frac {\frac {2\,a^2}{5}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a^2-2\,b^2\right )+\frac {4\,a\,b\,\mathrm {tan}\left (c+d\,x\right )}{3}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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